[ALGEBRA] Matrix representations

This page writes a linear map $T:V\to W$ between finite-dimensional spaces as a matrix. A basis gives each vector unique coordinates and a linear map is fixed by its values on a basis, so the matrix records the coordinate vectors of the images of the domain basis. The page then reads injectivity, surjectivity and invertibility of $T$ from its matrix.

The coordinate matrix

[DEFINITION]

Linear map

Let $V$ and $W$ be vector spaces over the same field $F$ . A function $T:V\to W$ is a linear map if

T(u+v)=T(u)+T(v)\qquad\text{for all }u,v\in V,

T(\omega u)=\omega T(u)\qquad\text{for all }\omega\in F,\ u\in V.

[DEFINITION]

Coordinate vector

Let $B=(u_1,\ldots,u_n)$ be an ordered basis for a vector space $V$ . If

u=x_1u_1+\cdots+x_nu_n,

then the coordinate vector of $u$ with respect to $B$ is the column

[u]_B= \begin{pmatrix} x_1\\ \vdots\\ x_n \end{pmatrix}.

[DEFINITION]

Basis

A set $B$ in a vector space $V$ is a basis for $V$ if it is linearly independent and spans $V$ .

[1]:Span
[2]:Linear independence
[3]:Eigenbasis
[4]:Diagonalisation
[5]:A system with zero initial derivative
[6]:Values on a basis determine a linear map
22 more

The matrix depends on the chosen bases, but the linear map $T$ does not. Changing a basis changes the coordinate columns used to write $T$ , not which vectors $T$ sends to which outputs.

Using the matrix takes three steps: write the input as a coordinate column in $B_V$ , multiply by $A$ and read the result as a coordinate column in $B_W$ .

[THEOREM]

Matrix representation of a linear map

Let $T:V\to W$ be a linear map, where $\dim(V)=n$ and $\dim(W)=m$ . For ordered bases

B_V=(v_1,\ldots,v_n),\qquad B_W=(w_1,\ldots,w_m),

form the coordinate columns

a_j=[T(v_j)]_{B_W}\qquad (j=1,\ldots,n).

Then

A=\begin{pmatrix}a_1&\cdots&a_n\end{pmatrix}

is the unique $m\times n$ matrix satisfying

[T(v)]_{B_W}=A[v]_{B_V}\qquad\text{for every }v\in V.

We call $A$ the matrix representation of $T$ relative to $B_V$ and $B_W$ and may write it as $[T]_{B_V}^{B_W}$ .

With this notation, the subscript names the domain basis and the superscript names the codomain basis, matching the direction $[v]_{B_V}\mapsto[T(v)]_{B_W}$ in the displayed coordinate identity.

The $j$ th column is the coordinate vector of $T(v_j)$ in the codomain basis. Thus the number of columns is the dimension of the domain and the number of entries in each column is the dimension of the codomain.

[1]:Linear map
[2]:Basis
[3]:Coordinate vector
[4]:Values on a basis determine a linear map
[5]:Diagonalisation
[6]:Diagonalisation
[7]:Characteristic polynomial
[8]:Eigenvalues and eigenvectors
10 more

[PROOF]

Write

v=c_1v_1+\cdots+c_nv_n.

Then

[v]_{B_V}= \begin{pmatrix} c_1\\ \vdots\\ c_n \end{pmatrix}.

Preservation of finite linear combinations gives

\begin{aligned} [T(v)]_{B_W} &=c_1[T(v_1)]_{B_W}+\cdots+c_n[T(v_n)]_{B_W}\\ &=c_1a_1+\cdots+c_na_n\\ &=A[v]_{B_V}. \end{aligned}

This proves the coordinate identity. If another matrix has the same identity, apply it to each $v_j$ . Since $[v_j]_{B_V}$ is the $j$ th standard coordinate vector, its $j$ th column must be $[T(v_j)]_{B_W}=a_j$ . Every column is therefore forced, so $A$ is unique.

[1]:Matrix representation of a linear map

The theorem turns the basis-data principle into a construction. The order of both bases matters: it fixes the order of the entries in each coordinate column and the order of the columns themselves.

Building and using the matrix

[METHOD]

Construct a matrix representation

For a linear map $T:V\to W$ :

choose an ordered basis $B_V=(v_1,\ldots,v_n)$ for $V$ and an ordered basis $B_W=(w_1,\ldots,w_m)$ for $W$ ;
evaluate $T(v_j)$ for each domain basis vector;
compute the codomain coordinate vector $[T(v_j)]_{B_W}$ for each image;
place these coordinate vectors, in the same order, as the columns of the $m\times n$ matrix.

[EXAMPLE]

Differentiation on polynomial spaces

For $n\ge 1$ , differentiation is the real-linear map

D:P_n(\mathbb R)\to P_{n-1}(\mathbb R),\qquad D(p)=p'.

