---
title: "Linear Maps"
book: "University Mathematics — Year 1"
subject: math
language: en
chapter: 20
exercises: 12
source: https://one-course.com/books/math/3/en/chapter/20-linear-maps
---

# Chapter 20 — Linear Maps

The [maps](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-map) worth studying between [vector spaces](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-def) are those compatible with the structure: *[linear maps](#def-b1-linmaps-def)*. Their two fundamental [subspaces](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-subspace) — [kernel](#def-b1-linmaps-kerim) and [image](#def-b1-linmaps-kerim) — measure injectivity and surjectivity, and in [finite dimension](https://one-course.com/books/math/3/en/chapter/19-finite-dimension#def-b1-findim-def) the rank–nullity theorem binds their sizes into one conservation law. [Projections](#def-b1-linmaps-projection) and symmetries, then [linear forms](#def-b1-linmaps-forms) and [hyperplanes](#def-b1-linmaps-forms), close the chapter.

## 20.1 Definitions and first properties

**Definition 20.1.**

Let $E, F$ be $K$-vector spaces. A [map](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-map) $u \colon E \to F$ is *linear* when

$$
\forall x, y \in E,\ \forall \lambda \in K, \qquad
u(x + \lambda y) = u(x) + \lambda u(y).
$$

Then $u(0) = 0$ and $u(\sum \lambda_i x_i) = \sum \lambda_i u(x_i)$. The [set](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets) $\mathcal{L}(E, F)$ of linear [maps](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-map) is itself a [vector space](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-def); composition of linear [maps](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-map) is linear, and bilinear in the pair. An *endomorphism* is a linear $E \to E$; an *isomorphism* is a [bijective](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-inj) linear [map](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-map) (its inverse is then automatically linear); $u^{-1}$ of an isomorphism, compositions of isomorphisms, are isomorphisms.

**Proof that the inverse is linear.** Let $u$ be [linear](#def-b1-linmaps-def) [bijective](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-inj), $y, y' \in F$ and $\lambda \in K$. Put $x = u^{-1}(y)$ and $x' = u^{-1}(y')$. Then

$$
u\bigl(x + \lambda x'\bigr) = u(x) + \lambda u(x')
= y + \lambda y' ,
$$

and applying $u^{-1}$ to both ends: $u^{-1}(y + \lambda y') = x
+ \lambda x' = u^{-1}(y) + \lambda\,u^{-1}(y')$. Nothing about $u^{-1}$ was computed: linearity transports through the defining property of $u$ alone — a pattern worth remembering, as structure often rides along bijections for free. ∎

**Proposition 20.2 (A linear map is known on a basis).**

Let $(e_1, \dots, e_n)$ be a [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) of $E$ and $(v_1, \dots, v_n)$ arbitrary vectors of $F$. There is exactly one [linear map](#def-b1-linmaps-def) $u \colon
E \to F$ with $u(e_i) = v_i$ for all $i$. Moreover:

$$
u \text{ injective} \iff (v_i) \text{ free};
\qquad
u \text{ surjective} \iff (v_i) \text{ generates } F .
$$

**Proof.** Existence/uniqueness: every $x$ has unique [coordinates](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#prop-b1-vspaces-coordinates) $x = \sum
\lambda_i e_i$ ([Proposition 18.15](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#prop-b1-vspaces-coordinates)); linearity forces $u(x) = \sum \lambda_i v_i$, and this formula does define a [linear map](#def-b1-linmaps-def).

Injectivity: by [Proposition 20.5](#prop-b1-linmaps-kernel) below, $u$ is [injective](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-inj) iff its [kernel](#def-b1-linmaps-kerim) is trivial. Now $u\bigl(\sum\lambda_i
e_i\bigr) = 0$ means exactly $\sum\lambda_i v_i = 0$. If $(v_i)$ is [free](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free), this forces every $\lambda_i = 0$, i.e. the [kernel](#def-b1-linmaps-kerim) reduces to $0$: [injective](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-inj). If $(v_i)$ is linked, a nontrivial relation $\sum\lambda_i v_i = 0$ produces the nonzero vector $\sum\lambda_i e_i$ in the [kernel](#def-b1-linmaps-kerim) ($(e_i)$ is [free](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free)): not [injective](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-inj). The two conditions match term for term.

Surjectivity: the [image](#def-b1-linmaps-kerim) of $u$ is the [set](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets) of all $\sum\lambda_i
v_i$, i.e. exactly $\operatorname{Vect}(v_1, \dots, v_n)$, which equals $F$ iff the family generates. ∎

**Definition 20.3 (Kernel and image).**

For $u \in \mathcal{L}(E, F)$:

$$
\ker u = \{x \in E : u(x) = 0\} \subseteq E,
\qquad
\operatorname{im} u = u(E) \subseteq F ,
$$

both [subspaces](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-subspace) (direct verification with the criterion).

**Method 20.4 (Kernel and image, in practice).**

*[Kernel](#def-b1-linmaps-kerim)*: write $u(x) = 0$ as a system on the [coordinates](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#prop-b1-vspaces-coordinates) (or coefficients) of $x$, solve, and parametrize — the [kernel](#def-b1-linmaps-kerim) comes out with a [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) attached ([Method 19.10](https://one-course.com/books/math/3/en/chapter/19-finite-dimension#met-b1-findim-computedim)). *[Image](#def-b1-linmaps-kerim)*: it is the [span](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-span) of the [images](#def-b1-linmaps-kerim) of *any* generating family of $E$ — usually a [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free), so $\operatorname{im} u =
\operatorname{Vect}\bigl(u(e_1), \dots, u(e_n)\bigr)$; then eliminate redundant [images](#def-b1-linmaps-kerim) to extract a [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free). *Shortcut*: compute whichever of the two is easier and get the other’s dimension for free by rank–nullity ([Theorem 20.7](#thm-b1-linmaps-ranknullity)); when a plausible candidate [subspace](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-subspace) for the [image](#def-b1-linmaps-kerim) is known, comparing dimensions upgrades the easy inclusion to equality ([Theorem 19.14](https://one-course.com/books/math/3/en/chapter/19-finite-dimension#thm-b1-findim-subspaces)). Both shortcuts are used in [Example 20.11](#ex-b1-linmaps-delta) below.

**Proposition 20.5.**

$u$ is [injective](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-inj) $\iff$ $\ker u = \{0\}$; $u$ is [surjective](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-inj) $\iff$ $\operatorname{im} u = F$.

**Proof.** As for [groups](https://one-course.com/books/math/3/en/chapter/7-algebraic-structures#def-b1-structures-group) ([Proposition 7.11](https://one-course.com/books/math/3/en/chapter/7-algebraic-structures#prop-b1-structures-kernel)): $u(x) = u(y) \iff
u(x - y) = 0 \iff x - y \in \ker u$. The second point is the definition. ∎

## 20.2 The rank–nullity theorem

**Definition 20.6.**

The *rank* of $u \in \mathcal{L}(E,
F)$ (with $E$ [finite-dimensional](https://one-course.com/books/math/3/en/chapter/19-finite-dimension#def-b1-findim-def)) is $\operatorname{rk} u = \dim
\operatorname{im} u$ — also the rank of the family $\bigl(u(e_1), \dots, u(e_n)\bigr)$ for any [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) $(e_i)$ of $E$.

**Theorem 20.7 (Rank–nullity).**

Let $E$ be [finite-dimensional](https://one-course.com/books/math/3/en/chapter/19-finite-dimension#def-b1-findim-def) and $u \in \mathcal{L}(E, F)$. Then

$$
\dim E = \dim \ker u + \operatorname{rk} u .
$$

More precisely, if $S$ is any [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) of $\ker u$ in $E$, then $u$ restricts to an *isomorphism* from $S$ onto $\operatorname{im} u$.

**Proof.** Let $S$ satisfy $E = \ker u \oplus S$ ([Theorem 19.14](https://one-course.com/books/math/3/en/chapter/19-finite-dimension#thm-b1-findim-subspaces)), and let $v \colon S \to
\operatorname{im} u$ be the restriction of $u$.

$v$ is [injective](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-inj): $\ker v = S \cap \ker u = \{0\}$.

$v$ is [surjective](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-inj): any $u(x)$ with $x = k + s$ ($k \in \ker u$, $s
\in S$) equals $u(s) = v(s)$.

So $v$ is an isomorphism; an isomorphism sends a [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) to a [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) ([Proposition 20.2](#prop-b1-linmaps-basis)), so $\dim S = \dim\operatorname{im}
u$, and $\dim E = \dim\ker u + \dim S$ concludes. ∎

**Example 20.8 (Building a map to specification).**

Construct $u \in \mathcal{L}(\R^3)$ with $\ker u =
\operatorname{Vect}(1,1,1)$ and $\operatorname{im} u = \{z =
0\}$. Sanity first: rank–nullity demands $1 + 2 = 3$ — consistent, so a solution may exist. Choose a [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) adapted to the [kernel](#def-b1-linmaps-kerim), say $\bigl((1,1,1),\ e_1,\ e_2\bigr)$ ([Example 19.7](https://one-course.com/books/math/3/en/chapter/19-finite-dimension#ex-b1-findim-completion)), and prescribe [images](#def-b1-linmaps-kerim) ([Proposition 20.2](#prop-b1-linmaps-basis)):

$$
u(1,1,1) = 0, \qquad u(e_1) = e_1, \qquad u(e_2) = e_2 .
$$

Then $\ker u \supseteq \operatorname{Vect}(1,1,1)$ and $\operatorname{im} u = \operatorname{Vect}(e_1, e_2) = \{z =
0\}$; rank–nullity forces $\dim\ker u = 1$, so the [kernel](#def-b1-linmaps-kerim) is exactly the prescribed line. Explicitly, decomposing $(x, y, z)
= z(1,1,1) + (x - z)e_1 + (y - z)e_2$:

$$
u(x, y, z) = (x - z,\ y - z,\ 0).
$$

The recipe generalizes: a [linear map](#def-b1-linmaps-def) with prescribed [kernel](#def-b1-linmaps-kerim) $N$ and [image](#def-b1-linmaps-kerim) $I$ exists exactly when $\dim N + \dim I = \dim E$ — necessity is rank–nullity, sufficiency is this construction.

