CommeUnJeu · L1 MPSI

Matrix representation of linear maps

⌚ ~83 min ▢ 10 blocks ✓ 32 exercises ➣ Prerequisites : Linear maps, Matrix calculus

Two algebraic objects have been developed so far in parallel: linear maps (chapter Linear maps), studied for their abstract properties (kernel, image, rank, isomorphism, $\mathcal{L}(E, F)$ as a vector space, $\mathcal{L}(E)$ as a ring), and matrices (chapter Matrix calculus), studied for their algebraic calculus ($M_{n, p}(\mathbb{K})$ as a vector space, matrix product, the group $\mathrm{GL}_n(\mathbb{K})$). This chapter glues the two together. In a finite-dimensional vector space equipped with a basis, every linear map is encoded by a unique matrix, and every matrix corresponds to a unique linear map. Composition of maps becomes the matrix product, isomorphism becomes matrix invertibility, the rank of a map becomes the rank of its matrix. From now on every linear-map question becomes a matrix calculation, and every matrix-calculus question lifts to a coordinate-free statement.
The chapter unfolds in five sections. The first section introduces the matrix of a vector, of a family of vectors, and of a linear map in given bases; it states the key formula $\mathrm{Mat}_{\mathcal{B}'}(u(x)) = \mathrm{Mat}_{\mathcal{B}, \mathcal{B}'}(u) \cdot \mathrm{Mat}_{\mathcal{B}}(x)$ and shows how to read off images from columns. The second section closes the dictionary: the map $u \mapsto \mathrm{Mat}_{\mathcal{B}, \mathcal{B}'}(u)$ is an isomorphism of vector spaces $\mathcal{L}(E, F) \to M_{n, p}(\mathbb{K})$, the composition of maps corresponds to the matrix product, and $u \in \mathrm{GL}(E)$ if and only if $\mathrm{Mat}_{\mathcal{B}}(u) \in \mathrm{GL}_n(\mathbb{K})$. The third section develops the rank of a matrix and proves the matrix-form rank theorem, the rank normal form theorem, and the equality of row rank and column rank. The fourth section explains block-matrix writing and the form a matrix takes in a basis adapted to a stable direct sum $E = F \oplus G$. The fifth section summarises the dictionary in one table.
The chapter deliberately stops at the dictionary itself. The general change-of-basis formulas $A' = Q^{-1} A P$ and $A' = P^{-1} A P$, the named equivalence relations « matrices équivalentes » and « matrices semblables », the trace of an endomorphism (which requires basis-independence and hence the change-of-basis formula), and the classification language $A \sim J_r$ are deferred to the next chapter Change of bases. Diagonalisation, trigonalisation, and any reduction theory are out of program. Throughout the chapter, $\mathbb{K} \in \{\mathbb{R}, \mathbb{C}\}$, all vector spaces are $\mathbb{K}$-vector spaces of finite dimension, $n, p, q$ denote positive integers, and $r$ denotes a non-negative integer, usually a rank, with $0 \le r \le \min(n, p)$ when relevant.

I Matrix of a linear map

The starting move is the encoding of a single vector by its column of coordinates in a chosen basis. Extending this to a family of vectors gives a matrix one column at a time. Applying the same idea to the images $u(e_1), \dots, u(e_p)$ of the basis vectors of a linear map $u$ produces the matrix of $u$ in two bases. The numerical content of the chapter is then condensed in one identity --- multiplying the matrix of $u$ by the column of $x$ produces the column of $u(x)$ --- so the action of $u$ is read directly off its matrix. The dependence on the chosen bases is essential: the same map admits different matrices according to the chosen source and target bases.

Definition — Matrix of a vector in a basis

Let $E$ be a $\mathbb{K}$-vector space of finite dimension $p$, equipped with a basis $\mathcal{B} = (e_1, \dots, e_p)$. Every vector $x \in E$ decomposes uniquely as $$ x = x_1 e_1 + \dots + x_p e_p, \qquad (x_1, \dots, x_p) \in \mathbb{K}^p. $$ The matrix of the vector $x$ in the basis $\mathcal{B}$ is the column matrix of its coordinates: $$ \mathrm{Mat}_{\mathcal{B}}(x) = \begin{pmatrix} x_1 \\ \vdots \\ x_p \end{pmatrix} \in M_{p, 1}(\mathbb{K}). $$

Example

Take $E = \mathbb{R}_n[X]$ and the canonical basis $\mathcal{B} = (1, X, X^2, \dots, X^n)$. The polynomial $P = 1 - X^2$ decomposes as $P = 1 \cdot 1 + 0 \cdot X + (-1) \cdot X^2 + 0 \cdot X^3 + \dots + 0 \cdot X^n$, so $$ \mathrm{Mat}_{\mathcal{B}}(P) = \begin{pmatrix} 1 \\ 0 \\ -1 \\ 0 \\ \vdots \\ 0 \end{pmatrix} \in M_{n+1, 1}(\mathbb{R}). $$

Example

The same vector has different columns in different bases. In $E = \mathbb{R}^2$, take $x = (3, 5)$, the canonical basis $\mathcal{B}_c = ((1, 0), (0, 1))$, and the basis $\mathcal{B}' = (e'_1, e'_2)$ with $e'_1 = (1, 1)$ and $e'_2 = (1, -1)$. In $\mathcal{B}_c$ the column is read directly: $$ \mathrm{Mat}_{\mathcal{B}_c}(x) = \begin{pmatrix} 3 \\ 5 \end{pmatrix}. $$ In $\mathcal{B}'$, the coordinates $(a, b)$ are determined by $x = a e'_1 + b e'_2$, that is $(3, 5) = (a + b, a - b)$, giving $a = 4$ and $b = -1$: $$ \mathrm{Mat}_{\mathcal{B}'}(x) = \begin{pmatrix} 4 \\ -1 \end{pmatrix}. $$

Definition — Matrix of a family of vectors

Let $E$ be a $\mathbb{K}$-vector space of finite dimension $p$ with basis $\mathcal{B} = (e_1, \dots, e_p)$, and let $\mathcal{F} = (x_1, \dots, x_q)$ be a family of $q$ vectors of $E$. Each vector $x_j$ admits unique coordinates $x_j = a_{1j} e_1 + \dots + a_{pj} e_p$ in $\mathcal{B}$. The matrix of the family $\mathcal{F}$ in the basis $\mathcal{B}$ is the matrix whose $j$-th column is the column of $x_j$: $$ \mathrm{Mat}_{\mathcal{B}}(\mathcal{F}) = \begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1q} \\ a_{21} & a_{22} & \cdots & a_{2q} \\ \vdots & \vdots & \ddots & \vdots \\ a_{p1} & a_{p2} & \cdots & a_{pq} \end{pmatrix} \in M_{p, q}(\mathbb{K}). $$

The labelled matrix below shows the convention: each column $j$ is the column of $x_j$ in $\mathcal{B}$.

Mnemonic. Columns = vectors of the family $\mathcal{F}$. Rows = coordinates in the basis $\mathcal{B}$. The size is $p \times q$: $p$ rows because the basis $\mathcal{B}$ has $p$ vectors, $q$ columns because the family $\mathcal{F}$ has $q$ vectors.

