---
title: "Finite Dimension"
book: "University Mathematics — Year 1"
subject: math
language: en
chapter: 19
exercises: 12
source: https://one-course.com/books/math/3/en/chapter/19-finite-dimension
---

# Chapter 19 — Finite Dimension

A space spanned by finitely many vectors carries a well-defined *dimension* — the common size of all its bases. The proofs below all flow from one combinatorial engine, the exchange lemma: a [free family](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) can never outnumber a generating one. With dimension come the tools used everywhere afterwards: the incomplete [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) theorem, the [rank](#def-b1-findim-rank) of a family, Grassmann’s formula.

## 19.1 Existence of bases

**Definition 19.1.**

$E$ is *finite-dimensional* when it has a finite generating family. (Otherwise *infinite-dimensional*: so is $K[X]$, whose finite families only [span](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-span) [polynomials](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def) of bounded degree.)

**Theorem 19.2 (Exchange lemma).**

Let $(g_1, \dots, g_n)$ generate $E$ and $(f_1, \dots, f_p)$ be [free](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) in $E$. Then $p \leq n$.

**Proof.** We prove by induction on $k \leq p$: *after renumbering the $g_j$, the family $(f_1, \dots, f_k, g_{k+1}, \dots, g_n)$ generates $E$* — which forces $k \leq n$ at each stage, and $p \leq n$ at the end.

$k = 0$: the hypothesis. Step: assume it for $k - 1$; then $f_k$ is a combination of $(f_1, \dots, f_{k-1},$ $g_k, \dots, g_n)$. In this combination some $g_j$ ($j \geq k$) has a nonzero coefficient — otherwise $f_k$ would be a combination of $f_1, \dots, f_{k-1}$, contradicting freeness. Renumber so that $j = k$, and solve for $g_k$: $g_k$ is a combination of $(f_1, \dots, f_k, g_{k+1}, \dots,
g_n)$. Every vector of $E$, expressed through the $(k-1)$-family, can then be re-expressed through the $k$-family: it generates. (If $k - 1 = n$, no $g$ remains and $f_k$ would be a combination of the $f_i$’s alone: impossible; so $k \leq n$.) ∎

**Example 19.3 (The exchange, watched once).**

In $\R^2$, take the generating family $\bigl(g_1, g_2\bigr) =
\bigl((1,0), (0,1)\bigr)$ and the [free family](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) $\bigl(f_1,
f_2\bigr) = \bigl((1,2), (3,4)\bigr)$. Step $1$: $f_1 = 1\cdot
g_1 + 2\cdot g_2$; the coefficient of $g_2$ is nonzero, so exchange $g_2$ for $f_1$: the family $\bigl(f_1, g_1\bigr)$ still generates ($g_2 = \frac12(f_1 - g_1)$). Step $2$: $f_2 =
(3,4) = 2\,f_1 + 1\cdot g_1$; the coefficient of the remaining $g_1$ is nonzero (it must be: $f_2$ is not a multiple of $f_1$), so exchange again: $\bigl(f_1, f_2\bigr)$ generates $\R^2$. Had there been a third [free](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) vector $f_3$, no $g$ would remain to absorb it — which is precisely how the lemma forbids $3$ [free](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) vectors in $\R^2$. The proof above is this bookkeeping done in general.

**Theorem 19.4 (Bases in finite dimension).**

Let $E \neq \{0\}$ be [finite-dimensional](#def-b1-findim-def).

1. From every finite generating family one can extract a [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) .
2. (Incomplete [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) theorem) Every [free family](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) extends to a [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) , using vectors of any chosen generating family.
3. All bases of $E$ are finite, with the same number of elements: the *dimension* $\dim E$ . (Convention: $\dim\{0\} = 0$ .)

**Proof.** (1) Discard, one at a time, any vector that is a combination of the others; the family stays generating, since in any expression using the discarded vector one may substitute its combination of the survivors. The process terminates — each step shrinks a finite family by one — and it halts exactly when no remaining vector is a combination of the others. The final family is still generating, and it is [free](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free): a nontrivial null combination would carry some nonzero coefficient, and dividing by it would solve for the corresponding vector in terms of the others, making it discardable after all — contradicting that the process had halted.

(2) Let $(f_1, \dots, f_p)$ be [free](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free), $(g_1, \dots, g_n)$ generating. Run through $g_1, \dots, g_n$, appending $g_j$ to the current family whenever it is not already in its [span](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-span) ([Proposition 18.19](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#prop-b1-vspaces-freecriteria) (2) keeps the family [free](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free)). The final family is [free](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free), and generating: every $g_j$ lies in its [span](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-span) — either it was appended, or it was already a combination.

(3) Two bases are each [free](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) and each generating: the exchange lemma gives both inequalities between their cardinalities. (Finiteness: a [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) is [free](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free), hence by exchange no larger than a finite generating family.) ∎

**Example 19.5.**

$\dim K^n = n$ (canonical [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free)); $\dim K_n[X] = n + 1$ (monomials); $\dim_\R \C = 2$; the solution space of $y'' + ay' + by = 0$ has dimension $2$ ([Theorem 5.10](https://one-course.com/books/math/3/en/chapter/5-linear-differential-equations#thm-b1-diffeq-homogeneous2): the solutions are parametrized [bijectively](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-inj) and linearly by $(\lambda, \mu) \in K^2$).

**Example 19.6 (The same set, two dimensions).**

The [set](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets) $\C^2$ of pairs of complex numbers is a $\C$-vector space of dimension $2$ (canonical [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) $e_1, e_2$) — and an $\R$-vector space of dimension $4$, with [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free)

$$
(1, 0),\quad (\iu, 0),\quad (0, 1),\quad (0, \iu):
$$

every $(z, w) = (a + \iu b,\ c + \iu d)$ has real [coordinates](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#prop-b1-vspaces-coordinates) $(a, b, c, d)$, uniquely. Dimension is not a property of the [set](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets) of vectors alone: it counts the degrees of freedom *relative to the allowed scalars*, and halving the scalar supply from $\C$ to $\R$ doubles the count. (The weekend problem exploits the extreme case of this sensitivity, with scalars shrunk all the way to $\Q$.)

**Example 19.7 (Running the completion algorithm).**

Complete the [free family](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) $\bigl((1,1,1)\bigr)$ into a [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) of $\R^3$ using the canonical vectors. Run the proof of [Theorem 19.4](#thm-b1-findim-bases) (2) on the generating family $(e_1,
e_2, e_3)$: is $e_1 \in \operatorname{Vect}(1,1,1)$? No (multiples of $(1,1,1)$ have equal [coordinates](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#prop-b1-vspaces-coordinates)) — append it. Is $e_2 \in
\operatorname{Vect}\bigl((1,1,1), e_1\bigr)$? A combination $\alpha(1,1,1) + \beta e_1$ has equal second and third [coordinates](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#prop-b1-vspaces-coordinates), and $e_2$ does not — append it. The family $\bigl((1,1,1), e_1, e_2\bigr)$ is [free](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) with $3$ vectors: stop, it is a [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) ([Proposition 19.8](#prop-b1-findim-twoofthree) will make this reflex official). Note the outcome depends on the order in which the $g_j$ are scanned: completion is an algorithm, not a formula.

**Proposition 19.8 (The two-out-of-three rule).**

Let $\dim E = n$ and $\mathcal{F}$ a family of exactly $n$ vectors of $E$. Then

$$
\mathcal{F} \text{ free} \iff \mathcal{F} \text{ generating} \iff
\mathcal{F} \text{ basis}.
$$

Moreover any [free family](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) has $\leq n$ vectors, any generating family $\geq n$.

**Proof.** The [cardinality](https://one-course.com/books/math/3/en/chapter/2-counting#def-b1-counting-card) bounds are the exchange lemma against a [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free). If $\mathcal{F}$ (size $n$) is [free](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) but not generating, some $x$ lies outside its [span](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-span); appending $x$ gives a [free family](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) of $n + 1$ vectors: impossible. If $\mathcal{F}$ is generating but not [free](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free), extracting a [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) ([Theorem 19.4](#thm-b1-findim-bases) (1)) gives a [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) of $< n$ vectors: impossible. ∎

**Example 19.9 (Two-out-of-three, saving half the work).**

Is $\bigl((1,1,0), (0,1,1), (1,0,1)\bigr)$ a [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) of $\R^3$? Count: three vectors, dimension three — so freeness alone decides. A null combination gives $a + c = 0$, $a + b = 0$, $b +
c = 0$; adding all three, $2(a + b + c) = 0$, and subtracting each original equation from $a + b + c = 0$ leaves $b = c = a =
0$: [free](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free), hence a [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free), with the generating half of the verification supplied by the theorem for free. Compare [Example 18.18](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#ex-b1-vspaces-testbasis), where the same double verification had to be done by hand — one chapter of theory converts into exactly that saving, on every [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) check for the rest of the book.

**Method 19.10 (Computing a dimension).**

Three standard routes, in decreasing order of frequency.