**Corollary 20.9.**

If $\dim E = \dim F$ (finite), then for $u \in \mathcal{L}(E, F)$:

$$
u \text{ injective} \iff u \text{ surjective} \iff u \text{
bijective}.
$$

In particular this holds for endomorphisms in [finite dimension](https://one-course.com/books/math/3/en/chapter/19-finite-dimension#def-b1-findim-def). (It fails in infinite dimension: on $K[X]$, the [derivative](https://one-course.com/books/math/3/en/chapter/14-differentiation#def-b1-derivative-def) is [surjective](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-inj) but not [injective](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-inj), and $P \mapsto XP$ is [injective](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-inj) but not [surjective](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-inj).)

**Proof.** [Injective](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-inj) $\iff \dim\ker u = 0 \iff \operatorname{rk} u = \dim E =
\dim F \iff \operatorname{im} u = F$ (a [subspace](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-subspace) of full dimension is everything, [Theorem 19.14](https://one-course.com/books/math/3/en/chapter/19-finite-dimension#thm-b1-findim-subspaces)) $\iff$ [surjective](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-inj). ∎

**Example 20.10 (Interpolation, structurally).**

Fix distinct $x_0, \dots, x_n$ and let $u \colon \R_n[X] \to
\R^{n+1}$, $P \mapsto (P(x_0), \dots, P(x_n))$: [linear](#def-b1-linmaps-def). Its [kernel](#def-b1-linmaps-kerim) is $\{P : \deg P \leq n,\ n+1 \text{ roots}\} = \{0\}$ ([Corollary 8.8](https://one-course.com/books/math/3/en/chapter/8-polynomials#cor-b1-poly-nroots)). Equal dimensions $n + 1$: $u$ is an isomorphism — existence *and* uniqueness of the Lagrange interpolant ([Theorem 8.23](https://one-course.com/books/math/3/en/chapter/8-polynomials#thm-b1-poly-lagrange)) in one line.

The same one-line pattern handles data mixing values and [derivatives](https://one-course.com/books/math/3/en/chapter/14-differentiation#def-b1-derivative-def): $v \colon \R_3[X] \to \R^4$, $P \mapsto \bigl(P(0),
P'(0), P(1), P'(1)\bigr)$ is [linear](#def-b1-linmaps-def), and its [kernel](#def-b1-linmaps-kerim) consists of degree-$\leq 3$ [polynomials](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def) with double roots at $0$ *and* $1$, i.e. divisible by $X^2(X-1)^2$ of degree $4$: only $P =
0$. Equal dimensions again: every quadruple of data $(P(0),
P'(0), P(1), P'(1))$ is realized by exactly one cubic — Hermite interpolation, granted by a [kernel](#def-b1-linmaps-kerim) computation before any formula is written (the weekend problem of [Chapter 22](https://one-course.com/books/math/3/en/chapter/22-determinants-and-linear-systems#ch-b1-det) meets its determinant).

**Example 20.11 (Rank–nullity at work: the difference operator).**

Let $\Delta \colon \R_n[X] \to \R_n[X]$, $P \mapsto P(X+1) -
P(X)$: [linear](#def-b1-linmaps-def). [Kernel](#def-b1-linmaps-kerim): if $\Delta P = 0$, then $P(0) = P(1) =
P(2) = \dots$, so $P - P(0)$ has infinitely many roots and vanishes ([Corollary 8.8](https://one-course.com/books/math/3/en/chapter/8-polynomials#cor-b1-poly-nroots)): $\ker\Delta$ is the line of constants. Rank–nullity: $\operatorname{rk}\Delta = (n + 1) - 1
= n$. Since $\deg \Delta P < \deg P$ for nonconstant $P$ (the top terms cancel), $\operatorname{im}\Delta \subseteq \R_{n-1}[X]$, which has dimension exactly $n$: the inclusion is an equality. Conclusion, with no computation of [preimages](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-map): *every* [polynomial](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def) $Q$ of degree $\leq n - 1$ is a difference $Q = P(X+1)
- P(X)$ — the discrete antiderivative exists. (Compare the weekend problem of [Chapter 18](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#ch-b1-vspaces), where $\Delta$ was inverted explicitly in the binomial [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free).)

**Example 20.12 (Rank bookkeeping along a composition).**

On $\R_2[X]$, compose the [derivative](https://one-course.com/books/math/3/en/chapter/14-differentiation#def-b1-derivative-def) $D(P) = P'$ ([rank](#def-b1-linmaps-rank) $2$: [image](#def-b1-linmaps-kerim) $\R_1[X]$, [kernel](#def-b1-linmaps-kerim) the constants) with itself. Then $D
\circ D = D^2$ [maps](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-map) $P \mapsto P''$, with [image](#def-b1-linmaps-kerim) $\R_0[X]$: [rank](#def-b1-linmaps-rank) $1$. Compare with the general bounds: the crude one gives $\operatorname{rk} D^2 \leq \min(2, 2) = 2$; the exact formula of [Exercise 20.12](#exo-b1-linmaps-12) accounts for the loss precisely,

$$
\operatorname{rk} D^2 = \operatorname{rk} D -
\dim\bigl(\ker D \cap \operatorname{im} D\bigr) = 2 - 1 = 1 ,
$$

since the constants ([kernel](#def-b1-linmaps-kerim) of the outer $D$) sit inside $\R_1[X]$ ([image](#def-b1-linmaps-kerim) of the inner $D$) with dimension $1$. [Rank](#def-b1-linmaps-rank) is lost exactly where the outer [kernel](#def-b1-linmaps-kerim) ambushes the inner [image](#def-b1-linmaps-kerim) — the sentence to remember when composition [ranks](#def-b1-linmaps-rank) misbehave.

**Example 20.13 (Kernel and image of the Euler operator).**

On $\R_n[X]$, let $u(P) = X\,P'$ ([linear](#def-b1-linmaps-def): differentiation and multiplication by $X$ are). *[Kernel](#def-b1-linmaps-kerim)*: $XP' = 0$ forces $P' =
0$ (a product of [polynomials](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def) vanishes only if one factor does), so $\ker u$ is the line of constants. *[Image](#def-b1-linmaps-kerim)*: on the monomial [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free),

$$
u(X^k) = k\,X^{k} \qquad (k = 0, 1, \dots, n),
$$

so $\operatorname{im} u = \operatorname{Vect}(X, 2X^2, \dots,
nX^n) = \operatorname{Vect}(X, X^2, \dots, X^n)$: the [polynomials](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def) with zero constant term. Check against rank–nullity: $\operatorname{rk} u = (n + 1) - 1 = n$, which is indeed the dimension found. Two remarks worth keeping. First, here $\operatorname{im} u \oplus \ker u = \R_n[X]$ — but that is a *happy accident* of this operator, not a theorem: for the shift-like $v(P) = P'$ on $\R_1[X]$, $\ker v = \operatorname{im}
v = \R_0[X]$ and the sum is not direct. Second, the relation $u(X^k) = kX^k$ says each monomial is merely rescaled by $u$ — a [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) adapted to the [map](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-map), the germ of the eigenvalue idea developed in the Year 2 volume.

## 20.3 Projections and symmetries

**Definition 20.14.**

Let $E = F \oplus G$. The *projection onto $F$ along $G$* [maps](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-map) $x = f + g$ (unique decomposition) to $p(x) = f$; the associated *symmetry* is $s(x)
= f - g$. Both are [linear](#def-b1-linmaps-def), and $s = 2p - \mathrm{id}$.

**Theorem 20.15 (Algebraic characterization).**

1. An endomorphism $p$ is a [projection](#def-b1-linmaps-projection) (onto some $F$ along some $G$ ) if and only if $p \circ p = p$ ; then $F =  \operatorname{im} p = \ker(p - \mathrm{id})$ and $G = \ker  p$ .
2. An endomorphism $s$ is a symmetry if and only if $s \circ s  = \mathrm{id}$ ; then $E = \ker(s - \mathrm{id}) \oplus  \ker(s + \mathrm{id})$ .

**Proof.** (1) A [projection](#def-b1-linmaps-projection) satisfies $p(f + g) = f$ and $p(f) = f$: $p^2 = p$. Conversely, let $p^2 = p$; set $F = \operatorname{im} p$, $G = \ker
p$. Every $x$ writes $x = p(x) + (x - p(x))$ with $p(x) \in F$ and $p\bigl(x - p(x)\bigr) = p(x) - p^2(x) = 0$: $E = F + G$. If $y \in
F \cap G$: $y = p(z)$ and $p(y) = 0$, so $y = p(z) = p^2(z) = p(y) =
0$: [direct sum](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-sum), and $p$ is the [projection](#def-b1-linmaps-projection) onto $F$ along $G$. Finally on $F$: $y = p(z)$ gives $p(y) = y$, so $F \subseteq
\ker(p - \mathrm{id})$, and conversely $p(y) = y$ puts $y$ in the [image](#def-b1-linmaps-kerim).