Example

Take the canonical basis $\mathcal{B} = (1, X, X^2, \dots, X^n)$ of $\mathbb{R}_n[X]$ and the family $\mathcal{F} = (P_0, P_1, \dots, P_n)$ where $P_i = (X + 1)^i$. The binomial expansion gives $P_i = \sum_{k = 0}^i \binom{i}{k} X^k$, so the column of $P_i$ stores the binomial coefficients $\binom{i}{0}, \binom{i}{1}, \dots, \binom{i}{i}, 0, \dots, 0$: $$ \mathrm{Mat}_{\mathcal{B}}(\mathcal{F}) = \begin{pmatrix} 1 & 1 & 1 & \cdots & 1 \\ 0 & 1 & 2 & \cdots & \binom{n}{1} \\ 0 & 0 & 1 & \cdots & \binom{n}{2} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & 1 \end{pmatrix}. $$ It is an upper triangular matrix with diagonal coefficients all equal to $1$.

Definition — Matrix of a linear map in given bases

Let $E$ and $F$ be two $\mathbb{K}$-vector spaces of finite dimensions $p$ and $n$, equipped respectively with bases $\mathcal{B} = (e_1, \dots, e_p)$ and $\mathcal{B}' = (f_1, \dots, f_n)$. Let $u \colon E \to F$ be a linear map. For each $j \in \llbracket 1, p \rrbracket$, the vector $u(e_j)$ decomposes uniquely on $\mathcal{B}'$: $$ u(e_j) = a_{1j} f_1 + \dots + a_{nj} f_n. $$ The matrix of $u$ in the bases $\mathcal{B}$ and $\mathcal{B}'$ is the matrix of $M_{n, p}(\mathbb{K})$ whose $j$-th column is the column of $u(e_j)$ in $\mathcal{B}'$: $$ \mathrm{Mat}_{\mathcal{B}, \mathcal{B}'}(u) = \mathrm{Mat}_{\mathcal{B}'}(u(e_1), \dots, u(e_p)) = \begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1p} \\ a_{21} & a_{22} & \cdots & a_{2p} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{np} \end{pmatrix}. $$ The columns are indexed by the source basis $\mathcal{B}$, the rows by the target basis $\mathcal{B}'$.
Endomorphism notation. When $E = F$ and $\mathcal{B} = \mathcal{B}'$ (a single basis is used for both source and target), the notation collapses to $\mathrm{Mat}_{\mathcal{B}}(u) \in M_n(\mathbb{K})$.

The labelled matrix below shows the convention: each column $j$ is the column of $u(e_j)$ in $\mathcal{B}'$.

Mnemonic. Columns = images of source basis vectors. Rows = coordinates in the target basis. The size is $n \times p$: $n$ rows because the target has dimension $n$, $p$ columns because the source has dimension $p$.

Proposition — Size of $\mathrm{Mat}_{\mathcal{B}\virgule \mathcal{B}'}(u)$

If $E$ and $F$ have dimensions $p$ and $n$ respectively, then $\mathrm{Mat}_{\mathcal{B}, \mathcal{B}'}(u) \in M_{n, p}(\mathbb{K})$ for any $u \in \mathcal{L}(E, F)$.

Example

The derivation $\Function{\Phi}{\mathbb{R}_n[X]}{\mathbb{R}_n[X]}{P}{P'}$ is an endomorphism. Take the canonical basis $\mathcal{B} = (1, X, X^2, \dots, X^n)$. We have $\Phi(1) = 0$, and for each $i \in \llbracket 1, n \rrbracket$, $\Phi(X^i) = i X^{i-1}$, so the column of $\Phi(X^i)$ in $\mathcal{B}$ has a single nonzero coefficient $i$ at row $i - 1$ if rows are counted from $0$. The matrix is the upper-shifted diagonal: $$ \mathrm{Mat}_{\mathcal{B}}(\Phi) = \begin{pmatrix} 0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 2 & \cdots & 0 \\ \vdots & & \ddots & \ddots & \vdots \\ 0 & & & 0 & n \\ 0 & 0 & \cdots & 0 & 0 \end{pmatrix} \in M_{n+1}(\mathbb{R}). $$

Example

The same linear map admits different matrices in different basis pairs. Consider $u \colon \mathbb{R}^2 \to \mathbb{R}^2$, $u(x, y) = (x + y, x - y)$.
In the canonical bases $\mathcal{B}_c = ((1, 0), (0, 1))$ at both source and target, $u(1, 0) = (1, 1)$ and $u(0, 1) = (1, -1)$, so $$ \mathrm{Mat}_{\mathcal{B}_c, \mathcal{B}_c}(u) = \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}. $$ Now keep the canonical basis at the source but switch the target basis to $\mathcal{B}' = ((1, 1), (1, -1))$. We need to express $u(1, 0) = (1, 1)$ and $u(0, 1) = (1, -1)$ in $\mathcal{B}'$. We read directly $(1, 1) = 1 \cdot (1, 1) + 0 \cdot (1, -1)$ and $(1, -1) = 0 \cdot (1, 1) + 1 \cdot (1, -1)$, so $$ \mathrm{Mat}_{\mathcal{B}_c, \mathcal{B}'}(u) = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I_2. $$ The two matrices encode the same linear map, viewed through different glasses.

Theorem — Matrix-vector formula

Let $E$ and $F$ be finite-dimensional $\mathbb{K}$-vector spaces with bases $\mathcal{B}$ and $\mathcal{B}'$, and let $u \in \mathcal{L}(E, F)$. For every $x \in E$, $$ \mathrm{Mat}_{\mathcal{B}'}(u(x)) = \mathrm{Mat}_{\mathcal{B}, \mathcal{B}'}(u) \cdot \mathrm{Mat}_{\mathcal{B}}(x). $$ Equivalently, writing $A = \mathrm{Mat}_{\mathcal{B}, \mathcal{B}'}(u)$, $X = \mathrm{Mat}_{\mathcal{B}}(x)$, and $Y = \mathrm{Mat}_{\mathcal{B}'}(u(x))$: $$ Y = A X. $$

Proof

Set $\mathcal{B} = (e_1, \dots, e_p)$, $\mathcal{B}' = (f_1, \dots, f_n)$ and $A = (a_{ij})$, that is $u(e_j) = \sum_{i=1}^n a_{ij} f_i$ for each $j$. Write $x = \sum_{j=1}^p x_j e_j$. Using the linearity of $u$: $$ \begin{aligned} u(x) &= u\Bigl(\sum_{j=1}^p x_j e_j\Bigr) && \text{(decomposition of $x$ in $\mathcal{B}$)} \\ &= \sum_{j=1}^p x_j u(e_j) && \text{(linearity of $u$)} \\ &= \sum_{j=1}^p x_j \Bigl(\sum_{i=1}^n a_{ij} f_i\Bigr) && \text{(decomposition of $u(e_j)$ in $\mathcal{B}'$)} \\ &= \sum_{i=1}^n \Bigl(\sum_{j=1}^p a_{ij} x_j\Bigr) f_i && \text{(exchange of finite sums)}. \end{aligned} $$ Hence the $i$-th coordinate of $u(x)$ in $\mathcal{B}'$ is $\sum_{j=1}^p a_{ij} x_j$, which is precisely the $i$-th coefficient of the matrix product $A X$: $$ \mathrm{Mat}_{\mathcal{B}'}(u(x)) = \begin{pmatrix} \sum_{j=1}^p a_{1j} x_j \\ \vdots \\ \sum_{j=1}^p a_{nj} x_j \end{pmatrix} = A X. $$