1. *Parametrize, then read off a [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free).* Solve the defining constraints, express the general element linearly in the surviving parameters, and check that the vectors multiplying the parameters are [free](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) : the dimension is the number of parameters. (Run below on a concrete [subspace](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-subspace) of $\R^4$ .)
2. *Exhibit a [bijective](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-inj) linear parametrization.* When the elements are determined by finitely many values — initial conditions of a recurrence ( [Exercise 19.10](#exo-b1-findim-10) ), coefficients of a solution formula ( [Example 19.5](#ex-b1-findim-dims) ) — the dimension is the number of those values.
3. *Use the formulas.* Grassmann for intersections and sums, [rank](#def-b1-findim-rank) for [spans](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-span) , and later rank–nullity for kernels and images: dimensions are usually *computed* , not guessed.

In all three routes, the two-out-of-three rule is the finisher: once the count matches, freeness or generation alone concludes.

**Example 19.11 (Route 1, in full).**

Dimension of $H = \{(x, y, z, t) \in \R^4 : x + y + z + t = 0
\text{ and } x = t\}$. Solve: $t = x$ and $y + z = -2x$, so $z =
-2x - y$ with $x, y$ [free](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free):

$$
(x,\ y,\ -2x - y,\ x) = x\,(1, 0, -2, 1) + y\,(0, 1, -1, 0) .
$$

The two vectors are [free](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) (look at the first two [coordinates](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#prop-b1-vspaces-coordinates): $(x, y) = (0,0)$), so they form a [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) of $H$ and $\dim H = 2$. The count was predictable — two independent linear constraints in $\R^4$ should each eat one dimension — but the parametrization proves it *and* hands over a [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free), which the prediction alone never does; [Exercise 19.9](#exo-b1-findim-9) makes the “each equation eats at most one dimension” slogan into a theorem.

**Example 19.12 (Route 2, in full).**

Dimension of $W = \{P \in \R_n[X] : P(1) = P(2) = 0\}$ (for $n
\geq 2$). By the factor theorem applied twice ([Theorem 8.7](https://one-course.com/books/math/3/en/chapter/8-polynomials#thm-b1-poly-factor); the roots $1$ and $2$ are distinct), $P \in W$ exactly when $P = (X - 1)(X - 2)\,Q$ with $\deg Q \leq
n - 2$. The correspondence $Q \mapsto (X-1)(X-2)Q$ is linear, reaches all of $W$, and is [injective](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-inj) (a product is zero only if $Q = 0$): $W$ is parametrized [bijectively](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-inj) and linearly by $\R_{n-2}[X]$, so

$$
\dim W = \dim \R_{n-2}[X] = n - 1 .
$$

A [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) comes with the parametrization: the images of the monomials, $\bigl((X-1)(X-2),\ (X-1)(X-2)X,\ \dots,\
(X-1)(X-2)X^{n-2}\bigr)$. Each new evaluation constraint at a fresh point costs exactly one dimension — the counting backbone of [Lagrange interpolation](https://one-course.com/books/math/3/en/chapter/8-polynomials#thm-b1-poly-lagrange) ([Theorem 8.23](https://one-course.com/books/math/3/en/chapter/8-polynomials#thm-b1-poly-lagrange)).

**Example 19.13 (An integral constraint costs one dimension too).**

Dimension of $H = \{P \in \R_2[X] : \int_0^1 P = 0\}$. Writing $P = a + bX + cX^2$, the constraint reads $a + \frac b2 + \frac
c3 = 0$; solve for $a$ and parametrize:

$$
P = b\Bigl(X - \frac12\Bigr) + c\Bigl(X^2 - \frac13\Bigr),
$$

so $H = \operatorname{Vect}\bigl(X - \frac12,\ X^2 -
\frac13\bigr)$, dimension $2$ (the two [polynomials](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def) have distinct degrees: [free](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free)). One linear condition — whether an evaluation, an [integral](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#thm-b1-integration-def), or any other linear recipe — removes at most one dimension, and exactly one as soon as the condition is not identically zero. [Chapter 20](https://one-course.com/books/math/3/en/chapter/20-linear-maps#ch-b1-linmaps) will name such recipes *linear forms* and their solution [sets](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets) *hyperplanes*; the [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) vectors found here reappear in [Chapter 23](https://one-course.com/books/math/3/en/chapter/23-euclidean-spaces#ch-b1-euclid) as the start of the Legendre family.

## 19.2 Subspaces, rank, Grassmann

**Theorem 19.14 (Subspaces).**

Let $E$ be [finite-dimensional](#def-b1-findim-def) and $F$ a [subspace](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-subspace). Then $F$ is [finite-dimensional](#def-b1-findim-def), $\dim F \leq \dim E$, with equality if and only if $F = E$. Moreover every [subspace](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-subspace) has a [supplementary](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-sum) [subspace](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-subspace).

**Proof.** [Free](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) families of $F$ have at most $\dim E$ vectors (exchange lemma in $E$: a [free family](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) of $F$ is in particular [free](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) in $E$, and $E$ has a finite generating family). Among the [free](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) families of $F$ pick one of *maximal* size $p$ — possible since the sizes are integers bounded by $\dim E$. It generates $F$: otherwise some $x \in F$ would lie outside its [span](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-span), and appending $x$ would give a [free family](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) of $F$ of size $p + 1$ ([Proposition 18.19](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#prop-b1-vspaces-freecriteria) (2)), contradicting maximality. Being [free](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) and generating, it is a [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) of $F$, and $\dim F = p \leq \dim E$. If $p = \dim E = n$: a [free family](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) of $n$ vectors of $E$ is a [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) of $E$ ([Proposition 19.8](#prop-b1-findim-twoofthree)), so $F
\supseteq \operatorname{span} = E$. [Supplementary](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-sum): complete a [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) $(f_1, \dots, f_p)$ of $F$ into a [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) $(f_1, \dots, f_p, g_{p+1},
\dots, g_n)$ of $E$ (incomplete [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) theorem); then $G =
\operatorname{Vect}(g_{p+1}, \dots, g_n)$ satisfies $E = F \oplus G$ (existence and uniqueness of decompositions = [coordinates](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#prop-b1-vspaces-coordinates) in the big [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free)). ∎

**Definition 19.15 (Rank of a family).**

The *rank* of a finite family of vectors is the dimension of its [span](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-span): $\operatorname{rk}(x_1, \dots, x_p) = \dim
\operatorname{Vect}(x_1, \dots, x_p) \leq \min(p, \dim E)$, with equality to $p$ iff the family is [free](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free).

**Example 19.16 (Computing a rank by elimination).**

[Rank](#def-b1-findim-rank) of $\bigl((1,2,3), (2,3,4), (3,4,5), (1,1,1)\bigr)$ in $\R^3$. The [span](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-span) is unchanged when one subtracts from a vector a combination of the others (both families [span](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-span) the same combinations): replace $(2,3,4)$ by $(2,3,4) - (1,2,3) = (1,1,1)$ and $(3,4,5)$ by $(3,4,5) - (1,2,3) = (2,2,2)$. The [span](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-span) is now $\operatorname{Vect}\bigl((1,2,3), (1,1,1), (2,2,2),
(1,1,1)\bigr) = \operatorname{Vect}\bigl((1,2,3), (1,1,1)\bigr)$, and these two vectors are not proportional: the [rank](#def-b1-findim-rank) is $2$. This subtract-and-discard procedure is systematized as Gaussian elimination in [Chapter 22](https://one-course.com/books/math/3/en/chapter/22-determinants-and-linear-systems#ch-b1-det).

**Example 19.17 (Summing by concatenation).**

Take, in $\R^3$,

$$
F = \operatorname{Vect}\bigl((1,2,3),\ (1,1,1)\bigr),
\qquad
G = \operatorname{Vect}\bigl((2,3,4)\bigr) .
$$

The sum $F + G$ is spanned by the concatenated family of all three generators, and

$$
(2, 3, 4) = (1, 2, 3) + (1, 1, 1)
$$

shows the third is redundant: $F + G = F$, of dimension $2$ — equivalently $G \subseteq F$, which the relation displays. Grassmann confirms: $\dim(F \cap G) = 2 + 1 - 2 = 1 = \dim G$. Sums are computed by concatenating generators and then reducing the pile by the [rank](#def-b1-findim-rank) algorithm; no new technique is ever needed.

**Theorem 19.18 (Grassmann’s formula).**

For [finite-dimensional](#def-b1-findim-def) [subspaces](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-subspace) $F, G$ of $E$:

$$
\dim(F + G) = \dim F + \dim G - \dim(F \cap G) .
$$

In particular $F + G$ is direct iff $\dim(F + G) = \dim F + \dim G$.

