University Mathematics — Year 2 · Bachelor Year 2

2Linear Algebra

The linear algebra of the Year 1 volume worked over $\R$ or $\C$ in finite dimension, and admitted the general determinant. This chapter upgrades all three restrictions: the theory is stated over an arbitrary field $K$ , the interplay between a space and its dual is developed systematically (dual bases, annihilators, transposes), and the determinant is finally constructed from alternating multilinear forms and the signature of Chapter 1 — discharging every admission of Year 1.

Throughout, $K$ is a field ( $\Q$ , $\R$ , $\C$ , or $\Z/p\Z$ — the theory does not care) and, unless stated, spaces are finite-dimensional over $K$ . The Year 1 results (bases, dimension, rank–nullity, matrices) transfer verbatim: their proofs never used anything but the field axioms.

2.1 Dual space

Definition 2.1 (Dual space, dual basis)

The dual of $E$ is $E^* = \mathcal{L}(E, K)$ , the space of linear forms. If $\mathcal{B} = (e_1, \dots, e_n)$ is a basis of $E$ , the coordinate forms $e_1^*, \dots, e_n^*$ defined by $e_i^*(e_j) = \delta_{ij}$ (Kronecker: $1$ if $i = j$ , else $0$ ) form the dual basis $\mathcal{B}^*$ of $E^*$ ; in particular $\dim E^* = \dim E$ , and

x = \sum_{i=1}^{n} e_i^*(x)\, e_i \quad (x \in E), \qquad \varphi = \sum_{i=1}^{n} \varphi(e_i)\, e_i^* \quad (\varphi \in E^*).

Proof that $\mathcal{B}^*$ is a basis. Free: applying a null combination $\sum \lambda_i e_i^* = 0$ to $e_j$ gives $\lambda_j = 0$ . Generating: for $\varphi \in E^*$ , the form $\varphi - \sum_i \varphi(e_i) e_i^*$ kills every $e_j$ , hence is zero (a linear map vanishing on a basis vanishes). The two display formulas are the same computations read forwards. ∎

Example 2.2

On $K_n[X]$ with basis $(1, X, \dots, X^n)$ : the dual basis is $P \mapsto \frac{P^{(k)}(0)}{k!}$ (Taylor coefficients). Another basis of the dual: the evaluations $P \mapsto P(x_i)$ at $n + 1$ distinct points — its “pre-dual” basis in $K_n[X]$ is exactly the family of Lagrange polynomials $L_i$ (Year 1 volume), since $L_i(x_j) = \delta_{ij}$ . Interpolation is duality.

Method 2.3 (Dual and antedual bases in practice)

To expand a form $\varphi$ on a basis $(e_i)$ of $E$ : the coordinates are the values $\varphi(e_i)$ — no system to solve. To find the basis $(u_j)$ of $E$ whose dual is a given basis $(\varphi_1, \dots, \varphi_n)$ of $E^*$ (the antedual): solve the $n$ linear systems

\varphi_i(u_j) = \delta_{ij} \qquad (1 \leq i \leq n),

one column $u_j$ at a time; in matrix terms, if the rows of $M$ list the coefficients of the $\varphi_i$ in a known basis of $E^*$ , the columns of $M^{-1}$ are the $u_j$ . Existence and uniqueness of the antedual are proved in this chapter’s weekend problem; the computation is always this inversion.

Example 2.4 (A dual basis of $\R^2$ , fully computed)

For the basis $b_1 = (1, 1)$ , $b_2 = (1, -1)$ of $\R^2$ : the dual basis $(b_1^*, b_2^*)$ must satisfy $b_i^*(b_j) = \delta_{ij}$ . Writing $b_1^*(x, y) = \alpha x + \beta y$ , the conditions $\alpha + \beta = 1$ and $\alpha - \beta = 0$ give

b_1^*(x, y) = \frac{x + y}{2}, \qquad\text{and likewise}\qquad b_2^*(x, y) = \frac{x - y}{2} .

Sanity checks: $b_1^*$ is not $e_1^* + e_2^*$ evaluated naively — the dual basis depends on the whole basis, not on each vector separately (replacing $b_2$ by $(0, 1)$ changes $b_1^*$ into $x \mapsto x$ ). And the expansion formula works: $(x, y) = \frac{x+y}2\,b_1 + \frac{x-y}2\,b_2$ , the even/odd decomposition of a pair — dual bases are coordinate extractors, and this one extracts symmetric and antisymmetric parts.

Definition 2.5 (Annihilator)

For a subspace $F \subseteq E$ , the annihilator is

F^{\circ} = \{\varphi \in E^* : \varphi|_F = 0\},

a subspace of $E^*$ .

Theorem 2.6 (Dimension of the annihilator)

$\dim F^{\circ} = \dim E - \dim F$ . Moreover $F \mapsto F^\circ$ reverses inclusions, and $F$ is recovered from its annihilator:

F = \{x \in E : \forall\varphi \in F^\circ,\ \varphi(x) = 0\}.

Consequently every subspace of dimension $p$ in dimension $n$ is the solution set of $n - p$ independent linear equations — and conversely.

Proof. Choose a basis $(e_1, \dots, e_p)$ of $F$ completed into a basis of $E$ . A form $\varphi = \sum \varphi(e_i) e_i^*$ annihilates $F$ iff its first $p$ coefficients vanish: $F^\circ = \operatorname{Vect}(e_{p+1}^*, \dots, e_n^*)$ , of dimension $n - p$ . Inclusion reversal is immediate. For the recovery: the right-hand side contains $F$ ; conversely, if $x \notin F$ , complete a basis of $F$ by $x$ and further vectors; the coordinate form of $x$ in this basis annihilates $F$ but not $x$ . The “equations” reading takes a basis $(\varphi_1, \dots, \varphi_{n-p})$ of $F^\circ$ : then $F = \bigcap \ker\varphi_j$ , an intersection of $n - p$ independent hyperplanes. ∎

Example 2.7 (An annihilator, both directions)

Let $F = \operatorname{Vect}\bigl((1, 2, 1),\ (1, 0, -1)\bigr) \subseteq \R^3$ . A form $\varphi = a\,e_1^* + b\,e_2^* + c\,e_3^*$ annihilates $F$ iff

a + 2b + c = 0 \qquad\text{and}\qquad a - c = 0 ,

i.e. $c = a$ and $b = -a$ : $F^\circ = \R\,(e_1^* - e_2^* + e_3^*)$ , of dimension $3 - 2 = 1$ as Theorem 2.6 requires. Reading it backwards: $F = \{(x, y, z) : x - y + z = 0\}$ — the plane recovered as the kernel of the single form spanning $F^\circ$ . Going from a spanning family to equations is computing an annihilator; going from equations to a parametrization is computing a pre-annihilator. (Check: both spanning vectors satisfy $x - y + z = 0$ .)

Definition 2.8 (Transpose map)

For $u \in \mathcal{L}(E, F)$ , the transpose $u^{\mathsf T} \in \mathcal{L}(F^*, E^*)$ is

u^{\mathsf T}(\psi) = \psi \circ u .

It satisfies $(v \circ u)^{\mathsf T} = u^{\mathsf T} \circ v^{\mathsf T}$ , and in dual bases, the matrix of $u^{\mathsf T}$ is the transposed matrix of $u$ — which finally explains the transpose of Year 1.

Example 2.9 (The transpose, entry by entry)

Let $u \colon \R^2 \to \R^3$ have matrix $A = \left(\begin{smallmatrix} 1 & 2\\ 0 & 1\\ 3 & 0\end{smallmatrix}\right)$ in the canonical bases. For $\psi = b_1f_1^* + b_2f_2^* + b_3f_3^* \in (\R^3)^*$ , compute $u^{\mathsf T}(\psi) = \psi \circ u$ on the basis of $\R^2$ :

(\psi \circ u)(e_1) = \psi(1, 0, 3) = b_1 + 3b_3, \qquad (\psi \circ u)(e_2) = \psi(2, 1, 0) = 2b_1 + b_2 .

So $u^{\mathsf T}(\psi) = (b_1 + 3b_3)\,e_1^* + (2b_1 + b_2)\,e_2^*$ , and in the dual bases the matrix of $u^{\mathsf T}$ is

\begin{pmatrix} 1 & 0 & 3\\ 2 & 1 & 0\end{pmatrix} = A^{\mathsf T} :

the abstract transpose is the flipped matrix, with no computation left to believe on faith. Note the mechanism: the $j$ -th column of $A$ became the $j$ -th row of the new matrix because $\psi \circ u$ reads $u$ ’s outputs through $\psi$ ’s coefficients.

Proposition 2.10

$\ker u^{\mathsf T} = (\operatorname{im} u)^{\circ}$ and $\operatorname{im} u^{\mathsf T} = (\ker u)^{\circ}$ . Consequently $\operatorname{rk}(u^{\mathsf T}) = \operatorname{rk}(u)$ : row rank equals column rank, proved structurally.

Proof. $\psi \in \ker u^{\mathsf T} \iff \psi \circ u = 0 \iff \psi$ kills $\operatorname{im} u$ : the first identity. For the second: $u^{\mathsf T}(\psi) = \psi \circ u$ kills $\ker u$ always, so $\operatorname{im} u^{\mathsf T} \subseteq (\ker u)^\circ$ ; dimensions match by rank–nullity and Theorem 2.6:

\operatorname{rk} u^{\mathsf T} = \dim F^* - \dim\ker u^{\mathsf T} = \dim F - \bigl(\dim F - \operatorname{rk} u\bigr) = \operatorname{rk} u = \dim (\ker u)^{\circ} . \qedhere

∎

Example 2.11 (Rank read on both sides)

Let

A = \begin{pmatrix} 1 & 2 & 0 & 1\\ 0 & 1 & 1 & 1\\ 1 & 3 & 1 & 2 \end{pmatrix} .