Example

Apply the theorem to the derivation $\Phi \colon \mathbb{R}_n[X] \to \mathbb{R}_n[X]$ and $P = 1 - X^2$. We have $\mathrm{Mat}_{\mathcal{B}}(P) = (1, 0, -1, 0, \dots, 0)^\top$ and $\mathrm{Mat}_{\mathcal{B}}(\Phi)$ is the matrix computed earlier. Then $$ \mathrm{Mat}_{\mathcal{B}}(\Phi(P)) = \mathrm{Mat}_{\mathcal{B}}(\Phi) \cdot \mathrm{Mat}_{\mathcal{B}}(P) = \begin{pmatrix} 0 & 1 & 0 & \cdots \\ 0 & 0 & 2 & \cdots \\ \vdots & & & \ddots \end{pmatrix} \begin{pmatrix} 1 \\ 0 \\ -1 \\ 0 \\ \vdots \end{pmatrix} = \begin{pmatrix} 0 \\ -2 \\ 0 \\ 0 \\ \vdots \end{pmatrix}, $$ which is the column of the polynomial $-2 X$, confirming $\Phi(P) = P' = -2 X$.

Method — Compute the matrix of a linear map

To compute $\mathrm{Mat}_{\mathcal{B}, \mathcal{B}'}(u)$ where $\mathcal{B} = (e_1, \dots, e_p)$ and $\mathcal{B}' = (f_1, \dots, f_n)$:

For each source basis vector $e_j$, compute the image $u(e_j) \in F$.
Decompose $u(e_j)$ uniquely on the target basis $\mathcal{B}'$: $u(e_j) = a_{1j} f_1 + \dots + a_{nj} f_n$.
Stack the columns side by side: the $j$-th column is $(a_{1j}, \dots, a_{nj})^\top$.

Decomposing on $\mathcal{B}'$, not on $\mathcal{B}$. The coordinates that go into the matrix are those of $u(e_j)$ in the target basis, never in the source basis.

Skills to practice

Computing the matrix of a linear map
Computing in non-canonical bases

II The dictionary $\mathcal{L}(E\virgule F)$ and $M_{n\virgule p}(\mathbb{K})$

The matrix encoding from the previous section turns out to be a bijection: every matrix corresponds to a unique linear map, and every linear map to a unique matrix. The correspondence is more than a bijection --- it respects the algebraic operations. Sum of maps becomes sum of matrices, composition of maps becomes the matrix product, and a map is invertible (an isomorphism) if and only if its matrix is invertible. The chapter reaches its central message: linear-algebra calculation in finite dimension reduces entirely to matrix calculation.

Definition — Linear map canonically associated to a matrix

Let $A \in M_{n, p}(\mathbb{K})$. The linear map canonically associated to $A$ is the map $$ \Function{u_A}{\mathbb{K}^p}{\mathbb{K}^n}{X}{AX} $$ where vectors of $\mathbb{K}^p$ and $\mathbb{K}^n$ are identified with column matrices of $M_{p, 1}(\mathbb{K})$ and $M_{n, 1}(\mathbb{K})$. The map $u_A$ is linear (immediate from the bilinearity of the matrix product), and one checks that $\mathrm{Mat}_{\mathcal{B}_c, \mathcal{B}'_c}(u_A) = A$ where $\mathcal{B}_c$ and $\mathcal{B}'_c$ are the canonical bases.

Example

The matrix of rotation by angle $\theta$ in the plane, $$ R = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix} \in M_2(\mathbb{R}), $$ is canonically associated to the rotation of $\mathbb{R}^2$: $$ \Function{u_R}{\mathbb{R}^2}{\mathbb{R}^2}{(x, y)}{(x \cos\theta - y \sin\theta, \; x \sin\theta + y \cos\theta)}. $$

Example

The shear matrix $$ S = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} $$ is canonically associated to $u_S \colon (x, y) \mapsto (x + y, y)$ on $\mathbb{R}^2$.

Proposition — Reading the image of basis vectors from columns

Let $A \in M_{n, p}(\mathbb{K})$ and let $u_A$ be its canonically associated linear map. For each $j \in \llbracket 1, p \rrbracket$, the image of the $j$-th canonical basis vector $e_j$ of $\mathbb{K}^p$ is the $j$-th column of $A$: $$ u_A(e_j) = \text{$j$-th column of } A. $$

Theorem — Dictionary isomorphism

Let $E$ and $F$ be finite-dimensional $\mathbb{K}$-vector spaces of dimensions $p$ and $n$, equipped with bases $\mathcal{B}$ and $\mathcal{B}'$. The map $$ \Function{\Phi_{\mathcal{B}, \mathcal{B}'}}{\mathcal{L}(E, F)}{M_{n, p}(\mathbb{K})}{u}{\mathrm{Mat}_{\mathcal{B}, \mathcal{B}'}(u)} $$ is an isomorphism of $\mathbb{K}$-vector spaces.

Proof

Set $\mathcal{B} = (e_1, \dots, e_p)$, $\mathcal{B}' = (f_1, \dots, f_n)$.

Linearity. Let $u, v \in \mathcal{L}(E, F)$ and $\lambda \in \mathbb{K}$. For each $j$, $(\lambda u + v)(e_j) = \lambda u(e_j) + v(e_j)$; decomposing on $\mathcal{B}'$ shows that the $j$-th column of $\mathrm{Mat}_{\mathcal{B}, \mathcal{B}'}(\lambda u + v)$ is $\lambda \cdot (\text{column $j$ of } \mathrm{Mat}_{\mathcal{B}, \mathcal{B}'}(u)) + \text{column $j$ of } \mathrm{Mat}_{\mathcal{B}, \mathcal{B}'}(v)$. Hence $\Phi_{\mathcal{B}, \mathcal{B}'}(\lambda u + v) = \lambda \Phi_{\mathcal{B}, \mathcal{B}'}(u) + \Phi_{\mathcal{B}, \mathcal{B}'}(v)$.
Injectivity. If $\Phi_{\mathcal{B}, \mathcal{B}'}(u) = 0$, then each $u(e_j) = 0$. Since the $e_j$ form a basis, every $x \in E$ is a linear combination of the $e_j$, hence $u(x) = 0$ by linearity. So $u = 0$.
Surjectivity. Let $A = (a_{ij}) \in M_{n, p}(\mathbb{K})$. Define $u \colon E \to F$ by setting $u(e_j) = \sum_{i=1}^n a_{ij} f_i$ for each $j$ and extending linearly --- this is well defined because $(e_j)$ is a basis. By construction $\mathrm{Mat}_{\mathcal{B}, \mathcal{B}'}(u) = A$, hence $\Phi_{\mathcal{B}, \mathcal{B}'}$ is surjective.