(2) The correspondence $s = 2p - \mathrm{id}$, $p = \frac{s +
\mathrm{id}}2$ is a bijection between endomorphisms, and under it

$$
s^2 = 4p^2 - 4p + \mathrm{id} = \mathrm{id}
\iff 4p^2 = 4p \iff p^2 = p :
$$

symmetries correspond exactly to [projections](#def-b1-linmaps-projection). Translating the [subspaces](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-subspace): $s(x) = x \iff p(x) = x$, so $\ker(s - \mathrm{id}) =
\operatorname{im} p = F$; and $s(x) = -x \iff 2p(x) = 0 \iff x
\in \ker p = G$, so $\ker(s + \mathrm{id}) = G$. The [direct sum](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-sum) $E = F \oplus G$ of point (1) becomes the announced decomposition into the fixed vectors and the reversed vectors of $s$. ∎

**Example 20.16 (A projection and its symmetry, explicitly).**

In $\R^2$, project onto $F = \operatorname{Vect}(1,1)$ along $G =
\operatorname{Vect}(0,1)$. Decompose $(x, y) = a(1,1) + b(0,1)$: the first coordinate gives $a = x$, the second $b = y - x$. Hence

$$
p(x, y) = (x, x),
\qquad
s(x, y) = 2p(x,y) - (x,y) = (x,\ 2x - y).
$$

Check the algebra: $p(p(x,y)) = p(x,x) = (x,x)$, and $s(s(x,y)) =
s(x, 2x - y) = (x, 2x - (2x - y)) = (x, y)$. Geometrically, $s$ is the “oblique reflection” across the line $y = x$ in the vertical direction: it fixes $F$ pointwise and reverses $G$. Had we projected onto the same $F$ along $G' =
\operatorname{Vect}(1,-1)$ instead, the formula would change to $p'(x,y) = \bigl(\frac{x+y}2, \frac{x+y}2\bigr)$: a [projection](#def-b1-linmaps-projection) is determined by its [image](#def-b1-linmaps-kerim) *and* its [kernel](#def-b1-linmaps-kerim), never by the [image](#def-b1-linmaps-kerim) alone.

![The projection onto F = Vect(1,1) along G = Vect(0,1) and its symmetry, on the point M = (2,\ 0.5): sliding vertically, M hits F at p(M) = (2,2) and lands at s(M) = 2p(M) - M = (2,\ 3.5), as far above F (measured along G) as M was below.](https://one-course.com/images/onecourse/chapters/math-3/b1-linmaps/fig-53ebff90a100.svg)

*The [projection](#def-b1-linmaps-projection) onto $F = \operatorname{Vect}(1,1)$ along $G = \operatorname{Vect}(0,1)$ and its symmetry, on the point $M
= (2,\ 0.5)$: sliding vertically, $M$ hits $F$ at $p(M) = (2,2)$ and lands at $s(M) = 2p(M) - M = (2,\ 3.5)$, as far above $F$ (measured along $G$) as $M$ was below.*

## 20.4 Linear forms and hyperplanes

**Definition 20.17.**

A *linear form* on $E$ is a [linear map](#def-b1-linmaps-def) $\varphi \colon E \to K$. A *hyperplane* of $E$ ($\dim E = n$) is a [subspace](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-subspace) of dimension $n - 1$.

**Example 20.18 (An evaluation form and its hyperplane).**

On $\R_2[X]$, the evaluation $\varphi(P) = P(2)$ is a [linear form](#def-b1-linmaps-forms), nonzero ($\varphi(1) = 1$). Its [kernel](#def-b1-linmaps-kerim) is the [hyperplane](#def-b1-linmaps-forms) of [polynomials](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def) vanishing at $2$, i.e. (factor theorem, [Theorem 8.7](https://one-course.com/books/math/3/en/chapter/8-polynomials#thm-b1-poly-factor)) the multiples of $X - 2$ within $\R_2[X]$:

$$
\ker\varphi = \operatorname{Vect}\bigl(X - 2,\ X(X - 2)\bigr),
\qquad \dim = 2 .
$$

In [coordinates](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#prop-b1-vspaces-coordinates) on $(1, X, X^2)$, $\varphi(a + bX + cX^2) = a +
2b + 4c$: every [linear form](#def-b1-linmaps-forms) on a [finite-dimensional](https://one-course.com/books/math/3/en/chapter/19-finite-dimension#def-b1-findim-def) space is, once a [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) is fixed, a fixed linear expression in the [coordinates](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#prop-b1-vspaces-coordinates) — forms are “row vectors”, as [Chapter 21](https://one-course.com/books/math/3/en/chapter/21-matrices#ch-b1-matrices) will make literal, and the coefficient row here, $(1, 2, 4)$, is a Vandermonde row: evaluation forms are how the interpolation theory of [Chapter 22](https://one-course.com/books/math/3/en/chapter/22-determinants-and-linear-systems#ch-b1-det)’s weekend problem enters linear algebra.

**Theorem 20.19.**

The [hyperplanes](#def-b1-linmaps-forms) of $E$ are exactly the [kernels](#def-b1-linmaps-kerim) of nonzero [linear forms](#def-b1-linmaps-forms). Two nonzero forms have the same [kernel](#def-b1-linmaps-kerim) if and only if they are proportional.

**Proof.** If $\varphi \neq 0$: $\operatorname{rk}\varphi = 1$ ([image](#def-b1-linmaps-kerim) a nonzero [subspace](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-subspace) of $K$), so $\dim\ker\varphi = n - 1$: a [hyperplane](#def-b1-linmaps-forms). Conversely, let $H$ be a [hyperplane](#def-b1-linmaps-forms), $(e_1, \dots, e_{n-1})$ a [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) of $H$ completed by $e_n$: the form “last coordinate” has [kernel](#def-b1-linmaps-kerim) $H$.

Proportional forms share their [kernel](#def-b1-linmaps-kerim). Conversely, suppose $\ker\varphi = \ker\psi = H$ and pick $a \notin H$: every $x$ writes $x = h + \lambda a$ (as $E = H \oplus Ka$), and

$$
\varphi(x) = \lambda \varphi(a), \qquad \psi(x) = \lambda\psi(a):
$$

so $\varphi = \frac{\varphi(a)}{\psi(a)}\,\psi$. ∎

**Example 20.20.**

In $K^n$, a [hyperplane](#def-b1-linmaps-forms) is a solution [set](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets) $\{a_1 x_1 + \dots + a_n
x_n = 0\}$ with not all $a_i$ zero — the familiar equation of a plane through the origin in $\R^3$. In function spaces, evaluation forms $P \mapsto P(1)$ or $f \mapsto \int_0^1 f$ define [hyperplanes](#def-b1-linmaps-forms) of $\R_n[X]$, of $C(\intcc{0}{1})$ (cf. [Exercise 19.6](https://one-course.com/books/math/3/en/chapter/19-finite-dimension#exo-b1-findim-6)).

**Example 20.21 (One hyperplane, worked three ways).**

Take $\varphi(x, y, z) = x - 2y + 3z$ on $\R^3$ and $H =
\ker\varphi$. *[Basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free)*: solve $x = 2y - 3z$:

$$
(2y - 3z,\ y,\ z) = y\,(2, 1, 0) + z\,(-3, 0, 1),
$$

two [free](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) vectors: $\dim H = 2$, a [hyperplane](#def-b1-linmaps-forms), as [Theorem 20.19](#thm-b1-linmaps-hyperplanes) predicts from $\varphi \neq
0$. *[Supplementary](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-sum) line*: any vector outside $H$ [spans](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-span) one, e.g. $a = (1, 0, 0)$ ($\varphi(a) = 1 \neq 0$); the decomposition of an arbitrary $v$ is explicit:

$$
v = \underbrace{\bigl(v - \varphi(v)\,a\bigr)}_{\in\,H}
+ \underbrace{\varphi(v)\,a}_{\in\,\operatorname{Vect}(a)},
$$

since $\varphi\bigl(v - \varphi(v)a\bigr) = \varphi(v) -
\varphi(v)\varphi(a) = 0$. *Proportionality*: if $\psi(x,y,z)
= -2x + 4y - 6z$, then $\psi = -2\varphi$ and both have [kernel](#def-b1-linmaps-kerim) $H$; conversely, any form vanishing on $H$ is a multiple of $\varphi$ ([Exercise 20.8](#exo-b1-linmaps-8)) — the equation of a [hyperplane](#def-b1-linmaps-forms) is unique up to scale, a fact used constantly for planes in geometry.

**Remark 20.22 (Common pitfalls).**

*[Kernel](#def-b1-linmaps-kerim) and [image](#def-b1-linmaps-kerim) live in different spaces*: $\ker u
\subseteq E$, $\operatorname{im} u \subseteq F$; the sum $\ker u
+ \operatorname{im} u$ only makes sense for endomorphisms, and even then it need not be direct ($u(x, y) = (y, 0)$ has $\ker u
= \operatorname{im} u$; [Exercise 20.7](#exo-b1-linmaps-7) characterizes when directness holds). *$u^2 = 0$ does not mean $u = 0$*: the same $u(x,y) =
(y, 0)$ squares to zero without vanishing — what $u^2 = 0$ really says is $\operatorname{im} u \subseteq \ker u$ ([Exercise 20.5](#exo-b1-linmaps-5)). *[Injective](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-inj) $\iff$ [surjective](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-inj) needs equal [finite dimensions](https://one-course.com/books/math/3/en/chapter/19-finite-dimension#def-b1-findim-def)*: on $K[X]$, the [derivative](https://one-course.com/books/math/3/en/chapter/14-differentiation#def-b1-derivative-def) is [surjective](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-inj) and not [injective](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-inj), $P \mapsto XP$ [injective](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-inj) and not [surjective](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-inj) ([Corollary 20.9](#cor-b1-linmaps-samedim)); and between spaces of *different* dimensions, one implication is simply impossible ($\operatorname{rk} u \leq \min(\dim E, \dim F)$). *Prescribing [images](#def-b1-linmaps-kerim) works on a [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free), not on any family*: demanding $u(1, 0) = a$, $u(0, 1) = b$, $u(1, 1) = c$ overdetermines $u$ unless $c = a + b$; a [linear map](#def-b1-linmaps-def) is [free](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) on a [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free), enslaved everywhere else. *[Rank](#def-b1-linmaps-rank) is not preserved by composition*: it can only drop, $\operatorname{rk}(vu) \leq \min(\operatorname{rk} u,
\operatorname{rk} v)$ ([Exercise 20.4](#exo-b1-linmaps-4)), with exact loss measured in [Exercise 20.12](#exo-b1-linmaps-12).