Column rank: the third row is the sum of the first two, so $\operatorname{rk} A \leq 2$ ; columns $1$ and $2$ are free: $\operatorname{rk} A = 2$ . The transpose’s kernel: solving $A^{\mathsf T}y = 0$ gives $y \in \R\,(1, 1, -1)$ , so $\ker A^{\mathsf T}$ has dimension $1 = 3 - 2$ : exactly $(\operatorname{im} A)^\circ$ under the identification of $(\R^3)^*$ with row vectors, as Proposition 2.10 asserts — the single relation “row $_3$ = row $_1$ + row $_2$ ” is the annihilator of the column space. Row rank ( $2$ free rows) and column rank agree not by accident but because both equal $\operatorname{rk} A = \operatorname{rk} A^{\mathsf T}$ .

Example 2.12 (Duality reads a quadrature rule)

Why does a rule like Simpson’s (Exercise 2.4) exist and why is it unique? Duality answers before any computation. On $E = \R_2[X]$ , the integral $P \mapsto \int_0^1 P$ is one specific vector of the three-dimensional dual $E^*$ ; the evaluations at $0$ , $\frac12$ , $1$ form a basis of $E^*$ ; hence the integral expands uniquely on them — that expansion is Simpson’s rule, coefficients included. A dimension count also calibrates expectations: on $\R_3[X]$ , four dimensions of forms cannot in general be spanned by three evaluations, so exactness on cubics is not owed by duality; that Simpson integrates cubics exactly anyway is a bonus symmetry (odd-degree cancellation around $\frac12$ ), to be checked by hand. Rules with $n + 1$ nodes are expansions of the integration form in an evaluation basis of $\R_n[X]^*$ : existence and uniqueness cost one dual-basis theorem; only the bonus degrees cost work.

2.2 Multilinear alternating forms

Definition 2.13

A map $f \colon E^n \to K$ is $n$ -linear when it is linear in each variable, and alternating when it vanishes whenever two arguments are equal. Alternating implies antisymmetric: swapping two arguments changes the sign (expand $f(\dots, x + y, \dots, x + y, \dots) = 0$ ); more generally, for $\sigma \in \mathfrak{S}_n$ ,

f(x_{\sigma(1)}, \dots, x_{\sigma(n)}) = \varepsilon(\sigma)\, f(x_1, \dots, x_n),

by decomposing $\sigma$ into transpositions (Theorem 1.21).

Theorem 2.14 (The fundamental theorem of determinants)

Let $\dim E = n$ and $\mathcal{B} = (e_1, \dots, e_n)$ a basis. The space of alternating $n$ -linear forms on $E$ has dimension $1$ : every such form is a multiple of

\det{}_{\mathcal{B}}(x_1, \dots, x_n) = \sum_{\sigma \in \mathfrak{S}_n} \varepsilon(\sigma) \prod_{i=1}^{n} a_{\sigma(i),\,i}, \qquad x_j = \sum_{i} a_{ij} e_i ,

and $\det_{\mathcal{B}}$ is the unique one taking the value $1$ on $\mathcal{B}$ .

Proof. Let $f$ be alternating $n$ -linear. Expanding each argument on $\mathcal{B}$ by multilinearity,

f(x_1, \dots, x_n) = \sum_{i_1, \dots, i_n} a_{i_1,1}\cdots a_{i_n,n}\, f(e_{i_1}, \dots, e_{i_n}).

Terms with a repeated index vanish (alternating); the surviving tuples $(i_1, \dots, i_n)$ are the injective ones, i.e. $i_k = \sigma(k)$ for a permutation $\sigma$ , and antisymmetry reorders $f(e_{\sigma(1)}, \dots, e_{\sigma(n)}) = \varepsilon(\sigma) f(e_1, \dots, e_n)$ . Hence

f = f(e_1, \dots, e_n) \cdot \det{}_{\mathcal{B}} :

every alternating form is that multiple, provided $\det_{\mathcal{B}}$ itself (the displayed sum) is alternating $n$ -linear and takes value $1$ on $\mathcal B$ . Multilinearity is clear (each summand is linear in each column). Value on $\mathcal B$ : the only nonzero term is $\sigma = \mathrm{id}$ . Alternating: suppose $x_j = x_k$ ( $j \neq k$ ), so that the coordinate columns satisfy $a_{i j} = a_{i k}$ for all $i$ . Pair each $\sigma$ with $\sigma' = \sigma\circ(j\,k)$ — an involution without fixed points on $\mathfrak{S}_n$ . The paired products coincide:

\prod_i a_{\sigma'(i),\,i} = a_{\sigma(k),\,j}\; a_{\sigma(j),\,k} \prod_{i \neq j,k} a_{\sigma(i),\,i} = a_{\sigma(k),\,k}\; a_{\sigma(j),\,j} \prod_{i \neq j,k} a_{\sigma(i),\,i} = \prod_i a_{\sigma(i),\,i},

using the equality of the columns $j$ and $k$ ; while $\varepsilon(\sigma') = -\varepsilon(\sigma)$ . Each pair contributes zero: the sum vanishes. ∎

Example 2.15 (Sarrus, derived and demolished)

For $n = 3$ the permutation formula has exactly $3! = 6$ terms. Listing $\mathfrak{S}_3$ by signature — $\mathrm{id}$ , $(1\,2\,3)$ , $(1\,3\,2)$ even; $(1\,2)$ , $(1\,3)$ , $(2\,3)$ odd — gives

\det A = a_{11}a_{22}a_{33} + a_{21}a_{32}a_{13} + a_{31}a_{12}a_{23} - a_{21}a_{12}a_{33} - a_{31}a_{22}a_{13} - a_{11}a_{32}a_{23} :

precisely the “diagonals” rule of Sarrus taught in school — now a theorem, with the mysterious signs identified as signatures. The demolition: for $n = 4$ there are $24$ permutations, of which only $8$ are picked up by any diagonal-drawing scheme; Sarrus has no degree- $4$ version, and cofactor expansion (Theorem 2.17 (4)) takes over. Counting terms is also a warning: the permutation formula has $n!$ summands, so it is a definition, not an algorithm — row reduction computes $\det$ in $O(n^3)$ operations instead.

Definition 2.16 (Determinants)

The determinant of a family in a basis is $\det_{\mathcal{B}}(x_1, \dots, x_n)$ ; the determinant of a matrix $A$ is the determinant of its columns in the canonical basis — the permutation formula above; the determinant of an endomorphism $u$ is the scalar $\det u$ such that

\det{}_{\mathcal{B}}\bigl(u(x_1), \dots, u(x_n)\bigr) = \det u \cdot \det{}_{\mathcal{B}}(x_1, \dots, x_n) \quad \text{for all } x_i

(the left side is alternating $n$ -linear, hence a multiple of $\det_\mathcal{B}$ by Theorem 2.14; the factor does not depend on $\mathcal{B}$ ).

Theorem 2.17 (The determinant calculus, proved)

$\det(uv) = \det u\,\det v$ ; $\;\det(AB) = \det A \det B$ .
$u$ is invertible $\iff \det u \neq 0$ ; a family is a basis $\iff$ its determinant in some basis is nonzero.
$\det(A^{\mathsf T}) = \det A$ .
Cofactor expansion along any row or column, as stated in the Year 1 volume, holds; similar matrices share their determinant.

Proof. (1) Apply the defining relation twice: $\det_{\mathcal B}(uv(x_i)) = \det u \cdot \det_{\mathcal B}(v(x_i)) = \det u \det v \cdot \det_{\mathcal B}(x_i)$ .

(2) If $u$ is invertible, $\det u \det u^{-1} = \det \mathrm{id} = 1 \neq 0$ . If not, the images $u(e_i)$ are linked; expressing one through the others and expanding, $\det_{\mathcal B}(u(e_i)) = 0$ (alternating kills repeated directions), so $\det u = 0$ . The basis criterion is the same statement for families.

(3) In the permutation formula, reindex each product by $j = \sigma(i)$ , i.e. $i = \tau(j)$ with $\tau = \sigma^{-1}$ : the factors are the same numbers in a different order, so

\prod_{i=1}^{n} a_{\sigma(i),\,i} = \prod_{j=1}^{n} a_{j,\,\tau(j)} ,

and $\varepsilon(\tau) = \varepsilon(\sigma)^{-1} = \varepsilon(\sigma)$ (values are $\pm1$ ; $\varepsilon$ is a morphism). Summing over $\sigma$ is the same as summing over $\tau$ (inversion is a bijection of $\mathfrak{S}_n$ ):

\det A = \sum_{\tau}\varepsilon(\tau)\prod_j a_{j,\tau(j)} = \det(A^{\mathsf T}),

the last sum being the permutation formula applied to the transposed entries $(A^{\mathsf T})_{ij} = a_{ji}$ .