Corollary — Dimension of $\mathcal{L}(E\virgule F)$

If $E$ and $F$ are finite-dimensional, then $$ \dim \mathcal{L}(E, F) = \dim(E) \times \dim(F). $$ Note. This formula was already established in the chapter Linear maps. The dictionary isomorphism gives a second, more concrete proof: $\mathcal{L}(E, F) \cong M_{n, p}(\mathbb{K})$ and $\dim M_{n, p}(\mathbb{K}) = n p$.

Example

$\dim \mathcal{L}(\mathbb{R}^2, \mathbb{R}^3) = 2 \times 3 = 6$. A linear map $\mathbb{R}^2 \to \mathbb{R}^3$ is described by $6$ scalars (the coefficients of its $3 \times 2$ matrix in canonical bases).

Theorem — Composition becomes matrix product

Let $E$, $F$, $G$ be three finite-dimensional $\mathbb{K}$-vector spaces with bases $\mathcal{B}$, $\mathcal{B}'$, $\mathcal{B}''$. Let $u \in \mathcal{L}(E, F)$ and $v \in \mathcal{L}(F, G)$. Then $$ \mathrm{Mat}_{\mathcal{B}, \mathcal{B}''}(v \circ u) = \mathrm{Mat}_{\mathcal{B}', \mathcal{B}''}(v) \cdot \mathrm{Mat}_{\mathcal{B}, \mathcal{B}'}(u). $$

Proof

Set $\mathcal{B} = (e_1, \dots, e_p)$. The columns of both matrices are read on the source basis vectors $e_j$.
For each $j$, the $j$-th column of $\mathrm{Mat}_{\mathcal{B}, \mathcal{B}''}(v \circ u)$ is the column of $(v \circ u)(e_j)$ in $\mathcal{B}''$, that is $\mathrm{Mat}_{\mathcal{B}''}(v(u(e_j)))$.
By the matrix-vector formula applied to $v$ and to the vector $u(e_j) \in F$: $$ \mathrm{Mat}_{\mathcal{B}''}(v(u(e_j))) = \mathrm{Mat}_{\mathcal{B}', \mathcal{B}''}(v) \cdot \mathrm{Mat}_{\mathcal{B}'}(u(e_j)). $$ The right-hand side is the product of $\mathrm{Mat}_{\mathcal{B}', \mathcal{B}''}(v)$ by the $j$-th column of $\mathrm{Mat}_{\mathcal{B}, \mathcal{B}'}(u)$, which is precisely the $j$-th column of $\mathrm{Mat}_{\mathcal{B}', \mathcal{B}''}(v) \cdot \mathrm{Mat}_{\mathcal{B}, \mathcal{B}'}(u)$.
Since both matrices have the same columns, they are equal.

Corollary — Endomorphism ring isomorphism

Let $E$ be a $\mathbb{K}$-vector space of finite dimension $n$ and let $\mathcal{B}$ be a basis of $E$. The map $$ \Function{\Phi_{\mathcal{B}}}{\mathcal{L}(E)}{M_n(\mathbb{K})}{u}{\mathrm{Mat}_{\mathcal{B}}(u)} $$ is a ring isomorphism: it is a bijection that respects sum, composition (becoming matrix product), and sends $\mathrm{Id}_E$ to $I_n$. The ring iso depends on the choice of basis $\mathcal{B}$: a different basis produces a different iso. This basis-dependence is precisely what motivates the next chapter Change of bases.

Method — Recover $u(x)$ via the matrix product

To compute the image of a vector $x \in E$ under a linear map $u \in \mathcal{L}(E, F)$ when matrices are at hand:

Pick bases $\mathcal{B}$ of $E$ and $\mathcal{B}'$ of $F$.
Write the matrix $A = \mathrm{Mat}_{\mathcal{B}, \mathcal{B}'}(u)$ and the column $X = \mathrm{Mat}_{\mathcal{B}}(x)$.
Compute $Y = A X$.
The image $u(x)$ is the vector of $F$ whose coordinates in $\mathcal{B}'$ are the entries of $Y$.

Theorem — Isomorphism $\Leftrightarrow$ invertible matrix

Let $E$ be a $\mathbb{K}$-vector space of finite dimension $n$, $\mathcal{B}$ a basis of $E$, and $u \in \mathcal{L}(E)$. Then $$ u \in \mathrm{GL}(E) \iff \mathrm{Mat}_{\mathcal{B}}(u) \in \mathrm{GL}_n(\mathbb{K}), $$ and in this case $$ \mathrm{Mat}_{\mathcal{B}}(u^{-1}) = (\mathrm{Mat}_{\mathcal{B}}(u))^{-1}. $$ Variant. More generally, if $\dim E = \dim F = n$ and $\mathcal{B}, \mathcal{B}'$ are bases of $E, F$, then $u \colon E \to F$ is an isomorphism if and only if $\mathrm{Mat}_{\mathcal{B}, \mathcal{B}'}(u) \in \mathrm{GL}_n(\mathbb{K})$.

Proof

Direct consequence of the ring isomorphism $\Phi_{\mathcal{B}} \colon \mathcal{L}(E) \to M_n(\mathbb{K})$. A ring isomorphism preserves invertible elements and inverses: an element is invertible in the source ring if and only if its image is invertible in the target ring, and the inverse is mapped to the inverse. Applied to $u$ and $\mathrm{Mat}_{\mathcal{B}}(u)$, this gives both equivalences.

Example

The derivation $\Phi \colon \mathbb{R}_n[X] \to \mathbb{R}_n[X]$ is not an isomorphism for $n \ge 1$. Its matrix in the canonical basis is nilpotent (a strictly upper triangular matrix); its kernel is $\mathbb{R}$ (the constants), of dimension $1$, hence $\Phi$ is not injective.

Example

Show that the linear map $u \colon \mathbb{R}^2 \to \mathbb{R}^2$, $u(x, y) = (x + y, x - y)$ is an automorphism and compute $u^{-1}$.

Answer

Injectivity via the kernel. Suppose $u(x, y) = (0, 0)$, that is $x + y = 0$ and $x - y = 0$. Adding the two equations gives $2 x = 0$, so $x = 0$, and then $y = 0$. Hence $\mathrm{Ker}\, u = \{0\}$.
From injectivity to bijectivity. Since $u$ is an endomorphism of the finite-dimensional space $\mathbb{R}^2$ and $u$ is injective, the rank theorem from the previous chapter gives $\dim \mathrm{Im}\, u = 2 - 0 = 2$, so $\mathrm{Im}\, u = \mathbb{R}^2$ and $u$ is surjective. Hence $u$ is an automorphism. No determinant computation is needed.
Inverse. Solve $u(x, y) = (X, Y)$, that is $x + y = X$ and $x - y = Y$. Adding gives $x = \frac{X + Y}{2}$, subtracting gives $y = \frac{X - Y}{2}$. So $$ u^{-1}(X, Y) = \Bigl(\frac{X + Y}{2}, \; \frac{X - Y}{2}\Bigr), \qquad \mathrm{Mat}_{\mathcal{B}_c}(u^{-1}) = \frac{1}{2} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}. $$ One checks $\mathrm{Mat}_{\mathcal{B}_c}(u) \cdot \mathrm{Mat}_{\mathcal{B}_c}(u^{-1}) = I_2$ directly.