**Remark 20.23 (Where these maps go).**

[Linear maps](#def-b1-linmaps-def) are about to become *matrices*: once bases are fixed, [Chapter 21](https://one-course.com/books/math/3/en/chapter/21-matrices#ch-b1-matrices) encodes every $u \in \mathcal{L}(E,
F)$ by a rectangular array, and composition becomes the matrix product — rank–nullity then drives the theory of [linear](#def-b1-linmaps-def) systems in [Chapter 22](https://one-course.com/books/math/3/en/chapter/22-determinants-and-linear-systems#ch-b1-det). [Projections](#def-b1-linmaps-projection) return in [Chapter 23](https://one-course.com/books/math/3/en/chapter/23-euclidean-spaces#ch-b1-euclid) in their most useful special case, the *orthogonal* [projection](#def-b1-linmaps-projection), where the [kernel](#def-b1-linmaps-kerim) is chosen perpendicular to the [image](#def-b1-linmaps-kerim). The weekend problem below pushes the projector algebra as far as first-year tools reach, up to Fitting’s lemma; the Year 2 volume goes further with the trace and with eigenvalue theory, for which projectors onto stable [subspaces](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-subspace) are the basic building blocks.

**Remark 20.24 (Perspectives inside Book 3: rank–nullity three more times).**

The conservation law $\dim E = \dim\ker u + \operatorname{rk} u$ will be re-read three times before the volume ends. In [Chapter 22](https://one-course.com/books/math/3/en/chapter/22-determinants-and-linear-systems#ch-b1-det) it becomes the shape of solution [sets](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets): a compatible system with $p$ unknowns and [rank](#def-b1-linmaps-rank) $r$ has a solution [set](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets) of dimension $p - r$ — [kernel](#def-b1-linmaps-kerim) dimension in disguise. In [Chapter 23](https://one-course.com/books/math/3/en/chapter/23-euclidean-spaces#ch-b1-euclid) it splits orthogonally, $\dim F + \dim
F^\perp = \dim E$, and powers every distance computation. In the weekend problem of [Chapter 25](https://one-course.com/books/math/3/en/chapter/25-functions-of-two-variables#ch-b1-multivar), it is the accountant of least squares: $n$ observations, $2$ fitted parameters, $n - 2$ residual dimensions, and the Pythagorean identity $\norm b^2 = \norm p^2 + \norm{b - p}^2$ is rank–nullity’s Euclidean shadow. One theorem, four costumes.

## 20.5 Exercises

**Exercise 20.1 ★.**

Which [maps](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-map) are [linear](#def-b1-linmaps-def)?

1. $\R^2 \to \R^2$ , $(x, y) \mapsto (x + y, x - 2y)$ ;
2. $\R^2 \to \R$ , $(x, y) \mapsto xy$ ;
3. $\R[X] \to \R[X]$ , $P \mapsto P' + XP$ ;
4. $\mathcal{F}(\R,\R) \to \R$ , $f \mapsto f(3)$ .

**Solution of Exercise 20.1.**

(1) [Linear](#def-b1-linmaps-def): [coordinates](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#prop-b1-vspaces-coordinates) are linear expressions. (2) Not [linear](#def-b1-linmaps-def): $u(2(1,1)) = 4 \neq 2 = 2u(1,1)$. (3) [Linear](#def-b1-linmaps-def): differentiation and multiplication by $X$ are, and sums of [linear maps](#def-b1-linmaps-def) are. (4) [Linear](#def-b1-linmaps-def): evaluation respects pointwise operations.

**Exercise 20.2 ★.**

Let $u \colon \R^3 \to \R^3$, $(x,y,z) \mapsto (x + y - z,\; 2x + y
+ z,\; 3x + 2y)$. Determine $\ker u$ ([basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free), dimension), $\operatorname{rk} u$, and a [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) of $\operatorname{im} u$. Is $u$ [injective](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-inj)? [surjective](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-inj)?

**Solution of Exercise 20.2.**

[Kernel](#def-b1-linmaps-kerim): solve $x + y - z = 0$, $2x + y + z = 0$, $3x + 2y = 0$. From the third, $y = -\frac{3x}{2}$; the first gives $z = x + y =
-\frac x2$; check in the second: $2x - \frac{3x}{2} - \frac x2 = 0$: satisfied. So $\ker u = \operatorname{Vect}\bigl((2, -3, -1)\bigr)$ (taking $x = 2$), dimension $1$.

Rank–nullity: $\operatorname{rk} u = 3 - 1 = 2$. [Image](#def-b1-linmaps-kerim): spanned by the [images](#def-b1-linmaps-kerim) of the canonical [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free), $u(e_1) = (1,2,3)$, $u(e_2) =
(1,1,2)$, $u(e_3) = (-1,1,0)$; the first two are [free](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free), and the [rank](#def-b1-linmaps-rank) is $2$: [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) $\bigl((1,2,3), (1,1,2)\bigr)$.

Not [injective](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-inj) ($\ker \neq \{0\}$), not [surjective](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-inj) ([rank](#def-b1-linmaps-rank) $2 < 3$): consistent with [Corollary 20.9](#cor-b1-linmaps-samedim).

**Exercise 20.3 ★.**

Let $u \colon \R_n[X] \to \R_n[X]$, $P \mapsto P - P'$. Prove that $u$ is an isomorphism: once via $\ker u$, once by exhibiting the inverse *(consider $P + P' + P'' + \dots$)*.

**Solution of Exercise 20.3.**

*[Kernel](#def-b1-linmaps-kerim):* $P = P'$ forces $\deg P = \deg P'$ unless $P = 0$; but $\deg P' < \deg P$ for $P \neq 0$: so $\ker u = \{0\}$, and $u$, an [injective](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-inj) endomorphism of the [finite-dimensional](https://one-course.com/books/math/3/en/chapter/19-finite-dimension#def-b1-findim-def) $\R_n[X]$, is an isomorphism ([Corollary 20.9](#cor-b1-linmaps-samedim)).

*Inverse:* let $v(P) = P + P' + P'' + \dots + P^{(n)}$ (a finite sum on $\R_n[X]$). Then

$$
v\bigl(u(P)\bigr) = \sum_{k=0}^{n} (P - P')^{(k)}
= \sum_{k=0}^{n} P^{(k)} - \sum_{k=0}^{n} P^{(k+1)}
= P - P^{(n+1)} = P ,
$$

telescoping, since $P^{(n+1)} = 0$. So $v = u^{-1}$.

**Exercise 20.4 ★.**

Let $u \in \mathcal{L}(E, F)$ and $v \in \mathcal{L}(F, G)$, spaces [finite-dimensional](https://one-course.com/books/math/3/en/chapter/19-finite-dimension#def-b1-findim-def). Prove:

$$
\operatorname{rk}(v \circ u) \leq
\min\bigl(\operatorname{rk} u,\ \operatorname{rk} v\bigr).
$$

**Solution of Exercise 20.4.**

$\operatorname{im}(v \circ u) = v(\operatorname{im} u) \subseteq
\operatorname{im} v$: [rank](#def-b1-linmaps-rank) $\leq \operatorname{rk} v$. And $v$ restricted to $\operatorname{im} u$ has [image](#def-b1-linmaps-kerim) $\operatorname{im}(vu)$ with rank–nullity inside $\operatorname{im} u$: $\operatorname{rk}(vu) \leq
\dim\operatorname{im} u = \operatorname{rk} u$.

**Exercise 20.5 ★★.**

Let $u$ be an endomorphism of $E$ ([finite-dimensional](https://one-course.com/books/math/3/en/chapter/19-finite-dimension#def-b1-findim-def)) with $u^2 =
0$. Prove that $\operatorname{im} u \subseteq \ker u$, hence $\operatorname{rk} u \leq \frac{\dim E}{2}$. For $E = \R^2$, give an example with equality.

**Solution of Exercise 20.5.**

$u^2 = 0$ means $u(u(x)) = 0$ for all $x$: every $u(x)$ lies in $\ker u$, i.e. $\operatorname{im} u \subseteq \ker u$. Then rank–nullity:

$$
\dim E = \dim\ker u + \operatorname{rk} u \geq 2\operatorname{rk} u .
$$

Equality example in $\R^2$: $u(x, y) = (y, 0)$: $u^2 = 0$, $\operatorname{rk} u = 1 = \frac{\dim E}{2}$.

**Exercise 20.6 ★★.**

Let $p, q$ be [projections](#def-b1-linmaps-projection) of $E$ with $p \circ q = q \circ p$. Prove that $p \circ q$ is a [projection](#def-b1-linmaps-projection), with

$$
\operatorname{im}(pq) = \operatorname{im} p \cap \operatorname{im}
q ,
\qquad
\ker (pq) = \ker p + \ker q .
$$

**Solution of Exercise 20.6.**

$(pq)^2 = pqpq = ppqq = pq$ (commutation): a [projection](#def-b1-linmaps-projection) ([Theorem 20.15](#thm-b1-linmaps-projchar)).