(4) Fix column $j$ and split $x_j = \sum_i a_{ij} e_i$ by linearity: $\det A = \sum_i a_{ij}\, \det(\dots, e_i, \dots)$ , and moving $e_i$ to the last position ( $n - i$ transpositions of rows, $n - j$ of columns, via (3)) identifies $\det(\dots, e_i, \dots) = (-1)^{i+j}\Delta_{ij}$ with the minor: exactly Year 1’s cofactor rule. Similarity: $\det(P^{-1}AP) = \det P^{-1}\det A \det P = \det A$ by (1). ∎

Example 2.18 (Cofactor expansion, executed)

Compute

\det\begin{pmatrix} 2 & 1 & 3\\ 0 & 4 & 1\\ 1 & 2 & 0 \end{pmatrix}

along the first column (two zeros’ worth of laziness: one). Signs follow the checkerboard $(-1)^{i+j}$ :

2\,\det\begin{pmatrix}4 & 1\\ 2 & 0\end{pmatrix} - 0 + 1\cdot\det\begin{pmatrix}1 & 3\\ 4 & 1\end{pmatrix} = 2(0 - 2) + (1 - 12) = -15 .

Cross-check by Sarrus (Example 2.15): $0 + 1 + 0 - 12 - 0 - 4 = -15$ . Strategy, not doctrine: expand along the line with the most zeros, and when none has any, make some first by row operations — one round of elimination costs less than two cofactor layers.

Example 2.19 (A determinant by the rules)

Let $J \in \mathcal{M}_n(K)$ be the all-ones matrix and $a \in K$ ; we compute $\det(aI_n + J)$ with the tools just proved. Every column of $aI_n + J$ sums the same way: add all rows to the first (the determinant is unchanged — adding a multiple of one row to another adds a repeated-direction term, killed by alternation). The first row becomes $(a + n, a + n, \dots, a + n)$ ; factor out $a + n$ by linearity in that row, then subtract the first column from every other column: what remains is triangular with diagonal $(1, a, \dots, a)$ . Hence

\det(aI_n + J) = (a + n)\,a^{\,n-1}.

The closing insight: the roots $a = 0$ (multiplicity $n - 1$ ) and $a = -n$ say that $J$ has eigenvalue $0$ with multiplicity $n - 1$ and eigenvalue $n$ once — the spectrum of the rank-one matrix $J$ , one chapter early (Chapter 3 will make this systematic).

Example 2.20 (A determinant by the permutation formula)

For a matrix with many zeros the formula is practical by itself: in

A = \begin{pmatrix} 0 & a & 0 & 0\\ 0 & 0 & b & 0\\ 0 & 0 & 0 & c\\ d & 0 & 0 & 0 \end{pmatrix},

the only permutation picking nonzero entries is the $4$ -cycle $\sigma = (1\,2\,3\,4)$ mapping column $1 \to$ row $4$ , etc.; $\varepsilon(\sigma) = (-1)^3 = -1$ , so $\det A = -abcd$ . (Check via three column swaps to reach a diagonal matrix.)

Example 2.21 (A Vandermonde by the product formula)

For the nodes $0, 1, 2$ (used by quadrature rules like Exercise 2.4’s), the Vandermonde determinant of Exercise 2.11 evaluates in one glance:

\det\begin{pmatrix} 1 & 1 & 1\\ 0 & 1 & 2\\ 0 & 1 & 4 \end{pmatrix} = (1 - 0)(2 - 0)(2 - 1) = 2 ,

and by direct expansion along the first column: $1\cdot(4 - 2) = 2$ : agreement. Nonvanishing for distinct nodes is the whole theory of interpolation in one determinant: the evaluation forms $P \mapsto P(a_i)$ are a basis of the dual exactly when this determinant is nonzero, i.e. always for distinct $a_i$ — Example 2.2 quantified.

2.3 Trace, revisited

Proposition 2.22

The trace $\operatorname{tr} \colon \mathcal{M}_n(K) \to K$ is the unique linear form with $\operatorname{tr}(AB) = \operatorname{tr}(BA)$ and $\operatorname{tr}(I_n) = n$ (for $\operatorname{char} K = 0$ ); the trace of an endomorphism is well defined via any matrix representation, and

\operatorname{tr}(u) = \sum_{i} e_i^*\bigl(u(e_i)\bigr)

in any basis — duality writes the trace basis-freely.

Proof. $\operatorname{tr}(AB) = \operatorname{tr}(BA)$ and basis-invariance were proved in Year 1. Uniqueness: a linear form $t$ with $t(AB) = t(BA)$ kills every commutator $AB - BA$ . We claim the commutators span the trace-zero hyperplane, of dimension $n^2 - 1$ . Two families of commutators suffice. The multiplication rule of the elementary matrices is $E_{ab}E_{cd} = \delta_{bc}E_{ad}$ . For $i \neq j$ it gives

E_{ii}E_{ij} - E_{ij}E_{ii} = E_{ij} - 0 = E_{ij}

(the second product is $E_{ij}E_{ii} = \delta_{ji}E_{ii} = 0$ since $j \neq i$ ): every off-diagonal $E_{ij}$ is a commutator. And

E_{ij}E_{ji} - E_{ji}E_{ij} = E_{ii} - E_{jj} .

The $E_{ij}$ ( $i \neq j$ , $n^2 - n$ of them) together with the $E_{11} - E_{jj}$ ( $j \geq 2$ , $n - 1$ of them) are $n^2 - 1$ linearly independent trace-zero matrices: they span the hyperplane $\ker\operatorname{tr}$ . So $t$ vanishes where $\operatorname{tr}$ does and factors through it: $t = c\operatorname{tr}$ ; then $t(I) = n$ forces $c = 1$ . The display: the $i$ -th diagonal entry of the matrix of $u$ is precisely $e_i^*(u(e_i))$ . ∎

Remark 2.23 (Common pitfalls)

(i) The determinant is $n$ -linear in the columns, not linear in the matrix: $\det(A + B) \neq \det A + \det B$ in general, and $\det(\lambda A) = \lambda^n\det A$ , not $\lambda\det A$ . (ii) Transposition reverses products: $(vu)^{\mathsf T} = u^{\mathsf T}v^{\mathsf T}$ ; forgetting the reversal wrecks every computation involving inverses. (iii) The annihilator $F^\circ$ lives in $E^*$ , not in $E$ : it becomes the familiar “orthogonal complement” only after an inner product identifies $E$ with $E^*$ (Chapter 12); no such identification is canonical. (iv) “Row rank equals column rank” does not mean row space equals column space — the two live in different spaces ( $K^n$ and $K^m$ ) and are related through Proposition 2.10, not equal. (v) The permutation formula is a proof device: for numbers, use row operations and cofactors (Example 2.15).

Example 2.24 (The trace pairing splits the matrix space)

On $\mathcal{M}_2(\R)$ with the pairing $\langle A, B\rangle = \operatorname{tr}(AB)$ of Exercise 2.9: decompose $M = \left(\begin{smallmatrix}1 & 4\\ 2 & 3\end{smallmatrix}\right)$ into symmetric and antisymmetric parts,

M = S + A, \qquad S = \tfrac12(M + M^{\mathsf T}) = \begin{pmatrix}1 & 3\\ 3 & 3\end{pmatrix}, \qquad A = \tfrac12(M - M^{\mathsf T}) = \begin{pmatrix}0 & 1\\ -1 & 0\end{pmatrix}.

Then $\operatorname{tr}(SA) = \operatorname{tr} \left(\begin{smallmatrix}-3 & 1\\ -3 & 3\end{smallmatrix}\right) = 0$ : the two parts are “orthogonal” for the trace pairing — an instance of the general fact (proved in the weekend problem of this chapter) that antisymmetric matrices form exactly the annihilator of the symmetric ones. Duality sees the decomposition $\mathcal{M}_n = \mathcal{S}_n \oplus \mathcal{A}_n$ before any inner product is chosen.

Remark 2.25 (Perspectives within this volume)

Watch the three constructions of this chapter change costume ahead. The transpose returns in Chapter 3: $u$ and $u^{\mathsf T}$ share eigenvalues with equal geometric multiplicities (this chapter’s weekend problem, question 15), which is why row and column analyses of a matrix never disagree. The determinant becomes a function of a parameter in Chapter 3 ( $\chi_u(X) = \det(X\,\mathrm{id} - u)$ ) and a Jacobian in Chapter 20, where its multilinearity turns into the change-of-variables factor. The trace seeds the similarity invariants: it is the second coefficient of $\chi_u$ , the sum of eigenvalues, and eventually the integral of the diagonal in Chapter 14-style identities. One linear-algebra chapter, three long shadows.

Remark 2.26 (Where this chapter is used)

The dual space is not an abstraction for its own sake: annihilators and transposes run the solvability theory of linear systems (this chapter’s weekend problem proves the finite-dimensional Fredholm alternative from them), nondegenerate pairings reappear as the polar form in Chapter 12 and the adjoint in Chapter 13, and the determinant built here powers the whole of Chapter 3. In the Year 3 volume the same duality, transported to infinite dimension, becomes the Riesz representation theorem and Fredholm theory on Hilbert spaces — with compactness replacing the dimension counts used here.

2.4 Exercises

Exercise 2.1 ★

In $\R^3$ , let $\varphi_1(x,y,z) = x + y$ , $\varphi_2 = y + z$ , $\varphi_3 = x + z$ . Prove that $(\varphi_1, \varphi_2, \varphi_3)$ is a basis of $(\R^3)^*$ and find the basis of $\R^3$ of which it is the dual.

Solution

Solution of Exercise 2.1.

Three forms in a $3$ -dimensional dual: freeness suffices. A relation $\alpha\varphi_1 + \beta\varphi_2 + \gamma\varphi_3 = 0$ evaluated at $(1,0,0), (0,1,0), (0,0,1)$ gives $\alpha + \gamma = 0$ , $\alpha + \beta = 0$ , $\beta + \gamma = 0$ , whence $\alpha = \beta = \gamma = 0$ .