Method — Show $u$ is an isomorphism and compute $u^{-1}$

Two complementary routes are available.
Vector route. Show $\mathrm{Ker}\, u = \{0\}$ by solving $u(x) = 0$. In a finite-dimensional endomorphism, injectivity implies bijectivity (rank theorem). Then compute $u^{-1}$ by solving $u(x) = y$ for arbitrary $y$.
Matrix route. Write the matrix $A = \mathrm{Mat}_{\mathcal{B}}(u)$. Apply the inversion algorithm of chapter Matrix calculus (Gauss-Jordan on $[A \,|\, I_n]$ to reach $[I_n \,|\, A^{-1}]$). The matrix is invertible if and only if $u$ is, by the equivalence above.
The two routes are equivalent; pick the one with lower computational cost on the case at hand.

Skills to practice

Showing isomorphisms via the matrix
Finding bases where the matrix has a prescribed form

III Rank of a matrix

Each matrix $A \in M_{n, p}(\mathbb{K})$ corresponds, via the canonical association, to a linear map $u_A \colon \mathbb{K}^p \to \mathbb{K}^n$. The notions of kernel, image and rank --- intrinsic to the linear map --- lift directly to the matrix: $\mathrm{Ker}\, A$ is the kernel of $u_A$, $\mathrm{Im}\, A$ is the image of $u_A$ (spanned by the columns of $A$), and $\mathrm{rg}\, A$ is their common rank. The two key results of the section are the matrix version of the rank theorem ($\dim \mathrm{Ker}\, A + \mathrm{rg}\, A = p$) and a structure theorem: every matrix of rank $r$ admits a normal form $P J_{n, p, r} Q$ with $P, Q$ invertible. This normal form is the operational tool that gives a determinant-free proof that the row rank equals the column rank.

Definition — Rank$\virgule$ kernel and image of a matrix

Let $A \in M_{n, p}(\mathbb{K})$ and let $u_A \colon \mathbb{K}^p \to \mathbb{K}^n$ be its canonically associated linear map. Define $$ \mathrm{Ker}\, A = \mathrm{Ker}\, u_A = \{ X \in M_{p, 1}(\mathbb{K}) : A X = 0 \}, \qquad \mathrm{Im}\, A = \mathrm{Im}\, u_A = \{ A X : X \in M_{p, 1}(\mathbb{K}) \}. $$ The rank of $A$ is the dimension of its image: $$ \mathrm{rg}\, A = \dim \mathrm{Im}\, A = \mathrm{rg}\, u_A. $$

Proposition — Rank $\equal$ rank of the column family

For $A \in M_{n, p}(\mathbb{K})$ with columns $C_1, \dots, C_p \in \mathbb{K}^n$: $$ \mathrm{Im}\, A = \mathrm{Vect}(C_1, \dots, C_p), \qquad \mathrm{rg}\, A = \mathrm{rg}(C_1, \dots, C_p) = \dim \mathrm{Vect}(C_1, \dots, C_p). $$

Proof

The image of $u_A$ is spanned by the images of the canonical basis vectors: $\mathrm{Im}\, u_A = \mathrm{Vect}(u_A(e_1), \dots, u_A(e_p))$. By the proposition « Reading the image of basis vectors from columns » above, $u_A(e_j) = C_j$. Hence $\mathrm{Im}\, A = \mathrm{Vect}(C_1, \dots, C_p)$, and $\mathrm{rg}\, A$ is the dimension of that span, which is the rank of the family.

Example

The matrices $$ A = \begin{pmatrix} 1 & 2 \\ 2 & 4 \end{pmatrix}, \qquad B = I_n $$ have $\mathrm{rg}\, A = 1$ (the two columns are colinear) and $\mathrm{rg}\, B = n$ (the columns form a basis).

Proposition — Rank of $\mathrm{Mat}(u)$ equals rank of $u$

Let $u \in \mathcal{L}(E, F)$ with $E, F$ finite-dimensional and let $\mathcal{B}, \mathcal{B}'$ be bases of $E, F$. Then $$ \mathrm{rg}(\mathrm{Mat}_{\mathcal{B}, \mathcal{B}'}(u)) = \mathrm{rg}(u). $$

Proof

The columns of $A = \mathrm{Mat}_{\mathcal{B}, \mathcal{B}'}(u)$ are the columns of $u(e_1), \dots, u(e_p)$ in $\mathcal{B}'$. Writing $\mathcal{B}' = (f_1, \dots, f_n)$, the basis-coordinate isomorphism $F \to \mathbb{K}^n$, $y \mapsto \mathrm{Mat}_{\mathcal{B}'}(y)$ sends $\mathrm{Im}\, u = \mathrm{Vect}(u(e_1), \dots, u(e_p))$ bijectively to $\mathrm{Vect}(C_1, \dots, C_p) = \mathrm{Im}\, A$. An isomorphism preserves dimension, hence $\mathrm{rg}\, A = \mathrm{rg}\, u$.

Theorem — Rank theorem$\virgule$ matrix version

Let $A \in M_{n, p}(\mathbb{K})$. Then $$ \dim \mathrm{Ker}\, A + \mathrm{rg}\, A = p. $$

Proof

Direct from the rank theorem of the chapter Linear maps applied to $u_A \colon \mathbb{K}^p \to \mathbb{K}^n$: $\dim \mathrm{Ker}\, u_A + \dim \mathrm{Im}\, u_A = \dim(\mathbb{K}^p) = p$. By definition of $\mathrm{Ker}\, A$ and $\mathrm{rg}\, A$, this is $\dim \mathrm{Ker}\, A + \mathrm{rg}\, A = p$.

Theorem — Rank normal form

Let $A \in M_{n, p}(\mathbb{K})$ of rank $r$. Define the rank normal form $$ J_{n, p, r} = \begin{pmatrix} I_r & 0 \\ 0 & 0 \end{pmatrix} \in M_{n, p}(\mathbb{K}), $$ where the top-left block is the identity $I_r$ and the other three blocks are zero blocks of suitable sizes. Then there exist $P \in \mathrm{GL}_n(\mathbb{K})$ and $Q \in \mathrm{GL}_p(\mathbb{K})$ such that $$ A = P \, J_{n, p, r} \, Q. $$ This theorem is presented here as an operational tool for computing ranks --- not yet as a classification statement under matrix equivalence. The named equivalence relation « $A$ and $B$ are equivalent » and the interpretation of $P, Q$ as change-of-basis matrices belong to the next chapter Change of bases. This theorem is structurally important, but in routine computations one usually computes the rank by Gauss elimination.