Take $n=4$ and let

T:P_4(\mathbb R)\to P_3(\mathbb R),\qquad T(p)=\frac{dp(x)}{dx},

and use the ordered standard polynomial bases

B_V=(1,x,x^2,x^3,x^4),\qquad B_W=(1,x,x^2,x^3).

Evaluate $T$ on each domain basis vector and find the coordinates of the result in $B_W$ :

\begin{array}{c|c|c} \text{domain vector} & \text{image} & \text{codomain coordinate vector}\\ \hline 1 & T(1)=0 & \begin{pmatrix}0\\0\\0\\0\end{pmatrix}\\[6pt] x & T(x)=1 & \begin{pmatrix}1\\0\\0\\0\end{pmatrix}\\[6pt] x^2 & T(x^2)=2x & \begin{pmatrix}0\\2\\0\\0\end{pmatrix}\\[6pt] x^3 & T(x^3)=3x^2 & \begin{pmatrix}0\\0\\3\\0\end{pmatrix}\\[6pt] x^4 & T(x^4)=4x^3 & \begin{pmatrix}0\\0\\0\\4\end{pmatrix} \end{array}

These coordinate vectors, in order, are the columns of the matrix representation:

A=[T]_{B_V}^{B_W} = \begin{pmatrix} 0&1&0&0&0\\ 0&0&2&0&0\\ 0&0&0&3&0\\ 0&0&0&0&4 \end{pmatrix}.

For

p=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4,

[p]_{B_V}= \begin{pmatrix} a_0\\a_1\\a_2\\a_3\\a_4 \end{pmatrix}, \qquad A[p]_{B_V}= \begin{pmatrix} a_1\\2a_2\\3a_3\\4a_4 \end{pmatrix} =[T(p)]_{B_W}.

This agrees with

T(p)=a_1+2a_2x+3a_3x^2+4a_4x^3.

The zero first column records $T(1)=0$ , so differentiation is not injective. The four pivot columns give rank $4=\dim(P_3(\mathbb R))$ , so it is surjective. In words, differentiation loses the constant direction but still reaches every polynomial of degree at most $3$ .

[routine]

With standard bases, a vector in $\mathbb R^n$ equals its own coordinate column and the matrix representation is the matrix map from the geometric examples. For these maps, operations on maps match operations on matrices.

[PROPOSITION]

Operations on matrix representations

Let $T_A:\mathbb R^n\to\mathbb R^m$ and $T_B:\mathbb R^n\to\mathbb R^m$ be the standard-coordinate matrix maps

T_A(x)=Ax,\qquad T_B(x)=Bx.

Then

(T_A+T_B)(x)=(A+B)x,\qquad (\omega T_A)(x)=(\omega A)x.

If $T_A:\mathbb R^p\to\mathbb R^m$ and $T_B:\mathbb R^n\to\mathbb R^p$ , then

(T_A\circ T_B)(y)=A(By)=(AB)y.

Thus the matrix of a sum, scalar multiple, or composition is respectively the sum, scalar multiple, or product of the matrices, with the product ordered to match the composition.

[1]:Matrix maps are linear

For non-standard bases, the same rules hold when the matrices use the same domain and codomain bases for a sum and a matching intermediate basis for a composition. Otherwise the coordinate columns refer to different bases and cannot be combined directly.

Reading map properties in coordinates

The coordinate identity transfers questions about the kernel and image of $T$ to its matrix, where rank-nullity applies to coordinate columns. This section defines injective, surjective and inverse maps and reads each property from the matrix.

[DEFINITION]

Injective map

A map $T:V\to W$ is injective, or one-to-one, when

T(v_1)=T(v_2)\quad\Longrightarrow\quad v_1=v_2.

[COROLLARY]

Injectivity and the kernel

A linear map $T:V\to W$ is injective if and only if

\ker(T)=\{0\}.

If $V$ is finite-dimensional, this is equivalent to $\operatorname{nullity}(T)=0$ .

If $T(v_1)=T(v_2)$ , then $T(v_1-v_2)=0$ . A trivial kernel therefore gives $v_1=v_2$ . Conversely, injectivity forces any vector with $T(v)=T(0)$ to be $v=0$ .

[1]:Kernel
[2]:Injective map
[3]:Properties read from a matrix representation

[DEFINITION]

Surjective map

A map $T:V\to W$ is surjective, or onto, when every $w\in W$ equals $T(v)$ for some $v\in V$ . Equivalently, $\operatorname{im}(T)=W$ .

[DEFINITION]

Bijective map

A map is bijective when it is both injective and surjective.

[DEFINITION]

Identity map

The identity map on $V$ is

\operatorname{id}_V:V\to V,\qquad \operatorname{id}_V(v)=v.

[1]:Linear map

[DEFINITION]

Inverse map

For a map $T:V\to W$ , a map $S:W\to V$ is an inverse of $T$ when

S\circ T=\operatorname{id}_V, \qquad T\circ S=\operatorname{id}_W.