Pre-dual basis $(u_1, u_2, u_3)$ : solve $\varphi_i(u_j) = \delta_{ij}$ . Writing $u_j = (x, y, z)$ : for $u_1$ : $x + y = 1$ , $y + z = 0$ , $x + z = 0$ gives $u_1 = \bigl(\tfrac12, \tfrac12, -\tfrac12\bigr)$ ; symmetrically $u_2 = \bigl(-\tfrac12, \tfrac12, \tfrac12\bigr)$ , $u_3 = \bigl(\tfrac12, -\tfrac12, \tfrac12\bigr)$ .

Exercise 2.2 ★

Compute by the permutation formula the determinants of

\begin{pmatrix} 0 & 0 & a\\ 0 & b & 0\\ c & 0 & 0 \end{pmatrix}, \qquad \begin{pmatrix} a & b & 0 & 0\\ c & d & 0 & 0\\ 0 & 0 & e & f\\ 0 & 0 & g & h \end{pmatrix},

and state the block-diagonal rule the second suggests.

Solution

Solution of Exercise 2.2.

First matrix: the only nonzero-product permutation sends $1 \mapsto 3$ , $2 \mapsto 2$ , $3 \mapsto 1$ — the transposition $(1\,3)$ , signature $-1$ : determinant $-abc$ .

Second: a permutation with nonzero product cannot mix the two blocks (an entry linking them is $0$ ), so it splits as a permutation of $\{1,2\}$ times one of $\{3,4\}$ , and the signature is the product of the two signatures: the sum factorizes as

(ad - bc)(eh - fg) .

General rule suggested (and true, same proof): the determinant of a block-diagonal matrix is the product of the determinants of the blocks.

Exercise 2.3 ★

Let $F = \{(x,y,z,t) \in \R^4 : x + y = z + t \text{ and } x = 2y\}$ . Give a basis of $F^\circ$ and check Theorem 2.6 on dimensions.

Solution

Solution of Exercise 2.3.

$F$ is defined by the two independent equations $\varphi_1(x,y,z,t) = x + y - z - t = 0$ and $\varphi_2 = x - 2y = 0$ : by Theorem 2.6 read backwards, $F^\circ = \operatorname{Vect}(\varphi_1, \varphi_2)$ — they lie in $F^\circ$ by construction, they are free (not proportional), and $\dim F^\circ = 4 - \dim F = 4 - 2 = 2$ since $\dim F = 2$ (two independent equations in $\R^4$ ). Basis: $(\varphi_1, \varphi_2)$ ; dimensions: $2 + 2 = 4$ , as the theorem demands.

Exercise 2.4 ★★

Let $a_0, \dots, a_n$ be distinct points of $K$ and $\varphi_i \colon P \mapsto P(a_i)$ on $K_n[X]$ . Prove that $(\varphi_0, \dots, \varphi_n)$ is a basis of $K_n[X]^*$ , identify its pre-dual basis, and expand the form $P \mapsto \int_0^1 P(t)\,\dd t$ (for $K = \R$ , $n = 2$ , $a_i = 0, \frac12, 1$ ) in this basis — recognizing Simpson’s rule.

Solution

Solution of Exercise 2.4.

The $\varphi_i$ are $n + 1$ forms on an $(n+1)$ -dimensional space: freeness suffices. If $\sum_i \lambda_i \varphi_i = 0$ , evaluate on the Lagrange polynomial $L_j$ of the nodes: $\lambda_j = 0$ . The pre-dual basis is $(L_0, \dots, L_n)$ , since $\varphi_i(L_j) = L_j(a_i) = \delta_{ij}$ .

For the integral form with nodes $0, \frac12, 1$ on $\R_2[X]$ : $\int_0^1 P = \sum_i c_i P(a_i)$ with $c_i = \int_0^1 L_i$ . Compute: $L_0 = 2(X - \tfrac12)(X - 1)$ , $\int_0^1 L_0 = \frac16$ ; $L_1 = -4X(X-1)$ , $\int_0^1 L_1 = \frac46$ ; $L_2 = 2X(X - \tfrac12)$ , $\int_0^1 L_2 = \frac16$ . Hence

\int_0^1 P = \frac{1}{6}\Bigl(P(0) + 4P\bigl(\tfrac12\bigr) + P(1)\Bigr) \quad (P \in \R_2[X]) :

Simpson’s rule, exact on quadratics — a statement about dual bases.

Exercise 2.5 ★★

Let $u \in \mathcal{L}(E)$ with $\dim E = n$ and $\operatorname{rk} u = 1$ . Prove that $u = \varphi(\cdot)\, a$ for a vector $a$ and a form $\varphi$ ; that $\operatorname{tr} u = \varphi(a)$ ; and that $u^2 = (\operatorname{tr} u)\, u$ . Deduce $\det(I + u) = 1 + \operatorname{tr} u$ .

Solution

Solution of Exercise 2.5.

$\operatorname{im} u = Ka$ for some $a \neq 0$ ; then $u(x) = \varphi(x)\,a$ where $\varphi(x)$ is the coordinate of $u(x)$ on $a$ — linear in $x$ . Trace: complete $a = e_1$ into a basis; the matrix of $u$ has columns $\varphi(e_j)\,e_1$ , so its only diagonal entry is $\varphi(e_1) = \varphi(a)$ : $\operatorname{tr} u = \varphi(a)$ . Then

u^2(x) = \varphi(x)\, u(a) = \varphi(x)\varphi(a)\, a = (\operatorname{tr} u)\, u(x).

Determinant, in two cases. If $\varphi(a) \neq 0$ : take any basis of the hyperplane $\ker\varphi$ and append $a$ . Then $u$ kills $\ker\varphi$ (there $u(x) = \varphi(x)a = 0$ ) and $u(a) = \varphi(a)\,a$ : the matrix of $I + u$ is diagonal, $(1, \dots, 1,\, 1 + \varphi(a))$ , so $\det(I + u) = 1 + \varphi(a) = 1 + \operatorname{tr} u$ . If $\varphi(a) = 0$ : then $a \in \ker\varphi$ ; take a basis of $\ker\varphi$ whose first vector is $a$ , and append a vector $b$ with $\varphi(b) = 1$ . Then $I + u$ fixes the basis of $\ker\varphi$ and sends $b \mapsto b + a$ : triangular with unit diagonal, $\det(I + u) = 1 = 1 + \operatorname{tr} u$ . Both cases agree with the formula.

Exercise 2.6 ★★

Prove that every hyperplane of $\mathcal{M}_n(K)$ ( $n \geq 2$ ) contains an invertible matrix. Hint: a hyperplane is $\{M : \operatorname{tr}(AM) = 0\}$ for some $A \neq 0$ (Exercise 2.9). If $A$ is scalar, exhibit an invertible matrix of zero trace; otherwise, find an invertible $M$ making $AM$ have zero diagonal — a permutation-like matrix does it.

Solution

Solution of Exercise 2.6.

By Exercise 2.9, the hyperplane is $H_A = \{M : \operatorname{tr}(AM) = 0\}$ with $A \neq 0$ .

If $A = \lambda I$ : $H_A$ is the zero-trace hyperplane; the matrix of the $n$ -cycle permutation (ones in positions $(i, i+1)$ and $(n, 1)$ ) is invertible (its determinant is $\pm 1$ by Example 2.20’s computation) and has zero trace.

If $A$ is not scalar: first find an invertible $P$ such that $B = P^{-1}AP$ has a nonzero off-diagonal entry $b_{ji}$ ( $j \neq i$ ). Indeed, if $A$ already has one, take $P = I$ ; if $A$ is diagonal with two distinct entries $d_1 \neq d_2$ , conjugating by the transvection $P = I + E_{12}$ produces the off-diagonal entry $d_1 - d_2 \neq 0$ (compute: $P^{-1}AP = A + (d_1 - d_2)E_{12}$ ); and a diagonal matrix with all entries equal is scalar, excluded. Now set $M' = I + tE_{ij}$ with $t = -\operatorname{tr}(B)/b_{ji}$ : then

\operatorname{tr}(BM') = \operatorname{tr} B + t\,b_{ji} = 0,

and $M'$ is invertible (triangular with unit diagonal). Undoing the conjugation, $M = PM'P^{-1}$ is invertible and $\operatorname{tr}(AM) = \operatorname{tr}(BM') = 0$ : $M \in H_A$ .

Exercise 2.7 ★★

(Derivative of the determinant) For $A \in \mathcal{M}_n(\R)$ , prove from multilinearity that

\frac{\dd}{\dd t}\Big|_{t=0} \det(I_n + tA) = \operatorname{tr} A ,

and deduce $\det(\eu^{tA}) = \eu^{t\operatorname{tr} A}$ assuming the differentiability of $t \mapsto \det(\eu^{tA})$ and the group property $\eu^{(s+t)A} = \eu^{sA}\eu^{tA}$ (established in Chapter 16).

Solution

Solution of Exercise 2.7.

$\det(I + tA)$ is, by the permutation formula, a polynomial in $t$ ; its constant term is $1$ ( $t = 0$ ). Its $t$ -coefficient: expand $\det$ as an alternating form of the columns $e_j + t\,c_j(A)$ ; by multilinearity, the terms linear in $t$ replace exactly one $e_j$ by $c_j(A)$ :

\sum_{j} \det(e_1, \dots, c_j(A), \dots, e_n) = \sum_j a_{jj} = \operatorname{tr} A ,

(the determinant with all canonical columns except $c_j(A)$ in slot $j$ picks the $j$ -th diagonal entry). Hence the derivative at $0$ is $\operatorname{tr} A$ .