**Proof.** Start from a [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) $(e_1, \dots, e_r)$ of $F \cap G$; complete it into a [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) $(e_1, \dots, e_r, f_1, \dots, f_s)$ of $F$ and into a [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) $(e_1, \dots, e_r, g_1, \dots, g_t)$ of $G$ ([Theorem 19.4](#thm-b1-findim-bases) (2)). We claim

$$
\mathcal{B} = (e_1, \dots, e_r, f_1, \dots, f_s, g_1, \dots, g_t)
$$

is a [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) of $F + G$; the formula follows by counting: $(r + s) +
(r + t) - r = r + s + t$.

$\mathcal{B}$ generates $F + G$: any $u + v$ ($u \in F$, $v \in G$) expands through it. Freeness: suppose $\sum \alpha_i e_i + \sum
\beta_j f_j + \sum \gamma_k g_k = 0$. The vector $w = \sum \gamma_k
g_k = -\sum\alpha_i e_i - \sum\beta_j f_j$ lies in $G \cap F$, so it expands on $(e_i)$ alone; but $w$ also expands on $(g_k)$, and in the [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) of $G$ these two expressions must coincide: all $\gamma_k = 0$ (and the $e$-coordinates match). The relation reduces to $\sum\alpha_i e_i + \sum\beta_j f_j = 0$, a relation in the [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) of $F$: all remaining coefficients vanish. ∎

**Example 19.19.**

Two distinct planes $F, G$ (dimension $2$) of $\R^3$ satisfy $F + G
= \R^3$ (their sum strictly contains a plane), so $\dim(F \cap G) =
2 + 2 - 3 = 1$: they always intersect along a line — no “parallel planes” through the origin.

**Example 19.20 (Grassmann in action, in R4\R^4R4).**

Let $F = \operatorname{Vect}\bigl(e_1,\ e_2,\ (1,1,1,0)\bigr)$ and $G = \operatorname{Vect}(e_3, e_4)$ in $\R^4$. Dimensions: $\dim F = 3$ (the third generator has a nonzero third coordinate, outside $\operatorname{Vect}(e_1, e_2)$) and $\dim G = 2$. Sum: $F + G$ contains $e_1, e_2, e_4$ and $e_3 = (1,1,1,0) - e_1 -
e_2$: it is all of $\R^4$. Grassmann then *computes* the intersection’s size with no elimination at all:

$$
\dim(F \cap G) = 3 + 2 - 4 = 1 .
$$

To identify the line, look inside $G$: a vector $(0, 0, c, d)$ lies in $F$ exactly when it is $a e_1 + b e_2 + \lambda(1,1,1,0)$, forcing $\lambda = -a = -b$ and $d = 0$: the intersection is $\operatorname{Vect}(e_3)$ — consistent, since $e_3$ was exhibited in $F$ above and lies in $G$ by definition. Typical division of labor: Grassmann predicts *how much* to look for, the linear system then finds *what*.

**Example 19.21 (Intersecting by equations).**

When both [subspaces](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-subspace) come as solution [sets](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets), intersecting is just stacking the equations. In $\R^3$: $F = \{x + y + z = 0\}$ and $G = \{x = y\}$ give

$$
F \cap G = \{x = y,\ 2x + z = 0\}
= \{(x,\ x,\ -2x)\} = \operatorname{Vect}(1, 1, -2),
$$

a line. Cross-check by Grassmann: $F + G = \R^3$ (the planes are distinct, so their sum strictly contains a plane), hence $\dim(F\cap G) = 2 + 2 - 3 = 1$. The two descriptions of a [subspace](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-subspace) — by equations, by generators — each make one operation trivial: equations intersect by stacking, generators sum by concatenating; converting between them is exactly what solving a linear system means ([Chapter 22](https://one-course.com/books/math/3/en/chapter/22-determinants-and-linear-systems#ch-b1-det)).

**Remark 19.22 (Common pitfalls).**

*Dimensions do not add along sums* unless the sum is direct: two planes of $\R^3$ have $\dim(F + G) = 3$, not $4$; always correct by the intersection term (Grassmann). *An inclusion needs the dimension *and* the inclusion*: $\dim F = \dim G$ alone never gives $F = G$ (two distinct lines of $\R^2$); the equality case of [Theorem 19.14](#thm-b1-findim-subspaces) requires $F \subseteq G$ first. *Counting parameters is not yet a proof*: “two equations in $\R^4$, so dimension $2$” fails when the equations are dependent ($x + y = 0$ and $2x + 2y = 0$ leave dimension $3$); only an actual parametrization or a [rank](#def-b1-findim-rank) computation decides. *Do not speak of the dimension of a [non-subspace](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-subspace)*: solution [sets](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets) of *inhomogeneous* systems miss $0$; their “dimension” is that of the associated homogeneous solution space ([Chapter 22](https://one-course.com/books/math/3/en/chapter/22-determinants-and-linear-systems#ch-b1-det) makes this precise). *Infinite dimension exists*: $K[X]$ contains [free](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) families of every size (the monomials), so no finite generating family can exist — [statements](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-statement) like the two-out-of-three rule are strictly [finite-dimensional](#def-b1-findim-def) and fail badly on $K[X]$ ([Corollary 20.9](https://one-course.com/books/math/3/en/chapter/20-linear-maps#cor-b1-linmaps-samedim) will show the same for injectivity/surjectivity). *[Rank](#def-b1-findim-rank) is about the [span](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-span), not the list*: repeating a vector, reordering, or rescaling by nonzero constants leaves the [rank](#def-b1-findim-rank) unchanged, and $\operatorname{rk} = p$ (the number of vectors) is a *property to prove* — it is exactly freeness. A family of $5$ vectors of [rank](#def-b1-findim-rank) $2$ carries three vectors’ worth of redundancy, which elimination ([Example 19.16](#ex-b1-findim-rankcomputation)) locates explicitly.

**Remark 19.23 (Where dimension goes to work).**

Dimension is the book’s favorite counting argument from here on. [Chapter 20](https://one-course.com/books/math/3/en/chapter/20-linear-maps#ch-b1-linmaps) proves the rank–nullity theorem, the functional version of Grassmann’s formula; [Chapter 21](https://one-course.com/books/math/3/en/chapter/21-matrices#ch-b1-matrices) computes [ranks](#def-b1-findim-rank) by row reduction; [Chapter 22](https://one-course.com/books/math/3/en/chapter/22-determinants-and-linear-systems#ch-b1-det) turns “$n$ vectors of $K^n$ form a [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free)” into one number being nonzero. The weekend problem below shows dimension doing *arithmetic*: counting dimensions over the [field](https://one-course.com/books/math/3/en/chapter/7-algebraic-structures#def-b1-structures-field) $\Q$ proves irrationality [statements](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-statement) that look untouchable by hand. In the Year 3 volume the same dimension counts, refined by [group](https://one-course.com/books/math/3/en/chapter/7-algebraic-structures#def-b1-structures-group) theory, decide which classical construction problems are solvable — that story is Galois theory.

**Remark 19.24 (Perspectives inside Book 3).**

Dimension is the conserved quantity of the rest of this volume, and it is worth naming the conservation laws in advance. [Chapter 20](https://one-course.com/books/math/3/en/chapter/20-linear-maps#ch-b1-linmaps) proves $\dim E = \dim\ker u +
\operatorname{rk} u$: what a linear [map](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-map) crushes plus what it keeps always totals the source. [Chapter 22](https://one-course.com/books/math/3/en/chapter/22-determinants-and-linear-systems#ch-b1-det) refines this into the structure of solution [sets](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets): $p$ unknowns minus $\operatorname{rk} A$ pivots leaves the dimension of the solution space, which Gaussian elimination exhibits as [free](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) parameters. [Chapter 23](https://one-course.com/books/math/3/en/chapter/23-euclidean-spaces#ch-b1-euclid) splits $\dim E = \dim F + \dim
F^\perp$ orthogonally, and the weekend problem of [Chapter 25](https://one-course.com/books/math/3/en/chapter/25-functions-of-two-variables#ch-b1-multivar) spends exactly this budget: $n$ data points, $2$ parameters fitted, $n - 2$ dimensions of residual. Whenever a count refuses to balance in a later chapter, the error is a forgotten kernel or a non-direct sum — come back to Grassmann’s formula first.

## 19.3 Exercises

**Exercise 19.1 ★.**

Give a [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) and the dimension of:

1. $F = \{(x,y,z) \in \R^3 : x + y + z = 0\}$ ;
2. $G = \{(x,y,z,t) \in \R^4 : x = y,\ z = 2t\}$ ;
3. $H = \{P \in \R_3[X] : P(1) = P'(1) = 0\}$ .

**Solution of Exercise 19.1.**

1. $z = -x - y$ : $F = \{(x, y, -x-y)\} =  \operatorname{Vect}\bigl((1,0,-1), (0,1,-1)\bigr)$ ; the two vectors are [free](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) ( [coordinates](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#prop-b1-vspaces-coordinates) ): $\dim F = 2$ .
2. $G = \{(x, x, 2t, t)\} =  \operatorname{Vect}\bigl((1,1,0,0), (0,0,2,1)\bigr)$ : [free](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) , $\dim G = 2$ .
3. $P(1) = P'(1) = 0$ means $(X-1)^2 \mid P$ ( [Proposition 8.11](https://one-course.com/books/math/3/en/chapter/8-polynomials#prop-b1-poly-multiplicity) ): $P = (X-1)^2(aX + b)$ . [Basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) $\bigl((X-1)^2, X(X-1)^2\bigr)$ , dimension $2$ .