Proof

Let $u_A \colon \mathbb{K}^p \to \mathbb{K}^n$ be the canonically associated linear map; by hypothesis $\mathrm{rg}\, u_A = r$.
Choice of source basis. Pick a basis $(g_{r+1}, \dots, g_p)$ of $\mathrm{Ker}\, u_A$ (dimension $p - r$ by the rank theorem). Complete it into a basis $\mathcal{B}_E = (g_1, \dots, g_r, g_{r+1}, \dots, g_p)$ of $\mathbb{K}^p$.
Choice of target basis. Set $h_i = u_A(g_i)$ for $i \in \llbracket 1, r \rrbracket$. We claim that $(h_1, \dots, h_r)$ is a basis of $\mathrm{Im}\, u_A$.
Free. If $\sum_{i=1}^r \lambda_i h_i = 0$, then $u_A\bigl(\sum_{i=1}^r \lambda_i g_i\bigr) = 0$, so $\sum_{i=1}^r \lambda_i g_i \in \mathrm{Ker}\, u_A = \mathrm{Vect}(g_{r+1}, \dots, g_p)$. Since $(g_1, \dots, g_p)$ is a basis, all $\lambda_i$ are zero.
Spanning. For any $x = \sum_{i=1}^p x_i g_i$, since $g_{r+1}, \dots, g_p \in \mathrm{Ker}\, u_A$, $u_A(x) = \sum_{i=1}^r x_i h_i$. So $\mathrm{Im}\, u_A \subset \mathrm{Vect}(h_1, \dots, h_r)$; the reverse inclusion is obvious.
Complete $(h_1, \dots, h_r)$ into a basis $\mathcal{B}_F = (h_1, \dots, h_r, h_{r+1}, \dots, h_n)$ of $\mathbb{K}^n$.
Conclusion via direct verification. Let $S \in \mathrm{GL}_p(\mathbb{K})$ be the matrix whose columns are the coordinates of $g_1, \dots, g_p$ in the canonical basis of $\mathbb{K}^p$, and let $P \in \mathrm{GL}_n(\mathbb{K})$ be the matrix whose columns are the coordinates of $h_1, \dots, h_n$ in the canonical basis of $\mathbb{K}^n$. Set $Q = S^{-1} \in \mathrm{GL}_p(\mathbb{K})$.
We verify directly $A = P \, J_{n, p, r} \, Q$. Let $X \in \mathbb{K}^p$ and set $Y = Q X = S^{-1} X$. Then $Y$ is the coordinate column of the vector $X$ in the basis $(g_1, \dots, g_p)$, so $X = \sum_{i=1}^p y_i g_i$. Since $g_{r+1}, \dots, g_p \in \mathrm{Ker}\, u_A$: $$ A X = u_A(X) = \sum_{i=1}^r y_i u_A(g_i) = \sum_{i=1}^r y_i h_i. $$ On the other hand, $J_{n, p, r} Y$ is the column $(y_1, \dots, y_r, 0, \dots, 0)^\top$, and multiplying by $P$ forms the same linear combination of the columns $h_1, \dots, h_n$: $$ P J_{n, p, r} Y = \sum_{i=1}^r y_i h_i. $$ Thus $A X = P J_{n, p, r} Q X$ for every $X \in \mathbb{K}^p$, hence $A = P J_{n, p, r} Q$.

Proposition — Invertible multiplication preserves rank

Let $A \in M_{n, p}(\mathbb{K})$, $P \in \mathrm{GL}_n(\mathbb{K})$, $Q \in \mathrm{GL}_p(\mathbb{K})$. Then $$ \mathrm{rg}(P A) = \mathrm{rg}\, A = \mathrm{rg}(A Q). $$

Proof

Left multiplication by $P$ invertible. Observe that $$ \mathrm{Im}(P A) = \{ P A X : X \in \mathbb{K}^p \} = P(\mathrm{Im}\, A). $$ Since $P$ is an isomorphism of $\mathbb{K}^n$, its restriction to $\mathrm{Im}\, A$ is an isomorphism from $\mathrm{Im}\, A$ onto $P(\mathrm{Im}\, A)$. Hence $$ \dim \mathrm{Im}(P A) = \dim \mathrm{Im}\, A, $$ so $\mathrm{rg}(P A) = \mathrm{rg}\, A$.
Right multiplication by $Q$ invertible. Since $Q$ is an automorphism of $\mathbb{K}^p$, $$ \mathrm{Im}(A Q) = \{ A Q X : X \in \mathbb{K}^p \} = \{ A Y : Y \in \mathbb{K}^p \} = \mathrm{Im}\, A. $$ Thus $\mathrm{rg}(A Q) = \mathrm{rg}\, A$.

Corollary — Row rank $\equal$ column rank

For every $A \in M_{n, p}(\mathbb{K})$, $$ \mathrm{rg}\, A = \mathrm{rg}\, A^\top. $$

Proof

By the rank normal form, write $A = P \, J_{n, p, r} \, Q$ with $P \in \mathrm{GL}_n(\mathbb{K})$ and $Q \in \mathrm{GL}_p(\mathbb{K})$, where $r = \mathrm{rg}\, A$. Transposing: $$ A^\top = Q^\top \, J_{n, p, r}^\top \, P^\top = Q^\top \, J_{p, n, r} \, P^\top, $$ since the transpose of $J_{n, p, r}$ (the $n \times p$ matrix with $I_r$ top-left) is $J_{p, n, r}$ (the $p \times n$ matrix with $I_r$ top-left). The matrices $P^\top \in \mathrm{GL}_n(\mathbb{K})$ and $Q^\top \in \mathrm{GL}_p(\mathbb{K})$ are invertible (the transpose of an invertible matrix is invertible). By the previous proposition, $$ \mathrm{rg}\, A^\top = \mathrm{rg}(J_{p, n, r}) = r = \mathrm{rg}\, A. $$

Theorem — Rank of a product

Let $A \in M_{n, m}(\mathbb{K})$ and $B \in M_{m, p}(\mathbb{K})$. Then $$ \mathrm{rg}(A B) \le \min(\mathrm{rg}\, A, \, \mathrm{rg}\, B). $$ Moreover:

If $A \in \mathrm{GL}_m(\mathbb{K})$ (square invertible, $n = m$), then $\mathrm{rg}(A B) = \mathrm{rg}\, B$.
If $B \in \mathrm{GL}_m(\mathbb{K})$ (square invertible, $m = p$), then $\mathrm{rg}(A B) = \mathrm{rg}\, A$.

Proof

$\mathrm{rg}(A B) \le \mathrm{rg}\, A$. Every column of $A B$ is a linear combination of columns of $A$ (since column $j$ of $A B$ is $A$ times column $j$ of $B$). Hence $\mathrm{Im}(A B) \subset \mathrm{Im}\, A$, so $\mathrm{rg}(A B) \le \mathrm{rg}\, A$.
$\mathrm{rg}(A B) \le \mathrm{rg}\, B$. Identify with linear maps: $u_{A B} = u_A \circ u_B$. Then $\mathrm{Im}(u_A \circ u_B) = u_A(\mathrm{Im}\, u_B)$, the image by $u_A$ of a subspace of dimension $\mathrm{rg}\, B$. The dimension of the image of a subspace by a linear map is at most the dimension of the subspace, so $\mathrm{rg}(A B) \le \mathrm{rg}\, B$.
Invertibility refinements. If $A$ invertible, the previous proposition gives $\mathrm{rg}(A B) = \mathrm{rg}\, B$. Symmetrically, if $B$ invertible, $\mathrm{rg}(A B) = \mathrm{rg}\, A$.

Method — Compute the rank of a matrix by Gauss elimination

The rank of a matrix can be computed by elementary row or column operations:

Apply elementary row operations to $A$ (swap rows, multiply a row by a nonzero scalar, add a multiple of one row to another) to bring $A$ into row-echelon form.
The rank of $A$ is the number of nonzero rows (equivalently, the number of pivots) in the echelon form.