Let $g(t) = \det(\eu^{tA})$ . The group property gives $g(s + t) = g(s)g(t)$ (multiplicativity of $\det$ ), $g$ is differentiable, and $g'(0) = \operatorname{tr} A$ by the above ( $\eu^{tA} = I + tA + O(t^2)$ ). A differentiable morphism $(\R, +) \to (\R^*, \times)$ satisfies $g' = g'(0)\,g$ (differentiate $g(s+t)$ in $s$ at $0$ ), so $g(t) = \eu^{t\operatorname{tr} A}$ by the uniqueness of solutions of $y' = cy$ with $y(0) = 1$ (Year 1 volume).

Exercise 2.8 ★★

(Circulant, $3 \times 3$ ) Let $j = \eu^{2\iu\pi/3}$ and

C = \begin{pmatrix} a & b & c\\ c & a & b\\ b & c & a \end{pmatrix} \in \mathcal{M}_3(\C).

Verify that the columns of the Vandermonde matrix of $1, j, j^2$ are eigenvectors of $C$ , and deduce

\det C = (a + b + c)(a + bj + cj^2)(a + bj^2 + cj).

Solution

Solution of Exercise 2.8.

Let $v_k = (1, j^k, j^{2k})^{\mathsf T}$ for $k = 0, 1, 2$ . Using $1 + j + j^2 = 0$ and $j^3 = 1$ :

C v_k = \begin{pmatrix} a + b j^k + c j^{2k}\\ c + a j^k + b j^{2k}\\ b + c j^k + a j^{2k} \end{pmatrix} = (a + b j^k + c j^{2k}) \begin{pmatrix} 1\\ j^k\\ j^{2k}\end{pmatrix},

(check the second row: $j^k(a + bj^k + cj^{2k}) = aj^k + bj^{2k} + cj^{3k} = c + aj^k + bj^{2k}$ ). So $v_k$ is an eigenvector with eigenvalue $\lambda_k = a + bj^k + cj^{2k}$ . The $v_k$ form a basis (Vandermonde of the distinct $1, j, j^2$ ), so $C$ is diagonalizable with these eigenvalues and

\det C = \lambda_0\lambda_1\lambda_2 = (a+b+c)(a + bj + cj^2)(a + bj^2 + cj).

Exercise 2.9 ★★★

Prove that every linear form $t$ on $\mathcal{M}_n(K)$ is $M \mapsto \operatorname{tr}(AM)$ for a unique $A$ : the map $A \mapsto \operatorname{tr}(A\,\cdot)$ is an isomorphism from $\mathcal{M}_n(K)$ onto its dual. Deduce the uniqueness statement of Proposition 2.22 again.

Solution

Solution of Exercise 2.9.

The map $\Theta \colon A \mapsto \operatorname{tr}(A\,\cdot)$ is linear from $\mathcal{M}_n(K)$ to its dual, between spaces of equal dimension $n^2$ : injectivity suffices. If $\operatorname{tr}(AM) = 0$ for all $M$ , take $M = E_{ji}$ : $\operatorname{tr}(A E_{ji}) = a_{ij} = 0$ for all $i, j$ : $A = 0$ . So $\Theta$ is an isomorphism.

Uniqueness of the trace (Proposition 2.22): a form $t$ killing all commutators is $\operatorname{tr}(A\,\cdot)$ for some $A$ with $\operatorname{tr}(A(MN - NM)) = 0$ for all $M, N$ , i.e. $\operatorname{tr}((AM - MA)N) = 0$ for all $N$ (cyclicity), i.e. $AM = MA$ for all $M$ (injectivity of $\Theta$ ): $A$ commutes with everything, hence is scalar ( $A$ commutes with all $E_{ij}$ forces off-diagonal entries $0$ and equal diagonal entries), so $t = c \operatorname{tr}$ .

Exercise 2.10 ★★★

Let $u, v \in \mathcal{L}(E)$ with $u \circ v - v \circ u = u$ . Prove that $u$ is nilpotent. Hint: show $\operatorname{tr}(u^k) = 0$ for all $k \geq 1$ (compute $u^k v - v u^k$ by induction), then use the following fact, to be proved via Newton’s identities or by induction on the dimension: an endomorphism of a $\C$ -vector space all of whose powers have zero trace is nilpotent. Work over $\C$ .

Solution

Solution of Exercise 2.10.

Work over $\C$ (a real matrix is nilpotent iff it is as a complex matrix: nilpotence is $u^n = 0$ ).

Step 1: $\operatorname{tr}(u^k) = 0$ for $k \geq 1$ . By induction, $u^k v - v u^k = k\, u^k$ : for $k = 1$ it is the hypothesis; for the step,

u^{k+1}v - vu^{k+1} = u^k(uv - vu) + (u^k v - v u^k)u = u^{k+1} + k\,u^{k+1} .

Taking traces: $0 = \operatorname{tr}(u^k v) - \operatorname{tr}(vu^k) = k \operatorname{tr}(u^k)$ , so $\operatorname{tr}(u^k) = 0$ .

Step 2: zero power traces imply nilpotence (over $\C$ ). Let $\lambda_1, \dots, \lambda_r$ be the distinct nonzero eigenvalues of $u$ with multiplicities $m_1, \dots, m_r$ (in the characteristic polynomial, which splits over $\C$ — Chapter 3). Power traces are $\operatorname{tr}(u^k) = \sum_i m_i \lambda_i^k$ (trigonalize: the diagonal of a triangular matrix’s $k$ -th power is the $k$ -th powers). The system $\sum_i m_i \lambda_i^k = 0$ for $k = 1, \dots, r$ is Vandermonde-invertible in the unknowns $m_i\lambda_i$ (matrix $(\lambda_i^{k-1})$ times diagonal $\lambda_i$ , all $\lambda_i \neq 0$ distinct): every $m_i \lambda_i = 0$ , impossible with $m_i \geq 1$ unless $r = 0$ . So $u$ has no nonzero eigenvalue: its characteristic polynomial is $(-X)^n$ , and Cayley–Hamilton (Chapter 3) gives $u^n = 0$ : nilpotent.

Exercise 2.11 ★★

(Vandermonde) For $a_0, \dots, a_n \in K$ , prove

\det\begin{pmatrix} 1 & 1 & \cdots & 1\\ a_0 & a_1 & \cdots & a_n\\ \vdots & \vdots & & \vdots\\ a_0^n & a_1^n & \cdots & a_n^n \end{pmatrix} = \prod_{0 \leq i < j \leq n} (a_j - a_i).

(View the determinant as a polynomial in $a_n$ : identify its degree, its roots, and its leading coefficient; induct.)

Solution

Solution of Exercise 2.11.

Write $V(a_0, \dots, a_n)$ for the determinant and induct on $n$ ; $V(a_0) = 1$ starts. Fix $a_0, \dots, a_{n-1}$ and view $D(T) = V(a_0, \dots, a_{n-1}, T)$ , the determinant with last column $(1, T, \dots, T^n)$ : expanding along that column, $D$ is a polynomial of degree $\leq n$ in $T$ whose $T^n$ -coefficient is the minor $V(a_0, \dots, a_{n-1})$ . Suppose first that $a_0, \dots, a_{n-1}$ are distinct. For each $T = a_i$ ( $i < n$ ) two columns coincide, so $D(a_i) = 0$ : with $n$ distinct roots and degree $\leq n$ ,

D(T) = V(a_0, \dots, a_{n-1}) \prod_{i=0}^{n-1}(T - a_i),

and $T = a_n$ plus the induction hypothesis give the product formula. If two of $a_0, \dots, a_{n-1}$ coincide, both sides are $0$ (repeated columns; a repeated factor), and the formula holds trivially.

Exercise 2.12 ★★★

Let $A, B, C, D \in \mathcal{M}_n(K)$ with $K$ infinite, and suppose $CD = DC$ . Prove that

\det\begin{pmatrix} A & B\\ C & D\end{pmatrix} = \det(AD - BC).

(Treat first $D$ invertible, multiplying on the right by $\left(\begin{smallmatrix} I & 0\\ -D^{-1}C & I\end{smallmatrix}\right)$ ; then replace $D$ by $D + tI$ and compare two polynomials in $t$ .)

Solution

Solution of Exercise 2.12.

$D$ invertible. Multiply on the right by the block matrix $T = \left(\begin{smallmatrix} I & 0\\ -D^{-1}C & I\end{smallmatrix}\right)$ , which is block-triangular with unit diagonal, $\det T = 1$ (its determinant, by the permutation formula, only picks the diagonal blocks — the block rule of Exercise 2.2):

\begin{pmatrix} A & B\\ C & D\end{pmatrix} T = \begin{pmatrix} A - BD^{-1}C & B\\ C - DD^{-1}C & D\end{pmatrix} = \begin{pmatrix} A - BD^{-1}C & B\\ 0 & D\end{pmatrix},

whose determinant is $\det(A - BD^{-1}C)\det D = \det\bigl((A - BD^{-1}C)D\bigr) = \det(AD - BD^{-1}CD)$ . Since $CD = DC$ , $BD^{-1}CD = BC$ : the determinant is $\det(AD - BC)$ .