**Exercise 19.2 ★.**

Compute the [rank](#def-b1-findim-rank) of the family $\bigl((1,1,1), (1,2,3), (3,5,7), (0,1,2)\bigr)$ in $\R^3$, and extract a [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) of its [span](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-span).

**Solution of Exercise 19.2.**

$(3,5,7) = (1,1,1) + 2(1,2,3)$ and $(0,1,2) = (1,2,3) - (1,1,1)$: both are combinations of the first two, which are [free](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) (not proportional). [Rank](#def-b1-findim-rank) $2$; [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) of the [span](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-span): $\bigl((1,1,1), (1,2,3)\bigr)$.

**Exercise 19.3 ★.**

Complete the [free family](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) $\bigl((1,1,0,0), (0,0,1,1)\bigr)$ into a [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) of $\R^4$ using canonical vectors, and justify.

**Solution of Exercise 19.3.**

Try appending $e_1 = (1,0,0,0)$ and $e_3 = (0,0,1,0)$. The family $\bigl((1,1,0,0), (0,0,1,1), e_1, e_3\bigr)$ is [free](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free): a null combination $\alpha(1,1,0,0) + \beta(0,0,1,1) + \gamma e_1 + \delta
e_3 = 0$ reads $(\alpha + \gamma, \alpha, \beta + \delta, \beta) =
0$, so $\alpha = \beta = 0$, then $\gamma = \delta = 0$. Four [free](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) vectors in dimension $4$: a [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) ([Proposition 19.8](#prop-b1-findim-twoofthree)).

**Exercise 19.4 ★.**

Prove that $\bigl(1, X, X(X-1), X(X-1)(X-2)\bigr)$ is a [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) of $\R_3[X]$, and find the [coordinates](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#prop-b1-vspaces-coordinates) of $X^3$ in it.

**Solution of Exercise 19.4.**

Degrees $0, 1, 2, 3$ pairwise distinct: [free](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) ([Proposition 18.19](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#prop-b1-vspaces-freecriteria)), four vectors in dimension $4$: [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free). For $X^3$: expand downward,

$$
X(X-1)(X-2) = X^3 - 3X^2 + 2X,
\qquad
X(X-1) = X^2 - X,
$$

so $X^3 - X(X-1)(X-2) = 3X^2 - 2X$; and $3X^2 - 2X = 3(X^2 - X) + X
= 3\,X(X-1) + X$. Hence

$$
X^3 = X(X-1)(X-2) + 3\,X(X-1) + 1\cdot X + 0\cdot 1 :
$$

[coordinates](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#prop-b1-vspaces-coordinates) $(0, 1, 3, 1)$ on $\bigl(1, X, X(X-1),
X(X-1)(X-2)\bigr)$. (These are Stirling numbers in disguise.)

**Exercise 19.5 ★★.**

Let $F$ and $G$ be [subspaces](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-subspace) of dimensions $4$ and $5$ of a space $E$ with $\dim E = 7$. What are the possible values of $\dim(F \cap
G)$? Give an instance realizing each value with $E = \R^7$.

**Solution of Exercise 19.5.**

Grassmann: $\dim(F \cap G) = 4 + 5 - \dim(F + G)$, and $F + G$ is a [subspace](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-subspace) of $E$ containing $G$: $5 \leq \dim(F+G) \leq 7$. Hence $\dim(F \cap G) \in \{2, 3, 4\}$. Realizations in $\R^7$ with canonical [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) $(e_1, \dots, e_7)$, taking $G =
\operatorname{Vect}(e_1, \dots, e_5)$:

- $F = \operatorname{Vect}(e_1, e_2, e_6, e_7)$ : $F + G = \R^7$ , intersection $\operatorname{Vect}(e_1, e_2)$ , dimension $2$ ;
- $F = \operatorname{Vect}(e_1, e_2, e_3, e_6)$ : intersection of dimension $3$ ;
- $F = \operatorname{Vect}(e_1, e_2, e_3, e_4) \subseteq G$ : dimension $4$ .

**Exercise 19.6 ★★.**

Let $H = \{P \in \R_n[X] : P(1) = 0\}$. Prove that $H$ is a hyperplane of $\R_n[X]$ (a [subspace](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-subspace) of dimension $n$), exhibit a [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) of $H$ *(think of the factor theorem: $P = (X-1)Q$)*, and give a [supplementary](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-sum) line.

**Solution of Exercise 19.6.**

$H$ is a [subspace](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-subspace) ([Exercise 18.1](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#exo-b1-vspaces-1) (4) verbatim). By the factor theorem ([Theorem 8.7](https://one-course.com/books/math/3/en/chapter/8-polynomials#thm-b1-poly-factor)), $P \in H \iff P =
(X-1)Q$ with $\deg Q \leq n - 1$: the [map](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-map) $Q \mapsto (X-1)Q$ is a linear bijection from $\R_{n-1}[X]$ onto $H$, so a [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) of $H$ is

$$
\bigl((X-1),\ (X-1)X,\ (X-1)X^2,\ \dots,\ (X-1)X^{n-1}\bigr),
\qquad \dim H = n .
$$

A [supplementary](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-sum) line: $\operatorname{Vect}(1)$ (constants). Indeed $H \cap \operatorname{Vect}(1) = \{0\}$ (a nonzero constant does not vanish at $1$) and dimensions add up to $n + 1$: by Grassmann, $H
\oplus \operatorname{Vect}(1) = \R_n[X]$.

**Exercise 19.7 ★★.**

Let $u_1, \dots, u_p$ be vectors of [rank](#def-b1-findim-rank) $r$. Prove that removing one vector yields a family of [rank](#def-b1-findim-rank) $r$ or $r - 1$, and that appending one vector yields [rank](#def-b1-findim-rank) $r$ or $r + 1$. Deduce that [rank](#def-b1-findim-rank) changes by at most $1$ under any single insertion or deletion.

**Solution of Exercise 19.7.**

Deletion: removing $u_p$, the [span](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-span) can only shrink; and it shrinks by at most one dimension, since adding $u_p$ back to a [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) of the smaller [span](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-span) gives a generating family of the larger with at most one extra vector. Symmetrically, appending a vector $v$: the new [span](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-span) contains the old with at most one extra generator, so its dimension is $r$ (if $v$ was already in the [span](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-span)) or $r + 1$ (otherwise, by [Proposition 18.19](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#prop-b1-vspaces-freecriteria) (2) a [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) extends). Both [statements](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-statement) together give the “[rank](#def-b1-findim-rank) is $1$-Lipschitz” conclusion.

**Exercise 19.8 ★★★.**

Let $F_1 \subseteq F_2 \subseteq \dots \subseteq F_k$ be [subspaces](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-subspace) of $E$ ($\dim E = n$) with $F_i \neq F_{i+1}$ for all $i$. Prove $k
\leq n + 1$. Deduce that a strictly increasing chain of [subspaces](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-subspace) of $\R^n$ has length at most $n + 1$, and exhibit one of maximal length.

**Solution of Exercise 19.8.**

Along a strictly increasing chain, dimensions strictly increase ($F_i \subseteq F_{i+1}$, $F_i \neq F_{i+1}$ and [Theorem 19.14](#thm-b1-findim-subspaces): equality of dimensions would force equality of spaces). So $\dim F_1 < \dim F_2 < \dots < \dim F_k$ is a strictly increasing sequence of integers in $\intint{0}{n}$: at most $n + 1$ values, $k \leq n + 1$. Maximal chain in $\R^n$:

$$
\{0\} \subsetneq \operatorname{Vect}(e_1) \subsetneq
\operatorname{Vect}(e_1, e_2) \subsetneq \dots \subsetneq \R^n ,
$$

of length exactly $n + 1$.

**Exercise 19.9 ★★★.**

Let $E$ be of dimension $n$ and $F$, $G$ two hyperplanes (dimension $n - 1$), $F \neq G$. Compute $\dim(F \cap G)$. Generalize: the intersection of $k$ hyperplanes has dimension $\geq n - k$.

**Solution of Exercise 19.9.**

$F + G$ strictly contains $F$ (since $G \not\subseteq F$), so $\dim(F + G) = n$ and Grassmann gives $\dim(F \cap G) = (n-1) +
(n-1) - n = n - 2$.