Why this works. Each elementary row operation amounts to multiplying $A$ on the left by an elementary invertible matrix. The previous proposition says invertible left-multiplication preserves the rank, so the rank of the echelon form equals the rank of $A$. The same argument shows column operations also preserve rank.
By the corollary $\mathrm{rg}\, A = \mathrm{rg}\, A^\top$, computing the rank via column operations is equally valid.

Method — Determine $\mathrm{Ker}\ A$ and $\mathrm{Im}\ A$

For $A \in M_{n, p}(\mathbb{K})$:

Kernel. Solve the homogeneous system $A X = 0$ by Gauss elimination on the rows of $A$. The general solution describes $\mathrm{Ker}\, A$; a basis is read by parametrising the free variables.
Image. The image is the span of the original columns of $A$. A safe method is: row-reduce $A$ to echelon form, identify the pivot columns, then take the corresponding columns of the original matrix $A$. These original pivot columns form a basis of $\mathrm{Im}\, A$. Alternatively, one may use elementary column operations: since they amount to multiplying $A$ on the right by an invertible matrix, they preserve the column space. In that case, the nonzero columns of the column-reduced matrix form another basis of $\mathrm{Im}\, A$.
Consistency check. The matrix-form rank theorem $\dim \mathrm{Ker}\, A + \mathrm{rg}\, A = p$ acts as a sanity check between the two computations.

Example

Compute the kernel, image, and rank of $$ A = \begin{pmatrix} 1 & 2 & -1 \\ 2 & -1 & 8 \\ 1 & 1 & 1 \end{pmatrix} \in M_3(\mathbb{R}). $$

Answer

Kernel. Solve $A X = 0$. Reduce by row operations: $$ \begin{pmatrix} 1 & 2 & -1 \\ 2 & -1 & 8 \\ 1 & 1 & 1 \end{pmatrix} \;\xrightarrow{L_2 \leftarrow L_2 - 2 L_1, \; L_3 \leftarrow L_3 - L_1}\; \begin{pmatrix} 1 & 2 & -1 \\ 0 & -5 & 10 \\ 0 & -1 & 2 \end{pmatrix} \;\xrightarrow{L_2 \leftarrow L_2 - 5 L_3}\; \begin{pmatrix} 1 & 2 & -1 \\ 0 & 0 & 0 \\ 0 & -1 & 2 \end{pmatrix}. $$ After row swap, the echelon form has $2$ pivots, so $\mathrm{rg}\, A = 2$. The kernel is the line of solutions $x + 2 y - z = 0$, $-y + 2 z = 0$. Setting $z = t$: $y = 2 t$, $x = z - 2 y = t - 4 t = -3 t$. So $\mathrm{Ker}\, A = \mathbb{R} \cdot (-3, 2, 1)^\top$, of dimension $1$.
Image. $\mathrm{rg}\, A = 2$ means $\mathrm{Im}\, A$ is a plane in $\mathbb{R}^3$. The first two columns $C_1 = (1, 2, 1)^\top$ and $C_2 = (2, -1, 1)^\top$ are linearly independent: if $\alpha C_1 + \beta C_2 = 0$, the first two coordinates give $\alpha + 2 \beta = 0$ and $2 \alpha - \beta = 0$, whose only solution is $\alpha = \beta = 0$. They form a basis of $\mathrm{Im}\, A$: $\mathrm{Im}\, A = \mathrm{Vect}(C_1, C_2)$.
Consistency. $\dim \mathrm{Ker}\, A + \mathrm{rg}\, A = 1 + 2 = 3 = p$. $\checkmark$

Skills to practice

Computing rank$\virgule$ kernel and image by Gauss
Using the rank normal form
Estimating ranks

IV Block writing and adapted bases

A matrix can be split into rectangular blocks. The arithmetic on these blocks --- sum, product --- follows the same formal rules as scalar arithmetic, provided sizes are compatible. The block view becomes particularly powerful when matrices are read on bases adapted to a direct-sum decomposition of the underlying vector space, especially when one or both summands are stable: an endomorphism $u$ that stabilises a subspace $F$ takes a triangular block form in a basis built from $F$ and a complement, and a block-diagonal form when both $F$ and its complement are stable. Projectors and symmetries, defined in the previous chapter, are the prototypical examples.

Definition — Block writing of a matrix

Let $A \in M_{n, p}(\mathbb{K})$. Choose a partition $n = n_1 + \dots + n_s$ of the row indices and a partition $p = p_1 + \dots + p_t$ of the column indices. The matrix $A$ admits a block writing $$ A = \begin{pmatrix} A_{11} & A_{12} & \cdots & A_{1t} \\ A_{21} & A_{22} & \cdots & A_{2t} \\ \vdots & \vdots & \ddots & \vdots \\ A_{s1} & A_{s2} & \cdots & A_{st} \end{pmatrix}, $$ where the block $A_{kl} \in M_{n_k, p_l}(\mathbb{K})$ gathers the coefficients of $A$ at row indices in the $k$-th group and column indices in the $l$-th group.

Proposition — Block product

Let $A \in M_{n, p}(\mathbb{K})$ and $B \in M_{p, q}(\mathbb{K})$, with row/column partitions chosen so that the column partition of $A$ matches the row partition of $B$. Writing $$ A = (A_{kl}), \qquad B = (B_{lm}), $$ the product $A B$ admits a block writing whose blocks follow the formula $$ (A B)_{km} = \sum_l A_{kl} \, B_{lm}, $$ formally identical to the scalar product rule, with blocks playing the role of coefficients.

Example

Take the $4 \times 4$ block matrices, partitioned into $2 \times 2$ blocks: $$ A = \begin{pmatrix} 1 & 0 \,\big|\, 2 & 1 \\ 0 & 1 \,\big|\, 1 & 0 \\ \hline 0 & 0 \,\big|\, 1 & 0 \\ 0 & 0 \,\big|\, 0 & 1 \end{pmatrix} = \begin{pmatrix} I_2 & C \\ 0 & I_2 \end{pmatrix}, \qquad B = \begin{pmatrix} I_2 & D \\ 0 & I_2 \end{pmatrix}, $$ with $C = \begin{pmatrix} 2 & 1 \\ 1 & 0 \end{pmatrix}$ and $D \in M_2(\mathbb{R})$ arbitrary. The block product rule gives $$ A B = \begin{pmatrix} I_2 \cdot I_2 + C \cdot 0 & \;\; I_2 \cdot D + C \cdot I_2 \\ 0 \cdot I_2 + I_2 \cdot 0 & \;\; 0 \cdot D + I_2 \cdot I_2 \end{pmatrix} = \begin{pmatrix} I_2 & D + C \\ 0 & I_2 \end{pmatrix}. $$

Theorem — Matrix in a basis adapted to a stable direct sum

Let $E$ be a $\mathbb{K}$-vector space of dimension $n$, and let $E = F \oplus G$ with $\dim F = r$ and $\dim G = n - r$. Let $\mathcal{B}_F$ be a basis of $F$, $\mathcal{B}_G$ a basis of $G$, and $\mathcal{B} = (\mathcal{B}_F, \mathcal{B}_G)$ the basis of $E$ obtained by concatenation. Let $u \in \mathcal{L}(E)$.