[Image](#def-b1-linmaps-kerim): $\operatorname{im}(pq) \subseteq \operatorname{im} p$, and $= \operatorname{im}(qp) \subseteq \operatorname{im} q$: contained in the intersection. Conversely, if $x \in \operatorname{im} p \cap
\operatorname{im} q$, then $p(x) = x$ and $q(x) = x$ (fixed points characterize the [image](#def-b1-linmaps-kerim) of a [projection](#def-b1-linmaps-projection)), so $pq(x) = x$: $x \in
\operatorname{im}(pq)$.

[Kernel](#def-b1-linmaps-kerim): $\ker p \subseteq \ker(qp) = \ker(pq)$ and likewise $\ker q
\subseteq \ker(pq)$: the sum is contained. Conversely, let $pq(x) =
0$, and write

$$
x = \underbrace{q(x)}_{\in\, \ker p} +
\underbrace{(x - q(x))}_{\in\, \ker q} :
$$

the first term satisfies $p(q(x)) = 0$, so it lies in $\ker p$; the second is in $\ker q$ since $q(x - q(x)) = q(x) - q^2(x) = 0$. Hence $x \in \ker p + \ker q$.

**Exercise 20.7 ★★.**

Let $u \in \mathcal{L}(E)$, $E$ [finite-dimensional](https://one-course.com/books/math/3/en/chapter/19-finite-dimension#def-b1-findim-def). Prove the equivalence of:

1. $E = \ker u \oplus \operatorname{im} u$ ;
2. $\ker u = \ker u^2$ ;
3. $\operatorname{im} u = \operatorname{im} u^2$ .

**Solution of Exercise 20.7.**

Note first the general inclusions $\ker u \subseteq \ker u^2$ and $\operatorname{im} u^2 \subseteq \operatorname{im} u$, and by rank–nullity, (2) $\iff$ (3) (equal [kernels](#def-b1-linmaps-kerim) $\iff$ equal [ranks](#def-b1-linmaps-rank) $\iff$ equal [images](#def-b1-linmaps-kerim), given the inclusions).

(1 $\Rightarrow$ 2): let $u^2(x) = 0$; then $u(x) \in \ker u \cap
\operatorname{im} u = \{0\}$, so $x \in \ker u$.

(2 $\Rightarrow$ 1): by Grassmann and rank–nullity, $\dim(\ker u +
\operatorname{im} u) = \dim\ker u + \operatorname{rk} u -
\dim(\ker u \cap \operatorname{im} u) = \dim E - \dim(\ker u \cap
\operatorname{im} u)$: the sum is $E$ iff the intersection is $\{0\}$. Let $y \in \ker u \cap \operatorname{im} u$: $y = u(x)$ and $u(y) = 0$, so $u^2(x) = 0$, so (by (2)) $u(x) = 0$: $y = 0$. Hence $E = \ker u \oplus \operatorname{im} u$.

**Exercise 20.8 ★★.**

Let $\varphi, \psi$ be [linear forms](#def-b1-linmaps-forms) on $E$ with $\ker\varphi
\subseteq \ker\psi$. Prove $\psi = \lambda\varphi$ for some $\lambda \in K$ (including the degenerate cases).

**Solution of Exercise 20.8.**

If $\varphi = 0$: then $\ker\varphi = E \subseteq \ker\psi$ forces $\psi = 0 = 0\cdot\varphi$. If $\varphi \neq 0$: $\ker\varphi$ is a [hyperplane](#def-b1-linmaps-forms); pick $a \notin \ker\varphi$. Set $\lambda =
\frac{\psi(a)}{\varphi(a)}$. The form $\psi - \lambda\varphi$ vanishes on $\ker\varphi$ (both do, by the inclusion) and at $a$: it vanishes on $\ker\varphi \oplus Ka = E$. So $\psi =
\lambda\varphi$.

**Exercise 20.9 ★★★.**

Let $u \in \mathcal{L}(E)$ with $\dim E = n$, and suppose $u^n = 0$ but $u^{n-1} \neq 0$ (a *maximally nilpotent* endomorphism). Pick $x$ with $u^{n-1}(x) \neq 0$; prove that $\bigl(x, u(x), \dots,
u^{n-1}(x)\bigr)$ is a [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) of $E$. *(Apply powers of $u$ to a null combination, starting with $u^{n-1}$.)*

**Solution of Exercise 20.9.**

Suppose $\lambda_0 x + \lambda_1 u(x) + \dots + \lambda_{n-1}
u^{n-1}(x) = 0$. Apply $u^{n-1}$: all terms with a factor $u^{\geq
n}$ die, leaving $\lambda_0 u^{n-1}(x) = 0$, so $\lambda_0 = 0$. Apply $u^{n-2}$ to the remaining relation: $\lambda_1 u^{n-1}(x) =
0$, so $\lambda_1 = 0$; and so on. The family is [free](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free); being of size $n = \dim E$, it is a [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) ([Proposition 19.8](https://one-course.com/books/math/3/en/chapter/19-finite-dimension#prop-b1-findim-twoofthree)). (In this [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free), $u$ acts as a shift — the model of maximal nilpotence.)

**Exercise 20.10 ★★★.**

Let $f \in \mathcal{L}(\R^n)$ with $f \circ f = -\mathrm{id}$.

1. Prove that $f$ is an isomorphism and that no $x \neq 0$ satisfies $f(x) = \lambda x$ with $\lambda \in \R$ .
2. Prove that $n$ is even. *Hint: pick $x_1 \neq 0$; show $\operatorname{Vect}(x_1, f(x_1))$ is a plane stable under $f$; choose $x_2$ outside it and iterate, proving that $\bigl(x_1, f(x_1), x_2, f(x_2), \dots\bigr)$ stays [free](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free).*

**Solution of Exercise 20.10.**

1. $f \circ f = -\mathrm{id}$ is [bijective](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-inj) , so $f$ is ( [Proposition 1.26](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#prop-b1-logic-comp) adapted: $f$ has the two-sided inverse $-f$ ). If $f(x) = \lambda x$ with $x \neq 0$ : applying $f$ , $-x = \lambda^2 x$ , so $\lambda^2 = -1$ : impossible in $\R$ .
2. Build the family greedily. Take $x_1 \neq 0$: $(x_1,  f(x_1))$ is [free](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) by (1). If $\operatorname{Vect}$ of the current family $\bigl(x_1, f(x_1), \dots, x_k,  f(x_k)\bigr)$, call it $V_k$ — a [subspace](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-subspace) stable under $f$ (each generator [maps](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-map) to another generator or its negative: $f(f(x_i)) = -x_i$) — is not all of $E$, choose $x_{k+1}  \notin V_k$. *Claim: the enlarged family is [free](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free).* Suppose $\alpha x_{k+1} + \beta f(x_{k+1}) + v = 0$ with $v  \in V_k$ and $(\alpha, \beta) \neq (0,0)$. Apply $f$: $\alpha f(x_{k+1}) - \beta x_{k+1} + f(v) = 0$ with $f(v)  \in V_k$. Eliminate $f(x_{k+1})$ between the two relations (multiply the first by $\alpha$, the second by $-\beta$, and add): $$(\alpha^2 + \beta^2)\, x_{k+1} \in V_k ,$$ and $\alpha^2 + \beta^2 \neq 0$ forces $x_{k+1} \in V_k$: contradiction. So the construction continues, adding vectors *two at a time*, until $V_k = E$: the final family is a [basis](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) of even size, and $n$ is even.

**Exercise 20.11 ★★.**

Let $u, v \in \mathcal{L}(E, F)$, spaces [finite-dimensional](https://one-course.com/books/math/3/en/chapter/19-finite-dimension#def-b1-findim-def). Prove the two-sided bound

$$
\abs{\operatorname{rk} u - \operatorname{rk} v}
\;\leq\; \operatorname{rk}(u + v)
\;\leq\; \operatorname{rk} u + \operatorname{rk} v .
$$

*(For the [upper bound](https://one-course.com/books/math/3/en/chapter/10-real-numbers#def-b1-reals-bounds), compare $\operatorname{im}(u+v)$ with $\operatorname{im} u + \operatorname{im} v$; for the lower one, apply the [upper bound](https://one-course.com/books/math/3/en/chapter/10-real-numbers#def-b1-reals-bounds) cleverly.)*

**Solution of Exercise 20.11.**

[Upper bound](https://one-course.com/books/math/3/en/chapter/10-real-numbers#def-b1-reals-bounds): for every $x$, $(u + v)(x) = u(x) + v(x) \in
\operatorname{im} u + \operatorname{im} v$, so

$$
\operatorname{rk}(u + v)
\leq \dim(\operatorname{im} u + \operatorname{im} v)
\leq \operatorname{rk} u + \operatorname{rk} v
$$

(Grassmann, [Theorem 19.18](https://one-course.com/books/math/3/en/chapter/19-finite-dimension#thm-b1-findim-grassmann)). Lower bound: apply the [upper bound](https://one-course.com/books/math/3/en/chapter/10-real-numbers#def-b1-reals-bounds) to the pair $(u + v, -v)$, whose sum is $u$:

$$
\operatorname{rk} u \leq \operatorname{rk}(u + v) +
\operatorname{rk}(-v) = \operatorname{rk}(u + v) +
\operatorname{rk} v,
$$

so $\operatorname{rk} u - \operatorname{rk} v \leq
\operatorname{rk}(u+v)$; exchanging $u$ and $v$ gives the absolute value.