General claim, by induction on $k$: the intersection $I_k$ of $k$ hyperplanes has $\dim I_k \geq n - k$. True for $k = 1$. Step: $I_{k+1}
= I_k \cap H_{k+1}$, and Grassmann inside $E$:

$$
\dim(I_k \cap H_{k+1}) = \dim I_k + (n - 1) - \dim(I_k + H_{k+1})
\geq \dim I_k + (n-1) - n \geq n - k - 1 . \qedhere
$$

**Exercise 19.10 ★★.**

Let $E$ be the [set](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets) of real sequences satisfying $u_{n+2} =
3u_{n+1} - 2u_n$ for all $n$.

1. Show that $E$ is a [subspace](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-subspace) of the space of sequences, and that a sequence of $E$ is entirely determined, linearly, by the pair $(u_0, u_1)$ ; deduce $\dim E = 2$ .
2. Check that the constant sequence $(1)$ and the geometric sequence $(2^n)$ lie in $E$ and form a [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) of $E$ .
3. Find the sequence of $E$ with $u_0 = 0$ , $u_1 = 1$ .

**Solution of Exercise 19.10.**

1. The condition $u_{n+2} - 3u_{n+1} + 2u_n = 0$ is linear and satisfied by the zero sequence: $E$ is a [subspace](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-subspace) . By induction, $u_0$ and $u_1$ determine every $u_n$ , and the dependence is linear (each step is a linear combination of the two previous values); conversely every pair $(a, b)$ arises from exactly one sequence of $E$ (define $u_n$ by the recurrence). As in [Example 19.5](#ex-b1-findim-dims) , $E$ is parametrized [bijectively](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-inj) and linearly by $(u_0, u_1) \in  \R^2$ : $\dim E = 2$ .
2. Constants: $3\cdot1 - 2\cdot1 = 1$ . Geometric: $3\cdot  2^{n+1} - 2\cdot 2^n = (6 - 2)2^n = 2^{n+2}$ . Both lie in $E$ . Freeness: $a\cdot 1 + b\cdot 2^n = 0$ for all $n$ gives, at $n = 0$ and $n = 1$ : $a + b = 0$ , $a + 2b = 0$ , so $a = b = 0$ . Two [free](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) vectors in dimension $2$ : a [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) ( [Proposition 19.8](#prop-b1-findim-twoofthree) ).
3. Solve $a + b = 0$ , $a + 2b = 1$ : $b = 1$ , $a = -1$ , so $u_n = 2^n - 1$ (the Mersenne sequence).

**Exercise 19.11 ★★.**

Let $E$ be of dimension $n$.

1. If $F, G$ are [subspaces](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-subspace) with $\dim F + \dim G > n$ , prove $F \cap G \neq \{0\}$ . Illustrate: two [subspaces](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-subspace) of dimensions $51$ and $50$ of $\R^{100}$ always share a nonzero vector.
2. If $H$ is a hyperplane and $F$ a [subspace](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-subspace) with $F \cap H =  \{0\}$ , prove $\dim F \leq 1$ .

**Solution of Exercise 19.11.**

1. Grassmann: $\dim(F \cap G) = \dim F + \dim G - \dim(F +  G) \geq \dim F + \dim G - n > 0$ , so $F \cap G \neq  \{0\}$ . With $n = 100$ : $51 + 50 - 100 = 1 > 0$ , the intersection contains a line.
2. If $F \cap H = \{0\}$ , the sum is direct and $\dim F +  (n - 1) = \dim(F \oplus H) \leq n$ , so $\dim F \leq 1$ . (Conversely a line not contained in $H$ does satisfy this: hyperplanes miss almost nothing.)

**Exercise 19.12 ★★★.**

(Common [supplementary](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-sum)) Let $F, G$ be [subspaces](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-subspace) of $E$ ([finite dimension](#def-b1-findim-def)) with $\dim F = \dim G$. Prove that $F$ and $G$ admit a *common* [supplementary](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-sum): there is a [subspace](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-subspace) $S$ with $E = F
\oplus S = G \oplus S$. *(Induct downward on $\dim F$: if $F
\neq G \neq E$, pick $x \notin F \cup G$ — [Exercise 18.12](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#exo-b1-vspaces-12) allows it — and consider $F \oplus
\operatorname{Vect}(x)$ and $G \oplus \operatorname{Vect}(x)$.)*

**Solution of Exercise 19.12.**

Downward induction on $d = \dim F = \dim G$, from $d = n$ to $d =
0$. If $d = n$: $F = G = E$ and $S = \{0\}$ works. Suppose the [statement](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-statement) holds for [subspace](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-subspace) pairs of dimension $d + 1 \leq n$, and let $\dim F = \dim G = d < n$.

If $F = G$: take for $S$ any [supplementary](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-sum) of $F$ ([Theorem 19.14](#thm-b1-findim-subspaces)). If $F \neq G$: both are proper, so by [Exercise 18.12](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#exo-b1-vspaces-12) there exists $x \notin F \cup G$. The sums $F' = F \oplus \operatorname{Vect}(x)$ and $G' = G \oplus
\operatorname{Vect}(x)$ are direct ($x \notin F$, $x \notin G$) and have dimension $d + 1$; by the induction hypothesis they admit a common [supplementary](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-sum) $S'$: $E = F' \oplus S' = G' \oplus
S'$. Set $S = \operatorname{Vect}(x) \oplus S'$ — direct, because $S' \cap \operatorname{Vect}(x) \subseteq S' \cap F' =
\{0\}$.

Then $F + S = F + \operatorname{Vect}(x) + S' = F' + S' = E$, and $F \cap S = \{0\}$: if $f = \lambda x + s'$ with $f \in F$, $s'
\in S'$, then $s' = f - \lambda x \in F' \cap S' = \{0\}$, so $f
= \lambda x$, forcing $\lambda = 0$ ($x \notin F$) and $f = 0$. Hence $E = F \oplus S$, and symmetrically $E = G \oplus S$.

## 19.4 Problem: Dedekind’s tower law

**Problem 19.1.**

Nothing in Chapters 18–19 used anything about the scalars beyond the [field](https://one-course.com/books/math/3/en/chapter/7-algebraic-structures#def-b1-structures-field) axioms ([Definition 7.22](https://one-course.com/books/math/3/en/chapter/7-algebraic-structures#def-b1-structures-field)): one may therefore take $K = \Q$ and measure [sets](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets) of *real numbers* with the yardstick of $\Q$-dimension. This problem computes the dimension of $\Q(\sqrt2, \sqrt3)$, proves *Dedekind’s [tower law](#pb-b1-findim-1)* $\dim_\Q M = \dim_\Q K \cdot \dim_K M$, and harvests irrationality theorems by pure dimension counting — no $\varepsilon$, no decimals, just bases.

**Part I — Rational scalars.**

1. Check that $\Q$ is a [field](https://one-course.com/books/math/3/en/chapter/7-algebraic-structures#def-b1-structures-field) and that every definition and proof of Chapters 18–19 uses only the [field](https://one-course.com/books/math/3/en/chapter/7-algebraic-structures#def-b1-structures-field) axioms of the scalars; conclude that $\R$ is a $\Q$ -vector space and that the exchange lemma, the [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) theorems and Grassmann’s formula hold over $\Q$ . Point out the one step of the proof of [Theorem 19.2](#thm-b1-findim-exchange) where division by a nonzero scalar is performed.
2. Show that $(1, \sqrt2)$ is [free](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) over $\Q$ but linked over $\R$ . Set $\Q(\sqrt2) = \operatorname{Vect}_\Q(1,  \sqrt2) = \{a + b\sqrt2 : a, b \in \Q\}$ ; what is $\dim_\Q \Q(\sqrt2)$ ?
3. Show that $\Q(\sqrt2)$ is stable under multiplication. For $u = a + b\sqrt2$ set $\sigma(u) = a - b\sqrt2$ and $N(u) = u\,\sigma(u) = a^2 - 2b^2$ . Show $\sigma(uv) =  \sigma(u)\sigma(v)$ , deduce $N(uv) = N(u)N(v)$ , and show $N(u) \neq 0$ whenever $u \neq 0$ .
4. Deduce that every nonzero $u \in \Q(\sqrt2)$ has its inverse in $\Q(\sqrt2)$ , namely $u^{-1} =  \sigma(u)/N(u)$ : $\Q(\sqrt2)$ is a subfield of $\R$ . Compute $\dfrac1{3 + 2\sqrt2}$ and $\dfrac1{1 + \sqrt2}$ .