If $F$ is stable under $u$ (that is, $u(F) \subset F$), then $$ \mathrm{Mat}_{\mathcal{B}}(u) = \begin{pmatrix} A & B \\ 0 & D \end{pmatrix} $$ with $A \in M_r(\mathbb{K})$, $D \in M_{n - r}(\mathbb{K})$, $B \in M_{r, n - r}(\mathbb{K})$. Here $A = \mathrm{Mat}_{\mathcal{B}_F}(u|_F)$ is the matrix of the restriction.
If both $F$ and $G$ are stable under $u$, then $B = 0$ and $$ \mathrm{Mat}_{\mathcal{B}}(u) = \begin{pmatrix} A & 0 \\ 0 & D \end{pmatrix} $$ (block-diagonal form).

Proof

Set $\mathcal{B}_F = (e_1, \dots, e_r)$, $\mathcal{B}_G = (e_{r+1}, \dots, e_n)$. Read the columns of $\mathrm{Mat}_{\mathcal{B}}(u)$.
First $r$ columns. For $j \in \llbracket 1, r \rrbracket$, the vector $e_j$ belongs to $F$, hence $u(e_j) \in F$ (since $F$ is stable). The decomposition $u(e_j) = a_{1j} e_1 + \dots + a_{rj} e_r + 0 \cdot e_{r+1} + \dots + 0 \cdot e_n$ has zeros at indices $i > r$. So the bottom-left block (rows $> r$, columns $\le r$) is zero. Its top part is precisely $\mathrm{Mat}_{\mathcal{B}_F}(u|_F)$.
Last $n - r$ columns. For $j \in \llbracket r + 1, n \rrbracket$, the vector $e_j$ belongs to $G$; without a stability hypothesis on $G$, the image $u(e_j)$ may have arbitrary components on both $\mathcal{B}_F$ (giving the block $B$) and $\mathcal{B}_G$ (giving the block $D$).
If additionally $G$ is stable under $u$, then $u(e_j) \in G$ for every $j > r$, so the decomposition has zeros at indices $i \le r$: the top-right block $B$ is zero.

The following diagram visualises the block decomposition of a matrix written in a basis adapted to $E = F \oplus G$ with both subspaces stable.

Reading aid. $A$ is the matrix of $u|_F$ in $\mathcal{B}_F$, $D$ is the matrix of $u|_G$ in $\mathcal{B}_G$.

Example

Let $p$ be a projector on $F$ parallel to $G$ (so $E = F \oplus G$, $p(x) = x$ for $x \in F$, $p(x) = 0$ for $x \in G$), with $\dim F = r$. In a basis $\mathcal{B} = (\mathcal{B}_F, \mathcal{B}_G)$ adapted to $F \oplus G$: $$ \mathrm{Mat}_{\mathcal{B}}(p) = \begin{pmatrix} I_r & 0 \\ 0 & 0 \end{pmatrix}. $$ The two stable subspaces are $F = \mathrm{Im}\, p$ and $G = \mathrm{Ker}\, p$.

Example

Let $s$ be a symmetry with respect to $F$ parallel to $G$ ($s(x) = x$ for $x \in F$, $s(x) = -x$ for $x \in G$), with $\dim F = r$, $\dim G = n - r$. In an adapted basis: $$ \mathrm{Mat}_{\mathcal{B}}(s) = \begin{pmatrix} I_r & 0 \\ 0 & -I_{n - r} \end{pmatrix}. $$

Example

The homothety $h_\lambda \colon x \mapsto \lambda x$ has matrix $\lambda I_n$ in any basis of $E$. Every line $\mathbb{K} \cdot x$ is stable under $h_\lambda$, so the decomposition $E = F \oplus G$ may be chosen freely.

Method — Read $\mathrm{Ker}\ u$ and $\mathrm{Im}\ u$ from a block form

Suppose $u$ has matrix $\mathrm{Mat}_{\mathcal{B}}(u) = \begin{pmatrix} A & 0 \\ 0 & 0 \end{pmatrix}$ with $A \in \mathrm{GL}_r(\mathbb{K})$ in a basis adapted to $E = F \oplus G$. Then:

$\mathrm{Im}\, u = F$ and $\dim \mathrm{Im}\, u = r = \mathrm{rg}\, u$.
$\mathrm{Ker}\, u = G$ and $\dim \mathrm{Ker}\, u = n - r$.

More generally, every direct sum decomposition $E = F \oplus G$ that block-diagonalises $u$ reveals the algebraic structure of $u$ directly.

Skills to practice

Block matrices and adapted bases

V Synthesis: the dictionary in one table

The chapter has built a one-to-one correspondence between linear-algebra objects (vectors, linear maps, kernel, image, rank) and matrix objects (column matrices, rectangular matrices, kernel of a matrix, image of a matrix, rank of a matrix), and has shown that this correspondence respects every algebraic operation. The two viewpoints are completely interchangeable in finite dimension. \renewcommand{\arraystretch}{1.3}

Vector-space language	Matrix language
$x \in E$	$X = \mathrm{Mat}_{\mathcal{B}}(x)$
$u \in \mathcal{L}(E, F)$	$A = \mathrm{Mat}_{\mathcal{B}, \mathcal{B}'}(u)$
$u(x)$	$A X$
$v \circ u$	$B A$
$u \in \mathrm{GL}(E)$	$A \in \mathrm{GL}_n(\mathbb{K})$
$\mathrm{rg}(u)$	$\mathrm{rg}(A)$
$\mathrm{Ker}\, u$	$\mathrm{Ker}\, A$
$\mathrm{Im}\, u$	column space of $A$

Method — Choose between linear-map formalism and matrix formalism

Both formalisms describe the same object; choose by the cost of the calculation at hand.

Matrix formalism is well suited when the data is explicit (numerical coefficients, polynomial basis) and a computation needs to be carried out: inverting, computing the image of a specific vector, finding a basis of the kernel by Gauss elimination.
Linear-map formalism is well suited when reasoning is abstract (statement about any basis, qualitative properties such as injectivity / surjectivity), or when the linear map has a natural definition that does not pass through coordinates (a projection, a derivation, an integration operator).

The dictionary lets one switch sides at will: prove the qualitative result on the linear-map side, then read it off the matrix side once a basis is chosen.

A question opens for the next chapter. Two bases of the same vector space produce two different matrices for the same linear map; how do these matrices relate? Quantifying that relationship is the topic of the chapter Change of bases, where the trace of an endomorphism --- a quantity that requires basis-independence --- will also be defined.

Skills to practice

True or false bank
Synthesis exercises

Vector-space language	Matrix language
\(x \in E\)	\(X = \mathrm{Mat}_{\mathcal{B}}(x)\)
\(u \in \mathcal{L}(E, F)\)	\(A = \mathrm{Mat}_{\mathcal{B}, \mathcal{B}'}(u)\)
\(u(x)\)	\(A X\)
\(v \circ u\)	\(B A\)
\(u \in \mathrm{GL}(E)\)	\(A \in \mathrm{GL}_n(\mathbb{K})\)
\(\mathrm{rg}(u)\)	\(\mathrm{rg}(A)\)
\(\mathrm{Ker}\, u\)	\(\mathrm{Ker}\, A\)
\(\mathrm{Im}\, u\)	column space of \(A\)