**Exercise 20.12 ★★★.**

(Frobenius’ inequality) Let $u \in \mathcal{L}(E, F)$, $w \in
\mathcal{L}(F, G)$ and $v \in \mathcal{L}(G, H)$, all spaces [finite-dimensional](https://one-course.com/books/math/3/en/chapter/19-finite-dimension#def-b1-findim-def). Prove the exact formula

$$
\operatorname{rk}(v \circ w) = \operatorname{rk} w -
\dim\bigl(\ker v \cap \operatorname{im} w\bigr),
$$

and deduce Frobenius’ inequality

$$
\operatorname{rk}(v \circ w) + \operatorname{rk}(w \circ u)
\;\leq\; \operatorname{rk} w + \operatorname{rk}(v \circ w \circ
u) .
$$

Check that the case $w = \mathrm{id}_F$ is Sylvester’s inequality, proved in matrix form in [Exercise 21.10](https://one-course.com/books/math/3/en/chapter/21-matrices#exo-b1-matrices-10).

**Solution of Exercise 20.12.**

*Exact formula.* Let $v'$ be the restriction of $v$ to the [subspace](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-subspace) $\operatorname{im} w$. Its [image](#def-b1-linmaps-kerim) is $v(w(F)) =
\operatorname{im}(v \circ w)$, and its [kernel](#def-b1-linmaps-kerim) is $\ker v \cap
\operatorname{im} w$. Rank–nullity for $v'$ on the space $\operatorname{im} w$:

$$
\operatorname{rk} w = \dim\operatorname{im} w
= \operatorname{rk}(v \circ w) + \dim(\ker v \cap
\operatorname{im} w) .
$$

*Frobenius.* Apply the exact formula twice, to $w$ and to $w \circ u$:

$$
\operatorname{rk} w - \operatorname{rk}(vw)
= \dim\bigl(\ker v \cap \operatorname{im} w\bigr),
\qquad
\operatorname{rk}(wu) - \operatorname{rk}(vwu)
= \dim\bigl(\ker v \cap \operatorname{im}(wu)\bigr) .
$$

Since $\operatorname{im}(w \circ u) \subseteq \operatorname{im}
w$, the second intersection is contained in the first, and its dimension is no larger:

$$
\operatorname{rk}(wu) - \operatorname{rk}(vwu)
\;\leq\; \operatorname{rk} w - \operatorname{rk}(vw) ,
$$

which rearranges into Frobenius’ inequality. With $w =
\mathrm{id}_F$ ([rank](#def-b1-linmaps-rank) $\dim F$, and $\operatorname{im}\,
\mathrm{id}_F = F$): $\operatorname{rk} v + \operatorname{rk} u
\leq \dim F + \operatorname{rk}(vu)$, Sylvester’s inequality — proved again, in matrix form, in [Exercise 21.10](https://one-course.com/books/math/3/en/chapter/21-matrices#exo-b1-matrices-10).

## 20.6 Problem: projector calculus and Fitting’s lemma

**Problem 20.1.**

[Projections](#def-b1-linmaps-projection) are the endomorphisms that [direct sums](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-sum) produce, and conversely: every identity $E = F_1 \oplus \dots \oplus F_k$ is secretly a family of projectors summing to the identity. This problem develops that dictionary — the algebra of one projector, of two, of $k$ — then applies the same stabilization ideas to an arbitrary endomorphism and proves *Fitting’s lemma*: every endomorphism of a [finite-dimensional](https://one-course.com/books/math/3/en/chapter/19-finite-dimension#def-b1-findim-def) space splits into a nilpotent part and an invertible part. Throughout, $E$ is a $K$-vector space of dimension $n$, and *projector* means $p \in \mathcal{L}(E)$ with $p^2 = p$ ([Theorem 20.15](#thm-b1-linmaps-projchar)).

**Part I — The algebra around one projector.** Let $p$ be a projector, $p \neq 0$, $p \neq \mathrm{id}$.

1. Show that $\mathrm{id} - p$ is a projector and identify $\operatorname{im}(\mathrm{id} - p)$ and $\ker(\mathrm{id} - p)$ .
2. Compute $(\lambda\,\mathrm{id} + \mu\,p)^2$ and determine all pairs $(\lambda, \mu) \in K^2$ for which $\lambda\,\mathrm{id} + \mu\,p$ is a projector.
3. Show that the plane $\operatorname{Vect}(\mathrm{id}, p)$ of $\mathcal{L}(E)$ is stable under composition, and that for every [polynomial](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def) $Q \in K[X]$, $$Q(p) = Q(0)\,\mathrm{id} + \bigl(Q(1) -  Q(0)\bigr)\,p .$$
4. Determine for which $(\lambda, \mu)$ the [map](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-map) $\lambda\,\mathrm{id} + \mu\,p$ is invertible, and give its inverse in the form $\alpha\,\mathrm{id} + \beta\,p$ . Interpret the answer through the action of $\lambda\,\mathrm{id} + \mu\,p$ on $\operatorname{im} p$ and on $\ker p$ .
5. Let $p'$ be another projector with the *same [image](#def-b1-linmaps-kerim)* $\operatorname{im} p' = \operatorname{im} p$ . Show that $p\,p' = p'$ and $p'\,p = p$ . What do these identities say about composing [projections](#def-b1-linmaps-projection) onto the same [subspace](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-subspace) along different [kernels](#def-b1-linmaps-kerim) ?

**Part II — Two projectors.** Let $p, q$ be projectors of $E$; assume the characteristic is not $2$ (true for $K = \R, \C$).

6. Suppose $p + q$ is a projector. Expanding $(p + q)^2$ , show $pq + qp = 0$ ; composing with $p$ on the left, then on the right, deduce $pq = qp$ , and conclude $pq = qp =  0$ .
7. Conversely, suppose $pq = qp = 0$. Show that $p + q$ is a projector, with $$\operatorname{im}(p + q) = \operatorname{im} p \oplus  \operatorname{im} q,  \qquad  \ker(p + q) = \ker p \cap \ker q .$$
8. Show that $p - q$ is a projector if and only if $pq = qp  = q$ . *(Apply questions 6–7 to $\mathrm{id} - p$ and $q$.)*
9. Show the geometric meaning of $pq = qp = q$ : it holds if and only if $\operatorname{im} q \subseteq  \operatorname{im} p$ and $\ker p \subseteq \ker q$ . (One then writes $q \leq p$ : “ $q$ projects onto less, along more”.)
10. Now let $p, q$ commute. Recall from [Exercise 20.6](#exo-b1-linmaps-6) that $pq$ is the projector onto $\operatorname{im} p \cap \operatorname{im} q$ along $\ker p + \ker q$. Show that $r = p + q - pq$ is a projector with $$\operatorname{im} r = \operatorname{im} p +  \operatorname{im} q,  \qquad  \ker r = \ker p \cap \ker q .$$ *(Consider $\mathrm{id} - r = (\mathrm{id} -  p)(\mathrm{id} - q)$.)*

**Part III — Decompositions of the identity.**

11. Let $E = F_1 \oplus \dots \oplus F_k$ and, for $x = x_1 +  \dots + x_k$ (unique decomposition, $x_i \in F_i$ ), set $p_i(x) = x_i$ . Show that each $p_i$ is a projector, that $p_i p_j = 0$ for $i \neq j$ , and that $p_1 + \dots + p_k  = \mathrm{id}$ ; identify $\operatorname{im} p_i$ and $\ker p_i$ .
12. Conversely, let $p_1, \dots, p_k \in \mathcal{L}(E)$ satisfy $p_1 + \dots + p_k = \mathrm{id}$ and $p_i p_j =  0$ for all $i \neq j$ . Show that each $p_i$ is a projector and that $E = \operatorname{im} p_1 \oplus  \dots \oplus \operatorname{im} p_k$ .
13. Two projectors with $p + q = \mathrm{id}$ : show that $pq  = qp = 0$ holds automatically.
14. Three projectors with $p + q + r = \mathrm{id}$ : show that $p + q$ is a projector, and deduce from question 6 that *all* pairwise products vanish — hence $E =  \operatorname{im} p \oplus \operatorname{im} q \oplus  \operatorname{im} r$ , with no hypothesis on the products.
15. For $k$ projectors with $p_1 + \dots + p_k =  \mathrm{id}$ : show first that for any [subspaces](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-subspace) , $\dim  (F_1 + \dots + F_k) \leq \dim F_1 + \dots + \dim F_k$ , with equality if and only if the sum is direct; then show $E = \operatorname{im} p_1 + \dots +  \operatorname{im} p_k$ , and prove that *if* moreover $\sum_i \operatorname{rk} p_i \leq n$ , the sum is direct and $p_i p_j = 0$ for $i \neq j$ .

**Part IV — Iterated [kernels](#def-b1-linmaps-kerim): Fitting’s lemma.** Let $u \in \mathcal{L}(E)$, $\dim E = n$.

16. Show the two chains, valid for every $k \geq 0$: $$\ker u^k \subseteq \ker u^{k+1},  \qquad  \operatorname{im} u^{k+1} \subseteq \operatorname{im}  u^k .$$
17. Show that if $\ker u^{r} = \ker u^{r+1}$ for some $r$ , then $\ker u^{k} = \ker u^{r}$ for all $k \geq r$ ; state and prove the analogous stabilization for [images](#def-b1-linmaps-kerim) .
18. Deduce that there is a smallest integer $r$ with $\ker  u^{r} = \ker u^{r+1}$ , that $r \leq n$ , and that [images](#def-b1-linmaps-kerim) stabilize at the same $r$ .
19. (Fitting’s lemma) Prove that $$E \;=\; \ker u^{r} \,\oplus\, \operatorname{im} u^{r} .$$
20. Show that both [subspaces](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-subspace) are stable under $u$ , that the restriction of $u$ to $\ker u^{r}$ is nilpotent, and that the restriction of $u$ to $\operatorname{im} u^{r}$ is an isomorphism of $\operatorname{im} u^{r}$ : every endomorphism is, on a canonical [direct sum](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-sum) , “nilpotent plus invertible”.
21. Let $\pi$ be the projector onto $\ker u^{r}$ along $\operatorname{im} u^{r}$ . Show that $\pi \circ u = u  \circ \pi$ .