General $D$ . Let $D_t = D + tI$ ; then $CD_t = D_tC$ still. Both

f(t) = \det\begin{pmatrix} A & B\\ C & D_t\end{pmatrix} \qquad\text{and}\qquad g(t) = \det(AD_t - BC)

are polynomial functions of $t$ . The polynomial $\det(D + tI)$ is monic of degree $n$ , hence has at most $n$ roots: for all but finitely many $t$ , $D_t$ is invertible and $f(t) = g(t)$ by the first case. Two polynomials over an infinite field agreeing at infinitely many points are equal: $f = g$ , and $t = 0$ concludes.

2.5 Problem: The Fredholm Alternative

When does the linear system $u(x) = b$ have a solution? The complete answer is a duality statement: exactly when $b$ is annihilated by every linear form that annihilates the image of $u$ — and those forms are computable, being the kernel of the transpose. This weekend problem builds the full dictionary of finite-dimensional duality (factorization of forms, biduality, annihilator calculus, the transpose), proves the finite-dimensional Fredholm alternative, and closes with the trace form and a characterization: the trace is the only linear invariant of similarity. Throughout, $E$ and $F$ are finite-dimensional $K$ -vector spaces, $n = \dim E$ .

Problem 2.1

Weekend problem — duality in finite dimension and the Fredholm alternative

Notation: for $S \subseteq E^*$ , the pre-annihilator is $S_\circ = \{x \in E : \varphi(x) = 0 \text{ for all } \varphi \in S\}$ ; annihilators $F^\circ$ and transposes $u^{\mathsf T}$ are those of Definition 2.5 and Definition 2.8.

Part I — The factorization lemma. Let $\varphi_1, \dots, \varphi_p, \varphi \in E^*$ .

Let $\Phi \colon E \to K^p$ , $x \mapsto (\varphi_1(x), \dots, \varphi_p(x))$ . Identify $\ker\Phi$ , show $\Phi^{\mathsf T}$ maps the coordinate forms of $K^p$ to the $\varphi_i$ , and deduce

$\dim \bigl(\ker\varphi_1 \cap \dots \cap \ker\varphi_p\bigr) = n - \dim \operatorname{Vect}(\varphi_1, \dots, \varphi_p).$
(Factorization lemma) Prove the equivalence:

$\varphi \in \operatorname{Vect}(\varphi_1, \dots, \varphi_p) \iff \ker\varphi_1 \cap \dots \cap \ker\varphi_p \subseteq \ker\varphi .$
Deduce: $(\varphi_1, \dots, \varphi_p)$ is free iff $\bigcap_i \ker\varphi_i$ has dimension $n - p$ ; and a subspace of codimension $p$ is an intersection of $p$ hyperplanes, never fewer.
In $\R^4$ , let $\varphi_1 = x + y - z$ , $\varphi_2 = y + z - t$ , $\psi = x + 2y - t$ and $\psi' = x + y + t$ . Decide, by the factorization lemma, whether $\psi$ and $\psi'$ belong to $\operatorname{Vect}(\varphi_1, \varphi_2)$ .
On $E = \R_2[X]$ , show that $\psi_0 \colon P \mapsto P(0)$ , $\psi_1 \colon P \mapsto P(1)$ , $\psi_2 \colon P \mapsto \int_0^1 P(t)\dd t$ form a basis of $E^*$ , compute the basis $(P_0, P_1, P_2)$ of $E$ of which it is the dual, and find the unique $P \in \R_2[X]$ with $P(0) = 1$ , $P(1) = 2$ , $\int_0^1 P = \frac32$ .

Part II — Biduality and the annihilator calculus.

Show that the evaluation map $J \colon E \to E^{**}$ , $J(x)(\varphi) = \varphi(x)$ , is linear and injective, hence an isomorphism in finite dimension.
(Double annihilator) Show $J(F) = F^{\circ\circ} := (F^\circ)^\circ$ for every subspace $F \subseteq E$ : under the identification $J$ , the annihilator of the annihilator is the subspace itself.
Prove the annihilator calculus: $(F + G)^\circ = F^\circ \cap G^\circ$ and $(F \cap G)^\circ = F^\circ + G^\circ$ .
Deduce (and reprove directly): two nonzero forms with the same kernel are proportional.
(Antedual basis) Show that for every basis $(\varphi_1, \dots, \varphi_n)$ of $E^*$ there is a unique basis $(u_1, \dots, u_n)$ of $E$ with $\varphi_i(u_j) = \delta_{ij}$ .

Part III — The transpose calculus.

Show that $u \mapsto u^{\mathsf T}$ is a linear bijection from $\mathcal{L}(E, F)$ onto $\mathcal{L}(F^*, E^*)$ , and that $(u^{-1})^{\mathsf T} = (u^{\mathsf T})^{-1}$ when $u$ is invertible.
(Naturality) Show that $u^{\mathsf T\mathsf T} \circ J_E = J_F \circ u$ : under the evaluation isomorphisms, the double transpose is $u$ .
Show: $u$ is surjective iff $u^{\mathsf T}$ is injective; $u$ is injective iff $u^{\mathsf T}$ is surjective.
For $u \in \mathcal{L}(E)$ : a subspace $F$ is stable under $u$ if and only if $F^\circ$ is stable under $u^{\mathsf T}$ .
Show that $\ker(u^{\mathsf T} - \lambda\, \mathrm{id}_{E^*}) = \bigl(\operatorname{im}(u - \lambda\, \mathrm{id}_E)\bigr)^\circ$ , and deduce that $u$ and $u^{\mathsf T}$ have the same eigenvalues with the same geometric multiplicities.

Part IV — The Fredholm alternative.

Prove that $\operatorname{im} u = (\ker u^{\mathsf T})_\circ$ for $u \in \mathcal{L}(E, F)$ , and deduce the Fredholm alternative in finite dimension: the equation $u(x) = b$ has a solution if and only if every $\psi \in F^*$ with $u^{\mathsf T}\psi = 0$ satisfies $\psi(b) = 0$ .
Matrix form: for $A \in \mathcal{M}_{m,n}(K)$ and $b \in K^m$ , exactly one of the following holds: (i) $Ax = b$ has a solution; (ii) there is $y \in K^m$ with $A^{\mathsf T}y = 0$ and $y^{\mathsf T}b = 1$ . Prove both the “at most one” and the “at least one”.
Find all $b \in \R^3$ for which the system

$x + y = b_1, \qquad y + z = b_2, \qquad x + 2y + z = b_3$

has a solution, by computing the kernel of the transposed matrix.
(A discrete Neumann problem) On $E = \R^n$ ( $n \geq 3$ ), define $L$ by $(Lx)_k = x_k - \frac12(x_{k-1} + x_{k+1})$ , indices modulo $n$ . Show $L^{\mathsf T} = L$ (canonical identifications), show $\ker L$ is the line of constant vectors (look at a maximal coordinate), and conclude: $Lx = b$ is solvable iff $\sum_k b_k = 0$ .

Part V — The trace form and the invariance theorem. Recall from Exercise 2.9 that $A \mapsto \operatorname{tr}(A\,\cdot)$ identifies $\mathcal{M}_n(K)$ with its dual. Assume $\operatorname{char} K = 0$ (e.g. $K = \Q, \R, \C$ ).

Under this identification, show that the annihilator of the subspace $\mathcal{S}_n$ of symmetric matrices is the subspace $\mathcal{A}_n$ of antisymmetric matrices, and conversely.
Show that the annihilator of the hyperplane $\mathfrak{sl}_n = \{M : \operatorname{tr} M = 0\}$ is the line $K I_n$ ; equivalently, a linear form vanishing on all trace-zero matrices is a multiple of the trace.
Show that every matrix of $\mathcal{M}_n(K)$ is the sum of two invertible matrices.
(The trace is the only linear similarity invariant) Let $t$ be a linear form on $\mathcal{M}_n(K)$ with $t(PMP^{-1}) = t(M)$ for every $M$ and every invertible $P$ . Show first $t(PX) = t(XP)$ for $P$ invertible, then $t(BX) = t(XB)$ for all $B$ , and conclude $t = c \operatorname{tr}$ for some $c \in K$ .
Show that $\operatorname{rk} u \leq r$ if and only if $u$ is a sum of $r$ maps of rank $\leq 1$ , i.e. $u = \sum_{i=1}^{r} \psi_i(\cdot)\,f_i$ with $\psi_i \in E^*$ , $f_i \in F$ ; deduce $\operatorname{rk}(u + v) \leq \operatorname{rk} u + \operatorname{rk} v$ .
(Synthesis) Draw up the dictionary proved in this problem: subspaces versus annihilators, sums versus intersections, maps versus transposes, solvability versus orthogonality to the transposed kernel, trace versus similarity. For each entry, cite the question that proved it, and state in one sentence what replaces the dimension counts when dimension becomes infinite (the Year 3 volume makes this precise on Hilbert spaces).

Solution

Solution of Problem 2.1.

1. $\Phi$ is linear with $\ker\Phi = \bigcap_i \ker\varphi_i$ (a $p$ -tuple vanishes iff each entry does). For the coordinate forms $\varepsilon_i$ of $K^p$ : $\Phi^{\mathsf T}(\varepsilon_i) = \varepsilon_i \circ \Phi = \varphi_i$ , so $\operatorname{im}\Phi^{\mathsf T} \supseteq \operatorname{Vect}(\varphi_i)$ ; conversely $\operatorname{im}\Phi^{\mathsf T}$ is spanned by the $\Phi^{\mathsf T}(\varepsilon_i)$ (the $\varepsilon_i$ span $(K^p)^*$ ). So $\operatorname{rk}\Phi = \operatorname{rk} \Phi^{\mathsf T} = \dim\operatorname{Vect}(\varphi_1, \dots, \varphi_p) =: r$ (Proposition 2.10), and rank–nullity gives $\dim\bigcap_i\ker\varphi_i = n - r$ .