**Part II — Adjoining $\sqrt3$.**

5. Prove that $\sqrt6 \notin \Q$ *(compare the exponent of $2$ on both sides of $6q^2 = p^2$, as in [Exercise 6.7](https://one-course.com/books/math/3/en/chapter/6-integer-arithmetic#exo-b1-arith-7))* , then that $\sqrt3 \notin  \Q(\sqrt2)$ *(square $\sqrt3 = a + b\sqrt2$ and discuss the cases $ab \neq 0$, $b = 0$, $a = 0$)* .
6. Let $M = \operatorname{Vect}_\Q(1, \sqrt2, \sqrt3,  \sqrt6)$ . Show that $M$ is stable under multiplication *(a table of the products of [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) vectors suffices)* .
7. Write $K = \Q(\sqrt2)$ . Show that $(1, \sqrt3)$ is [free](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) *over the [field](https://one-course.com/books/math/3/en/chapter/7-algebraic-structures#def-b1-structures-field) $K$* , and deduce that $M = K +  K\sqrt3$ is a $K$ -vector space of dimension $2$ with [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) $(1, \sqrt3)$ .
8. Prove that $(1, \sqrt2, \sqrt3, \sqrt6)$ is [free](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) over $\Q$ , hence $\dim_\Q M = 4$ . *(Group a null relation as $(a + b\sqrt2) + (c + d\sqrt2)\sqrt3 = 0$ and apply questions 7 then 2.)*

**Part III — The [tower law](#pb-b1-findim-1).** Let $\Q \subseteq K
\subseteq M \subseteq \R$ where $K$ and $M$ are subfields, $(e_1,
\dots, e_m)$ is a [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) of $K$ as a $\Q$-vector space, and $(f_1,
\dots, f_n)$ a [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) of $M$ as a $K$-vector space.

9. Show that the $mn$ products $(e_i f_j)$ generate $M$ over $\Q$ .
10. Show that the family $(e_i f_j)$ is [free](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) over $\Q$ . *(Reorganize a null $\Q$-combination as $\sum_j  \bigl(\sum_i \lambda_{ij} e_i\bigr) f_j$, whose inner coefficients live in $K$.)*
11. Conclude with *Dedekind’s [tower law](#pb-b1-findim-1)*: $$\dim_\Q M \;=\; \dim_\Q K \,\cdot\, \dim_K M ,$$ and check it on $\Q \subseteq \Q(\sqrt2) \subseteq M$ against questions 7 and 8.
12. ( [Fields](https://one-course.com/books/math/3/en/chapter/7-algebraic-structures#def-b1-structures-field) for free) Let $A \subseteq \R$ be a [finite-dimensional](#def-b1-findim-def) $\Q$ -subspace containing $1$ and stable under multiplication, and let $u \in A$ , $u \neq  0$ . Show that if $(a_1, \dots, a_d)$ is a [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) of $A$ , then $(u a_1, \dots, u a_d)$ is again a [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) of $A$ ; deduce that $u$ has an inverse *in $A$* : $A$ is a subfield of $\R$ . Which earlier questions does this recover?
13. Show that for every $x \in A$ (as in question 12, $\dim_\Q  A = d$ ) the family $(1, x, x^2, \dots, x^{d})$ is linked: every element of $A$ is a root of a nonzero [polynomial](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def) with rational coefficients, of degree at most $d$ .

**Part IV — One number generates everything.** Set $s = \sqrt2 + \sqrt3$.

14. Compute the [coordinates](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#prop-b1-vspaces-coordinates) of $s^2$ , $s^3$ and $s^4$ in the [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) $(1, \sqrt2, \sqrt3, \sqrt6)$ of $M$ .
15. Show that $(1, s, s^2, s^3)$ is [free](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) over $\Q$ , and deduce $\operatorname{Vect}_\Q(1, s, s^2, s^3) = M$ : every element of $M$ is a rational [polynomial](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def) in $s$ . Express $\sqrt2$ and $\sqrt3$ as such [polynomials](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def) .
16. Verify $s^4 - 10s^2 + 1 = 0$ , and show that $X^4 - 10X^2  + 1$ is the *[monic polynomial](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def) of least degree* vanishing at $s$ . Determine its four real roots.
17. Deduce: $1/s = 10s - s^3$ ; identify this number. Show that $a + b\sqrt2 + c\sqrt3 + d\sqrt6$ ( $a,b,c,d \in \Q$ ) is rational if and only if $b = c = d = 0$ ; in particular $\sqrt2 + \sqrt3 + \sqrt6$ is irrational (strengthening [Exercise 10.7](https://one-course.com/books/math/3/en/chapter/10-real-numbers#exo-b1-reals-7) ).

**Part V — The cube root stays outside.** Set $t
= 2^{1/3}$.

18. Show that $X^3 - 2$ has no rational root *(the rational-root criterion of [Exercise 8.5](https://one-course.com/books/math/3/en/chapter/8-polynomials#exo-b1-poly-5), or a valuation count)* ; deduce that $(1, t)$ is [free](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) over $\Q$ .
19. Show that $(1, t, t^2)$ is [free](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) over $\Q$ . *(If some nonzero $P \in \Q[X]$ of degree $\leq 2$ kills $t$, take one of least degree and divide $X^3 - 2$ by it, [Theorem 8.3](https://one-course.com/books/math/3/en/chapter/8-polynomials#thm-b1-poly-division); conclude that $X^3 - 2$ would have a rational root.)* Deduce that $A =  \operatorname{Vect}_\Q(1, t, t^2)$ has dimension $3$ , is stable under multiplication, and is a subfield of $\R$ .
20. Prove that $t \notin M$ : otherwise $M$ would be a [vector space](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-def) over the [field](https://one-course.com/books/math/3/en/chapter/7-algebraic-structures#def-b1-structures-field) $A$ , and the [tower law](#pb-b1-findim-1) would force $3 \mid 4$ . So $2^{1/3}$ is not a rational combination of $1, \sqrt2, \sqrt3, \sqrt6$ .
21. Deduce that $t \notin \Q(\sqrt2)$ , that $t + \sqrt2$ is irrational, and that no rationals $a, b$ satisfy $2^{1/3} = a + b\sqrt3$ .

**Part VI — All the subfields, and synthesis.**

22. Prove the [divisibility](https://one-course.com/books/math/3/en/chapter/6-integer-arithmetic#def-b1-arith-divides) of degrees: if $\Q \subseteq A  \subseteq B \subseteq \R$ are subfields with $\dim_\Q B$ finite, then $\dim_\Q A$ [divides](https://one-course.com/books/math/3/en/chapter/6-integer-arithmetic#def-b1-arith-divides) $\dim_\Q B$ . What are the possible dimensions of subfields of $M$ ?
23. Determine *all* subfields of $M$ of dimension $2$ over $\Q$ . *(Reduce to finding the $w = a + b\sqrt2  + c\sqrt3 + d\sqrt6 \in M \setminus \Q$ with $w^2 \in  \Q$: expand $w^2$ and annihilate the three irrational [coordinates](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#prop-b1-vspaces-coordinates).)*
24. Express $\dfrac1{1 + \sqrt2 + \sqrt3}$ in the [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) $(1,  \sqrt2, \sqrt3, \sqrt6)$ *(multiply by well-chosen conjugates)* .
25. Synthesis, in four sentences: why $\Q$ -dimension is an *arithmetic* invariant of a [set](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets) of reals; what finiteness of the dimension forces about every element (question 13); how the two-out-of-three rule produced inverses out of thin air (question 12); and what the [tower law](#pb-b1-findim-1) forbids (question 20). Name the theorem proved in Part III.

**Solution of Problem 19.1.**

**1.** $\Q$ contains $0 \neq 1$, is stable under addition, multiplication and opposites, and every nonzero rational has a rational inverse: a [field](https://one-course.com/books/math/3/en/chapter/7-algebraic-structures#def-b1-structures-field) ([Definition 7.22](https://one-course.com/books/math/3/en/chapter/7-algebraic-structures#def-b1-structures-field)). The definitions of [span](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-span), freeness, [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) and the proofs of Chapters 18–19 use only vector addition, distributivity and scalar arithmetic in a [field](https://one-course.com/books/math/3/en/chapter/7-algebraic-structures#def-b1-structures-field) — never an absolute value, an order or a limit. So $\R$, with its own addition and the multiplication $\Q \times \R \to \R$, is a $\Q$-vector space, and all the theorems apply. Division by a scalar occurs once in the proof of [Theorem 19.2](#thm-b1-findim-exchange): to “solve for $g_k$” one [divides](https://one-course.com/books/math/3/en/chapter/6-integer-arithmetic#def-b1-arith-divides) by its nonzero coefficient.

**2.** If $a + b\sqrt2 = 0$ with $a, b \in \Q$ not both zero: $b \neq 0$ would give $\sqrt2 = -a/b \in \Q$, contradicting the irrationality of $\sqrt2$ ([Exercise 6.7](https://one-course.com/books/math/3/en/chapter/6-integer-arithmetic#exo-b1-arith-7) with $p =
2$); so $b = 0$, then $a = 0$: [free](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) over $\Q$. Over $\R$ the relation $\sqrt2\cdot 1 + (-1)\cdot\sqrt2 = 0$ is nontrivial: linked. Hence $\dim_\Q \Q(\sqrt2) = 2$, with unique [coordinates](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#prop-b1-vspaces-coordinates) $(a, b)$.