**Part V — A worked case, and synthesis.**

22. Let $u(x, y, z) = (y, 0, z)$ on $\R^3$ . Compute $u^2$ and $u^3$ , determine the stabilization index $r$ , the [subspaces](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-subspace) $\ker u^{r}$ and $\operatorname{im} u^{r}$ , the Fitting projector $\pi$ , and verify on the formulas that $\pi u = u\pi$ and that $u$ is nilpotent on one factor, [bijective](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-inj) on the other.
23. Show the equivalences: $u$ nilpotent $\iff$ $\ker u^{r} =  E$ $\iff$ $\pi = \mathrm{id}$ ; and deduce that a nilpotent endomorphism of an $n$ -dimensional space always satisfies $u^{n} = 0$ (the nilpotency index never exceeds the dimension).
24. (Uniqueness) Suppose $E = A \oplus B$ with $A, B$ stable under $u$ , the restriction $u|_A$ nilpotent and $u|_B$ [bijective](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-inj) . Prove $A = \ker u^{r}$ and $B =  \operatorname{im} u^{r}$ : the Fitting decomposition is unique.
25. Synthesis, in four sentences: what dictionary Part III establishes between [direct sums](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-sum) and families of projectors; why question 14 needed no product hypotheses while question 15 needed a [rank](#def-b1-linmaps-rank) hypothesis (and which Year 2 tool, the trace, removes it); in what sense Fitting’s lemma is the stabilized version of [Exercise 20.7](#exo-b1-linmaps-7) ; and what the two factors of Fitting become in the eigenvalue theory of the Year 2 volume. Name the theorem proved in Part IV.

**Solution of Problem 20.1.**

**1.** $(\mathrm{id} - p)^2 = \mathrm{id} - 2p + p^2 =
\mathrm{id} - p$: a projector. If $y = x - p(x)$, then $p(y) =
p(x) - p^2(x) = 0$, and conversely $x \in \ker p$ gives $x =
(\mathrm{id} - p)(x)$: $\operatorname{im}(\mathrm{id} - p) = \ker
p$. And $(\mathrm{id} - p)(x) = 0 \iff p(x) = x \iff x \in
\operatorname{im} p$ (fixed points, [Theorem 20.15](#thm-b1-linmaps-projchar)): $\ker(\mathrm{id} - p) =
\operatorname{im} p$.

**2.** $(\lambda\,\mathrm{id} + \mu p)^2 =
\lambda^2\,\mathrm{id} + (2\lambda\mu + \mu^2)\,p$. The pair $(\mathrm{id}, p)$ is [free](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-free) in $\mathcal{L}(E)$: $p = c\,
\mathrm{id}$ would give $c^2 = c$, so $p = 0$ or $\mathrm{id}$, excluded. Identifying coefficients, the [map](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-map) is a projector iff $\lambda^2 = \lambda$ and $2\lambda\mu + \mu^2 = \mu$. For $\lambda = 0$: $\mu \in \{0, 1\}$. For $\lambda = 1$: $\mu^2 +
\mu = 0$, $\mu \in \{0, -1\}$. Exactly four projectors in the plane: $0$, $p$, $\mathrm{id}$, $\mathrm{id} - p$.

**3.** $(\lambda\,\mathrm{id} + \mu p)(\lambda'\,\mathrm{id}
+ \mu' p) = \lambda\lambda'\,\mathrm{id} + (\lambda\mu' +
\mu\lambda' + \mu\mu')\,p$: the plane is stable under composition. Since $p^k = p$ for every $k \geq 1$, for $Q =
\sum_k a_k X^k$:

$$
Q(p) = a_0\,\mathrm{id} + \Bigl(\sum_{k \geq 1} a_k\Bigr) p
= Q(0)\,\mathrm{id} + \bigl(Q(1) - Q(0)\bigr)\,p .
$$

**4.** On $\operatorname{im} p$ (where $p$ acts as the identity), $\lambda\,\mathrm{id} + \mu p$ multiplies by $\lambda
+ \mu$; on $\ker p$, by $\lambda$. As $E = \operatorname{im} p
\oplus \ker p$, the [map](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-map) is [bijective](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-inj) iff $\lambda \neq 0$ and $\lambda + \mu \neq 0$. Solving $\lambda\alpha = 1$, $\lambda\beta + \mu\alpha + \mu\beta = 0$ in the composition rule of question 3:

$$
(\lambda\,\mathrm{id} + \mu p)^{-1}
= \frac1\lambda\,\mathrm{id} -
\frac{\mu}{\lambda(\lambda + \mu)}\,p ,
$$

whose action is by $1/\lambda$ on $\ker p$ and $1/(\lambda +
\mu)$ on $\operatorname{im} p$, as it must be.

**5.** Write $F = \operatorname{im} p = \operatorname{im}
p'$. For every $x$, $p'(x) \in F$ and $p$ fixes $F$ pointwise: $p(p'(x)) = p'(x)$, i.e. $p\,p' = p'$; symmetrically $p'\,p =
p$. When two [projections](#def-b1-linmaps-projection) share their [image](#def-b1-linmaps-kerim), the one applied *first* decides: its output already lies in $F$, where the outer [projection](#def-b1-linmaps-projection) acts as the identity and changes nothing.

**6.** $(p + q)^2 = p^2 + pq + qp + q^2 = (p + q) + pq +
qp$, so $p + q$ projector forces $pq + qp = 0$. Compose on the left with $p$: $pq + pqp = 0$; on the right with $p$: $pqp + qp
= 0$. Subtracting, $pq = qp$; then $pq + qp = 2pq = 0$ and the characteristic is not $2$: $pq = qp = 0$.

**7.** With $pq = qp = 0$, the same expansion gives $(p +
q)^2 = p + q$. [Image](#def-b1-linmaps-kerim): $\operatorname{im}(p + q) \subseteq
\operatorname{im} p + \operatorname{im} q$ always. Conversely, for $x \in \operatorname{im} p$: $q(x) = q(p(x)) = 0$, so $(p +
q)(x) = p(x) = x$ and $x \in \operatorname{im}(p+q)$; same for $\operatorname{im} q$. Directness: $x \in \operatorname{im} p
\cap \operatorname{im} q$ gives $x = p(x) = p(q(x)) = 0$. [Kernel](#def-b1-linmaps-kerim): if $p(x) + q(x) = 0$, applying $p$ gives $p(x) +
p(q(x)) = p(x) = 0$, and applying $q$ gives $q(x) = 0$: $\ker(p
+ q) = \ker p \cap \ker q$ (the reverse inclusion is clear).

**8.** $p - q$ is a projector iff $\mathrm{id} - (p - q) =
(\mathrm{id} - p) + q$ is one (question 1 twice). By questions 6–7 applied to the projectors $\mathrm{id} - p$ and $q$, this holds iff $(\mathrm{id} - p)q = q(\mathrm{id} - p) = 0$, i.e. iff $pq = q$ and $qp = q$.

**9.** $pq = q$ means $p$ fixes every $q(x)$, i.e. $\operatorname{im} q \subseteq \ker(p - \mathrm{id}) =
\operatorname{im} p$. And $qp = q$ means $q\bigl((\mathrm{id} -
p)(x)\bigr) = 0$ for all $x$, i.e. $q$ vanishes on $\operatorname{im}(\mathrm{id} - p) = \ker p$: $\ker p \subseteq
\ker q$. Both steps are equivalences: the order $q \leq p$ says $q$ projects onto a smaller [image](#def-b1-linmaps-kerim), along a larger [kernel](#def-b1-linmaps-kerim).

**10.** Expanding, $(\mathrm{id} - p)(\mathrm{id} - q) =
\mathrm{id} - p - q + pq = \mathrm{id} - r$. The projectors $\mathrm{id} - p$ and $\mathrm{id} - q$ commute, so by [Exercise 20.6](#exo-b1-linmaps-6) their product $\mathrm{id} - r$ is the projector onto $\operatorname{im}(\mathrm{id} - p) \cap
\operatorname{im}(\mathrm{id} - q) = \ker p \cap \ker q$ along $\ker(\mathrm{id} - p) + \ker(\mathrm{id} - q) =
\operatorname{im} p + \operatorname{im} q$. By question 1, $r =
\mathrm{id} - (\mathrm{id} - r)$ is then the projector with $\operatorname{im} r = \operatorname{im} p + \operatorname{im}
q$ and $\ker r = \ker p \cap \ker q$.

**11.** $p_i$ is well defined (uniqueness of the decomposition) and [linear](#def-b1-linmaps-def) (the decomposition of $x + \lambda y$ is the sum of the decompositions, again by uniqueness). For $x_i
\in F_i$ the decomposition is $x_i$ itself, so $p_i(x_i) = x_i$: $p_i^2 = p_i$, and $p_j(x_i) = 0$ for $j \neq i$: $p_i p_j = 0$ ($p_j(x) \in F_j$). Summing the components, $\sum_i p_i =
\mathrm{id}$. Finally $\operatorname{im} p_i = F_i$ and $\ker
p_i = \bigoplus_{j \neq i} F_j$.

**12.** $p_i = p_i \circ \mathrm{id} = p_i\sum_j p_j =
p_i^2 + \sum_{j \neq i} p_i p_j = p_i^2$: each $p_i$ is a projector. Every $x = \mathrm{id}(x) = \sum_i p_i(x)$ lies in $\sum_i \operatorname{im} p_i$: the [images](#def-b1-linmaps-kerim) sum to $E$. Directness: suppose $y_1 + \dots + y_k = 0$ with $y_i \in
\operatorname{im} p_i$, so $p_i(y_i) = y_i$. Apply $p_j$: $p_j
(y_i) = p_j p_i (y_i) = 0$ for $i \neq j$, so $0 = p_j\bigl(\sum
y_i\bigr) = y_j$, for every $j$. Hence $E = \bigoplus_i
\operatorname{im} p_i$.