2. ( $\Leftarrow$ ) Keep a maximal free subfamily, say $\varphi_1, \dots, \varphi_r$ , spanning the same space (so the hypothesis still reads $\bigcap_{i \leq r}\ker\varphi_i \subseteq \ker\varphi$ : the intersection over all $i$ equals the one over $i \leq r$ , each discarded form being a combination). The map $\Psi = (\varphi_1, \dots, \varphi_r) \colon E \to K^r$ is surjective (question 1: its rank is $r$ ). If $\Psi(x) = \Psi(y)$ then $x - y \in \ker\Psi \subseteq \ker\varphi$ , so $\varphi(x) = \varphi(y)$ : $\varphi$ factors as $\varphi = \lambda \circ \Psi$ with $\lambda \colon K^r \to K$ well defined; $\lambda$ is linear because $\Psi$ is linear and surjective (for $t = \Psi(x)$ , $t' = \Psi(x')$ : $\lambda(t + \alpha t') = \varphi(x + \alpha x') = \lambda(t) + \alpha\lambda(t')$ ). Writing $\lambda = \sum c_i \varepsilon_i$ : $\varphi = \sum_{i \leq r} c_i\varphi_i$ . ( $\Rightarrow$ ) If $\varphi = \sum c_i \varphi_i$ , any $x$ killing every $\varphi_i$ kills $\varphi$ .

3. By question 1, $\dim\bigcap\ker\varphi_i = n - r$ with $r = \dim\operatorname{Vect}(\varphi_i) \leq p$ , and $r = p$ iff the family is free. A subspace $F$ of codimension $p$ : its annihilator has dimension $p$ (Theorem 2.6); a basis $(\varphi_1, \dots, \varphi_p)$ of $F^\circ$ gives $F = \bigcap_i\ker\varphi_i$ (the recovery formula). Fewer: an intersection of $q$ hyperplanes has dimension $\geq n - q > n - p$ by question 1.

4. Compute $\ker\varphi_1 \cap \ker\varphi_2$ : from $x + y - z = 0$ and $y + z - t = 0$ , parametrize by $(y, z)$ : $x = z - y$ , $t = y + z$ , giving the plane of vectors $(z - y,\; y,\; z,\; y + z)$ . On it, $\psi = x + 2y - t = (z - y) + 2y - (y + z) = 0$ : by the factorization lemma $\psi \in \operatorname{Vect}(\varphi_1, \varphi_2)$ — indeed $\psi = \varphi_1 + \varphi_2$ . But $\psi' = x + y + t = (z - y) + y + (y + z) = y + 2z$ is not identically zero there ( $y = 1, z = 0$ gives $1$ ): $\psi' \notin \operatorname{Vect}(\varphi_1, \varphi_2)$ .

5. Three forms on a $3$ -dimensional space: freeness suffices. If $a\psi_0 + b\psi_1 + c\psi_2 = 0$ , test on $1, X, X^2$ : $a + b + c = 0$ , $b + \frac c2 = 0$ , $b + \frac c3 = 0$ ; subtracting the last two gives $c = 0$ , then $b = 0$ , $a = 0$ . Antedual basis: writing $P = \alpha + \beta X + \gamma X^2$ and solving $\psi_i(P_j) = \delta_{ij}$ ( $P(0) = \alpha$ , $P(1) = \alpha + \beta + \gamma$ , $\int_0^1 P = \alpha + \frac\beta2 + \frac\gamma3$ ):

P_0 = 1 - 4X + 3X^2, \qquad P_1 = -2X + 3X^2, \qquad P_2 = 6X - 6X^2 .

(Check, e.g.: $\int_0^1 P_2 = 3 - 2 = 1$ , $P_2(0) = P_2(1) = 0$ .) The interpolation problem is solved by coordinates in the antedual basis:

P = 1\cdot P_0 + 2\cdot P_1 + \tfrac32\, P_2 = 1 + X

( $X$ -coefficient $-4 - 4 + 9 = 1$ , $X^2$ -coefficient $3 + 6 - 9 = 0$ ); indeed $P(0) = 1$ , $P(1) = 2$ , $\int_0^1 P = \frac32$ .

6. Linearity: for every $\varphi$ , $J(x + \alpha y)(\varphi) = \varphi(x + \alpha y) = J(x)(\varphi) + \alpha J(y)(\varphi)$ , i.e. $J(x + \alpha y) = J(x) + \alpha J(y)$ . Injectivity: if $x \neq 0$ , complete $x = e_1$ into a basis; the coordinate form $e_1^*$ has $J(x)(e_1^*) = 1 \neq 0$ . Since $\dim E^{**} = \dim E^* = \dim E$ , injective implies bijective.

7. Inclusion: for $x \in F$ and $\varphi \in F^\circ$ , $J(x)(\varphi) = \varphi(x) = 0$ , so $J(F) \subseteq F^{\circ\circ}$ . Dimensions (Theorem 2.6 twice):

\dim F^{\circ\circ} = \dim E^* - \dim F^\circ = n - (n - \dim F) = \dim F = \dim J(F),

$J$ being injective. Hence $J(F) = F^{\circ\circ}$ .

8. First identity: $\varphi$ kills $F + G$ iff it kills both $F$ and $G$ (it kills sums iff it kills the pieces): $(F+G)^\circ = F^\circ \cap G^\circ$ . Second: the inclusion $F^\circ + G^\circ \subseteq (F \cap G)^\circ$ is clear (each summand kills $F \cap G$ ). Dimensions, using the first identity and Grassmann:

\dim(F^\circ + G^\circ) = \dim F^\circ + \dim G^\circ - \dim(F^\circ \cap G^\circ) = (n - \dim F) + (n - \dim G) - \bigl(n - \dim(F + G)\bigr),

which by Grassmann in $E$ equals $n - \dim(F \cap G) = \dim(F \cap G)^\circ$ : equality.

9. Via the lemma: $\ker\psi \subseteq \ker\varphi$ with $p = 1$ gives $\varphi \in \operatorname{Vect}(\psi)$ , and $\varphi \neq 0$ makes the scalar nonzero. Directly: pick $x_0$ with $\psi(x_0) \neq 0$ ; every $x$ writes $x = \bigl(x - \frac{\psi(x)}{\psi(x_0)}x_0\bigr) + \frac{\psi(x)}{\psi(x_0)} x_0$ with the first term in $\ker\psi = \ker\varphi$ ; applying $\varphi$ : $\varphi(x) = \frac{\varphi(x_0)}{\psi(x_0)}\psi(x)$ .

10. Take the dual basis $(\varphi_1^*, \dots, \varphi_n^*)$ of $(\varphi_1, \dots, \varphi_n)$ inside $E^{**}$ (Definition 2.1 applied to $E^*$ ) and set $u_j = J^{-1}(\varphi_j^*)$ : a basis of $E$ ( $J$ is an isomorphism, question 6), with $\varphi_i(u_j) = J(u_j)(\varphi_i) = \varphi_j^*(\varphi_i) = \delta_{ij}$ . Uniqueness: the conditions $\varphi_i(u_j) = \delta_{ij}$ determine $J(u_j)$ on the basis $(\varphi_i)$ , hence determine $u_j$ .

11. Linearity: $(u + \alpha v)^{\mathsf T}\psi = \psi \circ (u + \alpha v) = u^{\mathsf T}\psi + \alpha\, v^{\mathsf T}\psi$ . Injectivity: if $u \neq 0$ , pick $x$ with $u(x) \neq 0$ and $\psi$ with $\psi(u(x)) \neq 0$ (question 6’s coordinate-form trick): $u^{\mathsf T}\psi \neq 0$ . The spaces $\mathcal{L}(E,F)$ and $\mathcal{L}(F^*, E^*)$ both have dimension $\dim E \dim F$ : bijective. If $u$ is invertible, the reversal rule $(vu)^{\mathsf T} = u^{\mathsf T}v^{\mathsf T}$ gives $u^{\mathsf T}(u^{-1})^{\mathsf T} = (u^{-1}u)^{\mathsf T} = \mathrm{id}_{E^*}$ and $(u^{-1})^{\mathsf T}u^{\mathsf T} = (uu^{-1})^{\mathsf T} = \mathrm{id}_{F^*}$ , so $(u^{\mathsf T})^{-1} = (u^{-1})^{\mathsf T}$ .

12. For $x \in E$ and $\psi \in F^*$ :

\bigl(u^{\mathsf T\mathsf T}(J_E x)\bigr)(\psi) = (J_E x)\bigl(u^{\mathsf T}\psi\bigr) = (u^{\mathsf T}\psi)(x) = \psi\bigl(u(x)\bigr) = \bigl(J_F(u(x))\bigr)(\psi).

As $\psi$ is arbitrary, $u^{\mathsf T\mathsf T} \circ J_E = J_F \circ u$ .

13. By Proposition 2.10: $\ker u^{\mathsf T} = (\operatorname{im} u)^\circ$ , so $u$ surjective $\iff \operatorname{im} u = F \iff (\operatorname{im}u)^\circ = \{0\}$ (Theorem 2.6) $\iff u^{\mathsf T}$ injective. And $\operatorname{im} u^{\mathsf T} = (\ker u)^\circ$ , so $u$ injective $\iff \ker u = \{0\} \iff (\ker u)^\circ = E^*$ $\iff u^{\mathsf T}$ surjective.