**3.** $(a + b\sqrt2)(c + d\sqrt2) = (ac + 2bd) + (ad +
bc)\sqrt2 \in \Q(\sqrt2)$. Then

$$
\sigma(uv) = (ac + 2bd) - (ad + bc)\sqrt2
= (a - b\sqrt2)(c - d\sqrt2) = \sigma(u)\sigma(v),
$$

so $N(uv) = uv\,\sigma(uv) = u\sigma(u)\,v\sigma(v) = N(u)N(v)$. If $N(u) = a^2 - 2b^2 = 0$ with $u \neq 0$: $b \neq 0$ would give $(a/b)^2 = 2$, a rational square root of $2$; so $b = 0$, then $a
= 0$ and $u = 0$: contradiction. Hence $N(u) \neq 0$ for $u \neq
0$.

**4.** $u \cdot \dfrac{\sigma(u)}{N(u)} = \dfrac{N(u)}{N(u)}
= 1$, and $\sigma(u)/N(u) \in \Q(\sqrt2)$: every nonzero element is invertible inside $\Q(\sqrt2)$, which is therefore a subfield of $\R$. Examples: $N(3 + 2\sqrt2) = 9 - 8 = 1$, so

$$
\frac1{3 + 2\sqrt2} = 3 - 2\sqrt2 ;
\qquad
N(1 + \sqrt2) = -1, \quad
\frac1{1 + \sqrt2} = \sqrt2 - 1 .
$$

**5.** If $\sqrt6 = p/q$ then $6q^2 = p^2$; the exponent of $2$ is $1 + 2v_2(q)$, odd, on the left, and $2v_2(p)$, even, on the right: impossible. Now suppose $\sqrt3 = a + b\sqrt2$ with $a, b \in \Q$. Squaring: $3 = a^2 + 2b^2 + 2ab\sqrt2$. If $ab
\neq 0$, then $\sqrt2 = (3 - a^2 - 2b^2)/(2ab) \in \Q$: impossible. If $b = 0$: $\sqrt3 = a \in \Q$, contradicting [Exercise 6.7](https://one-course.com/books/math/3/en/chapter/6-integer-arithmetic#exo-b1-arith-7) ($p = 3$). If $a = 0$: $\sqrt3 = b\sqrt2$, and multiplying by $\sqrt2$: $\sqrt6 = 2b \in \Q$: impossible. So $\sqrt3 \notin \Q(\sqrt2)$.

**6.** The products of [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) vectors are

$$
\sqrt2\,\sqrt3 = \sqrt6,\quad
\sqrt2\,\sqrt6 = 2\sqrt3,\quad
\sqrt3\,\sqrt6 = 3\sqrt2,\quad
(\sqrt2)^2 = 2,\ (\sqrt3)^2 = 3,\ (\sqrt6)^2 = 6,
$$

all in $M$. A product of two elements of $M$ expands by bilinearity into rational combinations of these: $M$ is stable under multiplication.

**7.** Let $x + y\sqrt3 = 0$ with $x, y \in K = \Q(\sqrt2)$. If $y \neq 0$, then $\sqrt3 = -x/y \in K$ (question 4: $K$ is a [field](https://one-course.com/books/math/3/en/chapter/7-algebraic-structures#def-b1-structures-field)), contradicting question 5. So $y = 0$, then $x = 0$: $(1,
\sqrt3)$ is [free](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) over $K$. It generates: $K + K\sqrt3 =
\{(a + b\sqrt2) + (c + d\sqrt2)\sqrt3\} = \operatorname{Vect}_\Q
(1, \sqrt2, \sqrt3, \sqrt6) = M$. Hence $\dim_K M = 2$.

**8.** A relation $a + b\sqrt2 + c\sqrt3 + d\sqrt6 = 0$ regroups as $(a + b\sqrt2) + (c + d\sqrt2)\sqrt3 = 0$ with coefficients in $K$; by question 7 both vanish, and by question 2 $a = b = 0$ and $c = d = 0$. So the family is [free](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) and $\dim_\Q M
= 4$.

**9.** Every $x \in M$ writes $x = \sum_j y_j f_j$ with $y_j
\in K$ ([basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) of $M$ over $K$), and each $y_j = \sum_i
\lambda_{ij} e_i$ with $\lambda_{ij} \in \Q$ ([basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) of $K$ over $\Q$); substituting, $x = \sum_{i,j} \lambda_{ij}\, e_i f_j$: the products generate $M$ over $\Q$.

**10.** Suppose $\sum_{i,j} \lambda_{ij}\, e_i f_j = 0$ with $\lambda_{ij} \in \Q$. Regroup: $\sum_j \bigl(\sum_i \lambda_{ij}
e_i\bigr) f_j = 0$, and the inner sums belong to $K$. Freeness of $(f_j)$ over $K$ gives $\sum_i \lambda_{ij} e_i = 0$ for each $j$; freeness of $(e_i)$ over $\Q$ then gives $\lambda_{ij} = 0$ for all $i, j$.

**11.** By questions 9 and 10, $(e_i f_j)_{i \leq m,\, j
\leq n}$ is a [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) of $M$ over $\Q$ with $mn$ elements:

$$
\dim_\Q M = m\,n = \dim_\Q K \cdot \dim_K M .
$$

Check: $\dim_\Q K = 2$ and $\dim_K M = 2$ (questions 2 and 7) give $\dim_\Q M = 4$, which is question 8.

**12.** The vectors $u a_1, \dots, u a_d$ belong to $A$ (stability). They are [free](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free): if $\sum_i \lambda_i\, u a_i = 0$, then $u \sum_i \lambda_i a_i = 0$ in $\R$, and $u \neq 0$ forces $\sum_i \lambda_i a_i = 0$, hence $\lambda_i = 0$ (the $a_i$ form a [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free)). So $(u a_1, \dots, u a_d)$ is a [free family](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) of $d$ vectors in $A$, $\dim_\Q A = d$: a [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) ([Proposition 19.8](#prop-b1-findim-twoofthree)). In particular $1 \in A$ decomposes as $1 = \sum_i \mu_i\, u a_i = u\,v$ with $v = \sum_i
\mu_i a_i \in A$: the inverse of $u$ lies in $A$. This recovers question 4 ($A = \Q(\sqrt2)$) and proves at one stroke that $M$ is a subfield of $\R$ (with question 6).

**13.** The $d + 1$ vectors $1, x, x^2, \dots, x^{d}$ all lie in $A$ (stability under products); a [free family](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) of $A$ has at most $d$ vectors ([Proposition 19.8](#prop-b1-findim-twoofthree)), so they are linked: there are rationals $\lambda_0, \dots, \lambda_d$, not all zero, with $\sum_k \lambda_k x^k = 0$. The [polynomial](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def) $P
= \sum_k \lambda_k X^k$ is nonzero, has rational coefficients, degree $\leq d$, and $P(x) = 0$.

**14.** $s^2 = 2 + 2\sqrt6 + 3 = 5 + 2\sqrt6$: [coordinates](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#prop-b1-vspaces-coordinates) $(5, 0, 0, 2)$. Then

$$
s^3 = s\,s^2 = (\sqrt2 + \sqrt3)(5 + 2\sqrt6)
= 5\sqrt2 + 2\sqrt{12} + 5\sqrt3 + 2\sqrt{18}
= 11\sqrt2 + 9\sqrt3 ,
$$

[coordinates](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#prop-b1-vspaces-coordinates) $(0, 11, 9, 0)$ (using $\sqrt{12} = 2\sqrt3$, $\sqrt{18} = 3\sqrt2$). Finally $s^4 = (s^2)^2 = 25 + 20\sqrt6 +
24 = 49 + 20\sqrt6$: [coordinates](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#prop-b1-vspaces-coordinates) $(49, 0, 0, 20)$.

**15.** Suppose $a + bs + cs^2 + ds^3 = 0$. Reading the four [coordinates](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#prop-b1-vspaces-coordinates) on $(1, \sqrt2, \sqrt3, \sqrt6)$:

$$
a + 5c = 0,\qquad b + 11d = 0,\qquad b + 9d = 0,\qquad 2c = 0 .
$$

So $c = 0$, then $a = 0$; subtracting the middle equations, $2d =
0$, then $b = 0$: $(1, s, s^2, s^3)$ is [free](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free). Four [free](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) vectors in $M$ of dimension $4$: a [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free), so $\operatorname{Vect}_\Q(1, s, s^2, s^3) = M$. From question 14, $s^3 - 9s = 2\sqrt2$ and $11s - s^3 = 2\sqrt3$:

$$
\sqrt2 = \frac{s^3 - 9s}{2},
\qquad
\sqrt3 = \frac{11s - s^3}{2} .
$$

**16.** $s^4 - 10s^2 + 1 = (49 + 20\sqrt6) - 10(5 +
2\sqrt6) + 1 = 0$. A [monic polynomial](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def) of degree $\leq 3$ vanishing at $s$ would produce a nontrivial null combination of $(1, s, s^2, s^3)$, contradicting question 15: $X^4 - 10X^2 + 1$ has least degree. Its roots: $X^2 = 5 \pm 2\sqrt6 = (\sqrt3 \pm
\sqrt2)^2$, so the four real roots are $\pm(\sqrt3 + \sqrt2)$ and $\pm(\sqrt3 - \sqrt2)$, i.e. $\pm\sqrt2 \pm \sqrt3$.