**13.** $q = \mathrm{id} - p$, and question 1 gives $pq =
p - p^2 = 0 = qp$ directly: for two projectors, summing to the identity already forces orthogonality of the pair.

**14.** $p + q = \mathrm{id} - r$ with $r$ a projector, and $(\mathrm{id} - r)$ is a projector (question 1): so $p + q$ is a projector, and question 6 gives $pq = qp = 0$. By symmetry ($q +
r = \mathrm{id} - p$ and $p + r = \mathrm{id} - q$), all pairwise products vanish, and question 12 concludes: $E =
\operatorname{im} p \oplus \operatorname{im} q \oplus
\operatorname{im} r$, automatically.

**15.** *Lemma.* By induction with Grassmann ([Theorem 19.18](https://one-course.com/books/math/3/en/chapter/19-finite-dimension#thm-b1-findim-grassmann)):

$$
\dim(F_1 + \dots + F_k) \leq \dim(F_1 + \dots + F_{k-1}) + \dim
F_k \leq \dots \leq \sum_i \dim F_i .
$$

If the total is an equality, each step is: $(F_1 + \dots +
F_{j-1}) \cap F_j = \{0\}$ for every $j$, and a relation $y_1 +
\dots + y_k = 0$ ($y_i \in F_i$) collapses from the right: $y_k
\in (F_1 + \dots + F_{k-1}) \cap F_k = \{0\}$, then $y_{k-1} =
0$, etc.: the sum is direct. Conversely a [direct sum](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-sum) has additive dimensions (concatenate bases). *Application:* $x = \sum_i p_i(x)$ shows $E = \sum_i \operatorname{im} p_i$, so $n \leq \sum_i \operatorname{rk} p_i$; the hypothesis gives equality, hence directness. Products: fix $j$ and $y \in
\operatorname{im} p_j$. Then $y = \sum_i p_i(y)$ with $p_i(y)
\in \operatorname{im} p_i$, while $y = y$ is also a decomposition (component $j$ alone); uniqueness forces $p_i(y)
= 0$ for $i \neq j$. Applied to $y = p_j(x)$: $p_i p_j = 0$.

**16.** If $u^k(x) = 0$ then $u^{k+1}(x) = u(0) = 0$. And $\operatorname{im} u^{k+1} = u^k\bigl(u(E)\bigr) \subseteq
u^k(E) = \operatorname{im} u^k$.

**17.** Assume $\ker u^{r} = \ker u^{r+1}$ and let $x \in
\ker u^{r+2}$: then $u(x) \in \ker u^{r+1} = \ker u^{r}$, so $u^{r+1}(x) = 0$: $x \in \ker u^{r+1}$. With question 16, $\ker u^{r+1} = \ker u^{r+2}$, and by induction all later [kernels](#def-b1-linmaps-kerim) coincide with $\ker u^{r}$. For [images](#def-b1-linmaps-kerim): rank–nullity gives $\dim\operatorname{im} u^k = n - \dim\ker u^k$, so the [image](#def-b1-linmaps-kerim) dimensions freeze exactly when the [kernel](#def-b1-linmaps-kerim) dimensions do, and with the inclusions of question 16, equal dimensions mean equal [subspaces](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-subspace) ([Theorem 19.14](https://one-course.com/books/math/3/en/chapter/19-finite-dimension#thm-b1-findim-subspaces)).

**18.** The sequence $\bigl(\dim\ker u^k\bigr)_k$ is nondecreasing with values in $\intint{0}{n}$; it cannot increase strictly $n + 1$ times, so some $\dim\ker u^{r} = \dim\ker
u^{r+1}$ with $r \leq n$, hence $\ker u^{r} = \ker u^{r+1}$ (inclusion plus equal dimension). Take $r$ smallest; question 17 freezes everything from $r$ on, [images](#def-b1-linmaps-kerim) included.

**19.** Intersection: let $x \in \ker u^{r} \cap
\operatorname{im} u^{r}$, say $x = u^{r}(y)$ with $u^{r}(x) =
0$. Then $u^{2r}(y) = 0$, and $\ker u^{2r} = \ker u^{r}$ (question 17), so $x = u^{r}(y) = 0$. Dimensions: rank–nullity for $u^{r}$ gives $\dim\ker u^{r} + \dim\operatorname{im} u^{r}
= n$; with trivial intersection, Grassmann makes the sum a [subspace](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-subspace) of dimension $n$: $E = \ker u^{r} \oplus
\operatorname{im} u^{r}$.

**20.** Stability: $u^{r}(u(x)) = u(u^{r}(x)) = 0$ for $x
\in \ker u^{r}$; and $u(u^{r}(y)) = u^{r}(u(y)) \in
\operatorname{im} u^{r}$. On $N = \ker u^{r}$: $(u|_N)^{r} = 0$ by definition of $N$: nilpotent. On $I = \operatorname{im}
u^{r}$: $\ker(u|_I) = \ker u \cap I \subseteq \ker u^{r} \cap I
= \{0\}$, so $u|_I$ is an [injective](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-inj) endomorphism of the [finite-dimensional](https://one-course.com/books/math/3/en/chapter/19-finite-dimension#def-b1-findim-def) $I$, hence [bijective](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-inj) ([Corollary 20.9](#cor-b1-linmaps-samedim)).

**21.** Let $x = a + b$ with $a \in N$, $b \in I$. Then $u(x) = u(a) + u(b)$ with $u(a) \in N$ and $u(b) \in I$ (question 20): this *is* the decomposition of $u(x)$, so $\pi(u(x)) = u(a) = u(\pi(x))$: $\pi u = u\pi$.

**22.** $u^2(x,y,z) = u(y, 0, z) = (0, 0, z)$ and $u^3(x,y,z) = u(0,0,z) = (0,0,z) = u^2(x,y,z)$. [Kernels](#def-b1-linmaps-kerim): $\ker u
= \{y = z = 0\} = \operatorname{Vect}(e_1)$, $\ker u^2 = \{z =
0\} = \operatorname{Vect}(e_1, e_2)$, $\ker u^3 = \ker u^2$: stabilization at $r = 2$. [Images](#def-b1-linmaps-kerim): $\operatorname{im} u =
\operatorname{Vect}(e_1, e_3)$, $\operatorname{im} u^2 =
\operatorname{Vect}(e_3)$. Fitting: $\R^3 =
\operatorname{Vect}(e_1, e_2) \oplus \operatorname{Vect}(e_3)$, and $\pi(x, y, z) = (x, y, 0)$. Check: $\pi u(x,y,z) = \pi(y, 0,
z) = (y, 0, 0)$ and $u\pi(x,y,z) = u(x, y, 0) = (y, 0, 0)$: equal. On the first factor $u(x, y, 0) = (y, 0, 0)$, whose square is $0$: nilpotent; on the second $u(0,0,z) = (0,0,z)$: the identity, [bijective](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-inj).

**23.** If $u^m = 0$ then $\ker u^m = E$; since [kernels](#def-b1-linmaps-kerim) are frozen from $r$ on, $\ker u^{r} = \ker u^{\max(m, r)} = E$. Conversely $\ker u^{r} = E$ means $u^{r} = 0$. And $\ker u^{r} =
E \iff$ the Fitting projector is onto $E$ along $\{0\}$, i.e. $\pi = \mathrm{id}$. Finally $r \leq n$ (question 18) gives: every nilpotent endomorphism satisfies $u^{n} = 0$ — the nilpotency index never exceeds the dimension.

**24.** Let $m$ be a nilpotency index of $u|_A$: $A
\subseteq \ker u^{m} \subseteq \ker u^{\max(m,r)} = \ker u^{r}$. Since $u|_B$ is [bijective](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-inj), $B = u(B) = u^{k}(B) \subseteq
\operatorname{im} u^{k}$ for every $k$, in particular $B
\subseteq \operatorname{im} u^{r}$. Then

$$
n = \dim A + \dim B \leq \dim\ker u^{r} +
\dim\operatorname{im} u^{r} = n :
$$

both inclusions are equalities of dimensions, hence of [subspaces](https://one-course.com/books/math/3/en/chapter/18-vector-spaces#def-b1-vspaces-subspace): $A = \ker u^{r}$, $B = \operatorname{im} u^{r}$.

**25.** (i) Part III is a dictionary: splittings $E = F_1
\oplus \dots \oplus F_k$ correspond exactly to families of projectors with $\sum p_i = \mathrm{id}$ and $p_i p_j = 0$, the $F_i$ being the [images](#def-b1-linmaps-kerim). (ii) For $k = 3$ the complements $\mathrm{id} - p_i$ are themselves projectors, which closed the argument with no extra hypothesis; for general $k$ one needs $\sum_i \operatorname{rk} p_i \leq n$, an inequality the Year 2 trace gives for free ($\operatorname{tr} p = \operatorname{rk}
p$ for a projector, and traces add up to $\operatorname{tr}
\mathrm{id} = n$). (iii) [Exercise 20.7](#exo-b1-linmaps-7) is Fitting’s lemma in the already-stabilized case $r \leq 1$; in general one lets the [kernel](#def-b1-linmaps-kerim) and [image](#def-b1-linmaps-kerim) chains freeze, which takes at most $n$ steps. (iv) In the Year 2 volume, applied to $u - \lambda\,
\mathrm{id}$, the nilpotent factor becomes the generalized eigenspace at $\lambda$ and the projectors of Part III become the spectral projectors of the reduction theory. The theorem of Part IV is *Fitting’s lemma*.