14. If $u(F) \subseteq F$ and $\varphi \in F^\circ$ : $(u^{\mathsf T}\varphi)(x) = \varphi(u(x)) = 0$ for $x \in F$ , so $u^{\mathsf T}\varphi \in F^\circ$ . Conversely, if $u(F) \not\subseteq F$ , pick $x \in F$ with $u(x) \notin F$ ; by the recovery formula of Theorem 2.6 there is $\varphi \in F^\circ$ with $\varphi(u(x)) \neq 0$ : then $(u^{\mathsf T}\varphi)(x) \neq 0$ although $x \in F$ , so $u^{\mathsf T}\varphi \notin F^\circ$ : $F^\circ$ not stable.

15. $u^{\mathsf T} - \lambda\,\mathrm{id}_{E^*} = (u - \lambda\,\mathrm{id}_E)^{\mathsf T}$ (transposition is linear and $\mathrm{id}^{\mathsf T} = \mathrm{id}$ ), so its kernel is $(\operatorname{im}(u - \lambda\,\mathrm{id}))^\circ$ (Proposition 2.10), of dimension

n - \operatorname{rk}(u - \lambda\,\mathrm{id}) = \dim\ker(u - \lambda\,\mathrm{id})

by rank–nullity. In particular one kernel is nonzero iff the other is: same eigenvalues, same geometric multiplicities.

16. Inclusion: if $b = u(x)$ and $u^{\mathsf T}\psi = 0$ , then $\psi(b) = \psi(u(x)) = (u^{\mathsf T}\psi)(x) = 0$ : so $\operatorname{im} u \subseteq (\ker u^{\mathsf T})_\circ$ . Dimensions: for a subspace $S \subseteq F^*$ , $S_\circ = J_F^{-1}(S^\circ)$ (unwind: $y \in S_\circ$ iff every $\psi \in S$ kills $y$ iff $J_F(y) \in S^\circ$ ), so $\dim S_\circ = \dim F - \dim S$ . With $S = \ker u^{\mathsf T}$ :

\dim(\ker u^{\mathsf T})_\circ = \dim F - \dim\ker u^{\mathsf T} = \operatorname{rk} u^{\mathsf T} = \operatorname{rk} u :

equality of dimensions, hence $\operatorname{im} u = (\ker u^{\mathsf T})_\circ$ . Restated: $b \in \operatorname{im} u$ iff $\psi(b) = 0$ for every $\psi$ with $u^{\mathsf T}\psi = 0$ — the Fredholm alternative.

17. Identify $(K^m)^*$ with $K^m$ by $y \mapsto \psi_y$ , $\psi_y(v) = y^{\mathsf T}v$ ; then $(u^{\mathsf T}\psi_y)(x) = y^{\mathsf T}Ax = (A^{\mathsf T}y)^{\mathsf T}x$ , so $u^{\mathsf T}\psi_y = \psi_{A^{\mathsf T}y}$ : the transpose is the transposed matrix. At most one: if $Ax = b$ and $A^{\mathsf T}y = 0$ , then $y^{\mathsf T}b = y^{\mathsf T}Ax = (A^{\mathsf T}y)^{\mathsf T}x = 0 \neq 1$ . At least one: if (i) fails, question 16 provides $\psi_y$ with $A^{\mathsf T}y = 0$ and $y^{\mathsf T}b \neq 0$ ; rescale $y$ to make it $1$ .

18. $A = \left(\begin{smallmatrix} 1 & 1 & 0\\ 0 & 1 & 1\\ 1 & 2 & 1\end{smallmatrix}\right)$ (third row = first + second, so $A$ is singular). Solve $A^{\mathsf T}y = 0$ : $y_1 + y_3 = 0$ , $y_1 + y_2 + 2y_3 = 0$ , $y_2 + y_3 = 0$ give $y_1 = y_2 = -y_3$ : the line spanned by $y = (1, 1, -1)$ . Fredholm: solvable iff $y^{\mathsf T}b = b_1 + b_2 - b_3 = 0$ , i.e. $b_3 = b_1 + b_2$ — visibly the right condition, since the third equation is the sum of the first two.

19. The matrix of $L$ has $1$ on the diagonal and $-\frac12$ in positions $(k, k\pm1)$ (mod $n$ ): symmetric, so $L^{\mathsf T} = L$ under the identification of question 17. Kernel: if $Lx = 0$ then each $x_k = \frac12(x_{k-1} + x_{k+1})$ . Let $k_0$ maximize $x_k$ ; the average of the two neighbours, both $\leq x_{k_0}$ , equals $x_{k_0}$ only if both equal $x_{k_0}$ ; propagating around the cycle, $x$ is constant. Conversely constants are killed. So $\ker L^{\mathsf T} = \ker L = \R(1, \dots, 1)$ , and the Fredholm alternative reads: $Lx = b$ solvable iff $(1,\dots,1)^{\mathsf T} b = \sum_k b_k = 0$ — the discrete compatibility condition: a “heat distribution” on a ring can be realized by a potential iff its total flux vanishes.

20. If $A$ is antisymmetric and $S$ symmetric:

\operatorname{tr}(AS) = \operatorname{tr}\bigl((AS)^{\mathsf T}\bigr) = \operatorname{tr}(S^{\mathsf T}A^{\mathsf T}) = -\operatorname{tr}(SA) = -\operatorname{tr}(AS),

so $2\operatorname{tr}(AS) = 0$ and ( $\operatorname{char} K \neq 2$ ) $\operatorname{tr}(AS) = 0$ : $\mathcal{A}_n \subseteq \mathcal{S}_n^\circ$ (identifying the dual with matrices). Dimensions: $\dim\mathcal{S}_n^\circ = n^2 - \frac{n(n+1)}2 = \frac{n(n-1)}2 = \dim\mathcal{A}_n$ : equality. Exchanging roles (same computation), $\mathcal{A}_n^\circ = \mathcal{S}_n$ .

21. $\operatorname{tr}(I_nM) = \operatorname{tr} M = 0$ for $M \in \mathfrak{sl}_n$ : the line $KI_n$ lies in the annihilator, whose dimension is $n^2 - (n^2 - 1) = 1$ : equality. Translated by the isomorphism $A \mapsto \operatorname{tr}(A\,\cdot)$ : a form vanishing on $\mathfrak{sl}_n$ is $\operatorname{tr}(\lambda I_n\,\cdot) = \lambda\operatorname{tr}$ .

22. Let $M \in \mathcal{M}_n(K)$ . The polynomial $t \mapsto \det(M - tI)$ is nonzero of degree $n$ , so it has at most $n$ roots; $K$ has characteristic $0$ , hence is infinite: pick $\lambda \neq 0$ that is not a root. Then $M = (M - \lambda I) + \lambda I$ writes $M$ as a sum of two invertible matrices.

23. Step 1: for invertible $P$ and arbitrary $X$ , apply invariance to $M = XP$ : $t(P(XP)P^{-1}) = t(XP)$ , i.e. $t(PX) = t(XP)$ . Step 2: fix $X$ ; both sides of $t(BX) = t(XB)$ are linear in $B$ and agree on invertible $B$ ; by question 22 every $B$ is a sum of two invertibles, so they agree everywhere. Step 3: $t$ kills every commutator $XB - BX$ ; the commutators span $\mathfrak{sl}_n$ (shown in the proof of Proposition 2.22), so $t$ vanishes on $\mathfrak{sl}_n$ and question 21 gives $t = c\operatorname{tr}$ . (Conversely every $c\operatorname{tr}$ is similarity-invariant: the trace is the linear similarity invariant.)

24. If $\operatorname{rk} u = r' \leq r$ : take a basis $(f_1, \dots, f_{r'})$ of $\operatorname{im} u$ and write $u(x) = \sum_{i=1}^{r'} \psi_i(x) f_i$ ; each coordinate $\psi_i(x)$ of $u(x)$ is linear in $x$ (composition of $u$ with a coordinate form), so $u$ is a sum of $r' \leq r$ rank- $\leq1$ maps (pad with zeros). Conversely, if $u = \sum_{i=1}^r \psi_i(\cdot)f_i$ , then $\operatorname{im} u \subseteq \operatorname{Vect}(f_1, \dots, f_r)$ : $\operatorname{rk} u \leq r$ . Subadditivity: write $u$ with $\operatorname{rk} u$ terms and $v$ with $\operatorname{rk} v$ terms; the sum has $\operatorname{rk} u + \operatorname{rk} v$ terms, so $\operatorname{rk}(u + v) \leq \operatorname{rk} u + \operatorname{rk} v$ .

25. The dictionary: a subspace $F$ corresponds to $F^\circ$ with complementary dimension (Theorem 2.6), and back again by biduality (questions 6–7); sums exchange with intersections (question 8); a map $u$ corresponds to $u^{\mathsf T}$ with $\ker u^{\mathsf T} = (\operatorname{im}u)^\circ$ , $\operatorname{im}u^{\mathsf T} = (\ker u)^\circ$ , equal ranks, exchanged injectivity/surjectivity, matched stable subspaces and eigenvalues (questions 11–15); the equation $u(x) = b$ is solvable iff $b$ is orthogonal to $\ker u^{\mathsf T}$ (questions 16–19); and on $\mathcal{M}_n$ the trace pairing realizes the whole dictionary concretely, with the trace as the unique linear similarity invariant (questions 20–23) and rank as the minimal length of a decomposition into elementary tensors (question 24). In infinite dimension the dimension counts fail and are replaced by closedness hypotheses on images and by completeness — on Hilbert spaces this becomes the Riesz representation theorem and the Fredholm theory of compact operators, proved honestly in the Year 3 volume.