**17.** From $s^4 - 10s^2 + 1 = 0$: $s\,(10s - s^3) = 1$, so

$$
\frac1s = 10s - s^3 = 10(\sqrt2 + \sqrt3) - (11\sqrt2 + 9\sqrt3)
= \sqrt3 - \sqrt2 ,
$$

consistent with $(\sqrt3 + \sqrt2)(\sqrt3 - \sqrt2) = 1$. If $a +
b\sqrt2 + c\sqrt3 + d\sqrt6 = r \in \Q$, then $(a - r) + b\sqrt2
+ c\sqrt3 + d\sqrt6 = 0$, and freeness (question 8) forces $b = c
= d = 0$ (and $a = r$). For $\sqrt2 + \sqrt3 + \sqrt6$ the [coordinates](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#prop-b1-vspaces-coordinates) $(0, 1, 1, 1)$ are not of this form: irrational — a [statement](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-statement) stronger than [Exercise 10.7](https://one-course.com/books/math/3/en/chapter/10-real-numbers#exo-b1-reals-7), obtained without any squaring tricks.

**18.** A rational root $p/q$ (lowest terms) of $X^3 - 2$ satisfies $p^3 = 2q^3$, and the criterion of [Exercise 8.5](https://one-course.com/books/math/3/en/chapter/8-polynomials#exo-b1-poly-5) gives $p \mid 2$, $q \mid 1$: candidates $\pm1, \pm2$, whose cubes are $\pm1, \pm8 \neq 2$. No rational root; in particular $t = 2^{1/3} \notin \Q$, so $(1, t)$ is [free](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) over $\Q$ (as in question 2).

**19.** Suppose $(1, t, t^2)$ linked: some nonzero $P \in
\Q[X]$ with $\deg P \leq 2$ has $P(t) = 0$; choose such a $P$ of *least* degree $d \geq 1$. By question 18, $d \neq 1$, so $d
= 2$. Euclidean division ([Theorem 8.3](https://one-course.com/books/math/3/en/chapter/8-polynomials#thm-b1-poly-division)): $X^3 - 2
= PQ + R$ with $Q, R \in \Q[X]$, $\deg R < 2$. Evaluating at $t$: $0 = P(t)Q(t) + R(t) = R(t)$, so $R$ vanishes at $t$; minimality of $d$ forces $R = 0$. Then $X^3 - 2 = PQ$ with $\deg Q = 1$: the rational root of $Q$ is a rational root of $X^3 - 2$, contradicting question 18. Hence $(1, t, t^2)$ is [free](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) and $\dim_\Q A = 3$. Stability: $t^3 = 2$ reduces every product of $1, t, t^2$ to a combination of them ($t\cdot t^2 = 2$, $t^2\cdot
t^2 = 2t$); $A$ contains $1$: by question 12, $A$ is a subfield of $\R$.

**20.** Suppose $t \in M$. Then $t^2 \in M$ (stability), so $A \subseteq M$, and $M$ is a [vector space](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-def) over the [field](https://one-course.com/books/math/3/en/chapter/7-algebraic-structures#def-b1-structures-field) $A$: the axioms are those of $\R$-arithmetic, restricted. It is [finite-dimensional](#def-b1-findim-def) over $A$ (a finite $\Q$-generating family generates a fortiori over $A \supseteq \Q$). The [tower law](#pb-b1-findim-1) for $\Q \subseteq A \subseteq M$ gives

$$
4 = \dim_\Q M = \dim_\Q A \cdot \dim_A M = 3\,\dim_A M ,
$$

impossible: $3$ does not divide $4$. So $2^{1/3} \notin M$: no rational combination of $1, \sqrt2, \sqrt3, \sqrt6$ equals $2^{1/3}$.

**21.** $\Q(\sqrt2) \subseteq M$, so $t \notin \Q(\sqrt2)$. If $t + \sqrt2 = r \in \Q$, then $t = r - \sqrt2 \in \Q(\sqrt2)$: contradiction — $2^{1/3} + \sqrt2$ is irrational. If $t = a +
b\sqrt3$, then $t \in \operatorname{Vect}_\Q(1, \sqrt3) \subseteq
M$: contradiction again.

**22.** $B$ is a [vector space](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-def) over the [field](https://one-course.com/books/math/3/en/chapter/7-algebraic-structures#def-b1-structures-field) $A$ (restriction of scalars), [finite-dimensional](#def-b1-findim-def) since $\dim_\Q B$ is finite. The [tower law](#pb-b1-findim-1) for $\Q \subseteq A \subseteq B$ gives $\dim_\Q B = \dim_\Q A \cdot \dim_A B$: the left factor [divides](https://one-course.com/books/math/3/en/chapter/6-integer-arithmetic#def-b1-arith-divides). Subfields of $M$ therefore have $\Q$-dimension $1$, $2$ or $4$: dimension $1$ is $\Q$ itself, dimension $4$ is $M$.

**23.** Let $F \subseteq M$ be a subfield with $\dim_\Q F =
2$, and $w \in F \setminus \Q$: $(1, w)$ is [free](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free), hence a [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) of $F$. By question 13 ($d = 2$), $w^2 = \alpha + \beta w$ for rationals $\alpha, \beta$. Setting $v = w - \beta/2 \in F
\setminus \Q$:

$$
v^2 = w^2 - \beta w + \frac{\beta^2}4 = \alpha + \frac{\beta^2}4
\;\in\; \Q,
$$

and $F = \operatorname{Vect}(1, v)$. Now write $v = a + b\sqrt2 +
c\sqrt3 + d\sqrt6$ and expand:

$$
v^2 = (a^2 + 2b^2 + 3c^2 + 6d^2)
+ (2ab + 6cd)\sqrt2 + (2ac + 4bd)\sqrt3 + (2ad + 2bc)\sqrt6 .
$$

The three irrational [coordinates](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#prop-b1-vspaces-coordinates) vanish: $ab + 3cd = ac + 2bd =
ad + bc = 0$. If $a \neq 0$: $b = -3cd/a$, and the second condition becomes $c(a^2 - 6d^2) = 0$; since $a^2 = 6d^2$ with $d
\neq 0$ would make $\sqrt6 = \abs{a/d}$ rational, either $c = 0$ or $d = 0$, and in both cases the remaining conditions force $b =
c = d = 0$, i.e. $v \in \Q$: excluded. So $a = 0$, and the conditions read $3cd = 2bd = bc = 0$: at most one of $b, c, d$ is nonzero. Hence $v$ is a rational multiple of $\sqrt2$, $\sqrt3$ or $\sqrt6$, and

$$
F \in \bigl\{\Q(\sqrt2),\ \Q(\sqrt3),\ \Q(\sqrt6)\bigr\} ,
$$

each of which is indeed a $2$-dimensional subfield (stability as in question 6, inverses by question 12): exactly three quadratic subfields.

**24.** $(1 + \sqrt2 + \sqrt3)(1 + \sqrt2 - \sqrt3) = (1 +
\sqrt2)^2 - 3 = 2\sqrt2$, so

$$
\frac1{1 + \sqrt2 + \sqrt3}
= \frac{1 + \sqrt2 - \sqrt3}{2\sqrt2}
= \frac{\sqrt2 + 2 - \sqrt6}{4}
= \frac12 + \frac14\sqrt2 + 0\cdot\sqrt3 - \frac14\sqrt6 .
$$

(Check: $(1 + \sqrt2 + \sqrt3)(2 + \sqrt2 - \sqrt6) = 4$ after expansion.)

**25.** (i) $\Q$-dimension attaches an *integer* to each subfield of $\R$, and the [tower law](#pb-b1-findim-1) makes these integers multiply along inclusions: dimension behaves like an arithmetic invariant, and [divisibility](https://one-course.com/books/math/3/en/chapter/6-integer-arithmetic#def-b1-arith-divides) constraints become impossibility proofs. (ii) [Finite dimension](#def-b1-findim-def) forces every element to satisfy a nonzero rational [polynomial](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def) equation of degree at most the dimension: finiteness means algebraicity. (iii) The two-out-of-three rule turned “multiplication by $u$ [maps](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-map) a [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) to a [free family](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) of maximal size” into surjectivity, producing $u^{-1}$ with no formula: inverses came from counting. (iv) The [tower law](#pb-b1-findim-1) forbids a $3$-dimensional [field](https://one-course.com/books/math/3/en/chapter/7-algebraic-structures#def-b1-structures-field) inside a $4$-dimensional one, which is why $2^{1/3}$ cannot be reached from $\sqrt2$ and $\sqrt3$. The theorem of Part III is *Dedekind’s [tower law](#pb-b1-findim-1)*, the opening move of Galois theory, developed in the Year 3 volume.