The first composition is the left-inverse condition and the second is the right-inverse condition. An inverse satisfies both.

Do not confuse the transpose with the inverse: $A^{\mathsf T}$ transposes rows and columns, while $A^{-1}$ reverses the map when a two-sided inverse exists. Taking reciprocals of individual entries does not generally produce $A^{-1}$ . For an orthogonal matrix $Q$ , the special identity $Q^{-1}=Q^{\mathsf T}$ follows from its orthonormal columns; see the symmetric-matrix diagonalisation.

[COROLLARY]

Properties read from a matrix representation

Let $A=[T]_{B_V}^{B_W}$ be the matrix representation of $T:V\to W$ over a scalar field $\mathbb F$ , with $\dim(V)=n$ and $\dim(W)=m$ . Then

\ker(A)=\{[v]_{B_V}:v\in\ker(T)\}

and

\operatorname{im}(A)=\{[w]_{B_W}:w\in\operatorname{im}(T)\}.

Since a linear map is injective exactly when its kernel is trivial and surjective exactly when its image is the codomain, these identities give the coordinate tests $\ker(A)=\{0\}$ and $\operatorname{im}(A)=\mathbb F^m$ . Here $\ker(A)$ and $\operatorname{im}(A)$ are sets of coordinate columns; the displayed equalities identify them with the corresponding abstract subspaces through the coordinate maps.

Consequently,

\operatorname{nullity}(A)=\operatorname{nullity}(T), \qquad \operatorname{rank}(A)=\operatorname{rank}(T).

Hence $T$ is injective if and only if $\operatorname{nullity}(A)=0$ and $T$ is surjective if and only if $\operatorname{rank}(A)=m$ . If $m=n$ , then $T$ is bijective if and only if $A$ is invertible.

[1]:Matrix representation of a linear map
[2]:Kernel
[3]:Image
[4]:Rank-nullity theorem
[5]:Rank-nullity theorem for vector spaces
[6]:Injectivity and the kernel
[7]:Bijectivity in equal dimensions
[8]:Inverse maps in coordinates

[EXAMPLE]

Integration gives an injective map

Define

I:P_n(\mathbb R)\to P_{n+1}(\mathbb R), \qquad I(p)(x)=\int_0^x p(t)\,dt.

The integral is linear in $p$ , so $I$ is a linear map. If $I(p)$ is the zero polynomial, differentiating the identity $I(p)(x)=0$ gives $p(x)=0$ for every $x$ , so $\ker(I)=\{0\}$ . Thus $I$ is injective. Its rank is $\dim(P_n(\mathbb R))=n+1$ , which is smaller than $\dim(P_{n+1}(\mathbb R))=n+2$ , so it is not surjective. Concretely, every polynomial $I(p)$ vanishes at $x=0$ , so the constant polynomial $1$ is not in the image.

[routine]

[1]:Injective map
[2]:Rank-nullity theorem for vector spaces

Invertibility in equal dimensions

When the domain and codomain have the same finite dimension, rank-nullity makes injectivity and surjectivity equivalent. A bijective linear map has an inverse and in coordinates that inverse is given by the inverse matrix.

[PROPOSITION]

Bijectivity in equal dimensions

For a linear map $T:V\to W$ between finite-dimensional vector spaces:

if $T$ is bijective, then $\dim(V)=\dim(W)$ ;
if $\dim(V)=\dim(W)$ and $T$ is injective, then $T$ is surjective;
if $\dim(V)=\dim(W)$ and $T$ is surjective, then $T$ is injective.

Indeed, if $T$ is bijective, then its kernel is trivial and its image is all of $W$ . Rank-nullity therefore gives $\dim(V)=\operatorname{rank}(T)=\dim(W)$ . If the dimensions are equal and $T$ is injective, then nullity is $0$ , so rank equals $\dim(V)=\dim(W)$ and $T$ is surjective. The surjective case is dual: rank equals $\dim(W)=\dim(V)$ , so nullity is $0$ and $T$ is injective.

[THEOREM]

Equivalent invertibility conditions

For a linear map $T:V\to W$ between finite-dimensional spaces, the following conditions are equivalent when $\dim(V)=\dim(W)$ :

$T$ is invertible;
$T$ is bijective;
$T$ has a left inverse;
$T$ has a right inverse.

A left inverse makes $T$ injective and a right inverse makes $T$ surjective. In equal dimensions either property makes $T$ bijective, so the one-sided inverse is then the two-sided inverse of $T$ .

[THEOREM]

Inverse maps in coordinates

Let $V$ and $W$ be finite-dimensional vector spaces, let $T:V\to W$ be bijective and fix ordered bases $B_V$ of $V$ and $B_W$ of $W$ . Then $T^{-1}:W\to V$ is linear. If

A=[T]_{B_V}^{B_W},

then $A$ is square and

[T^{-1}]_{B_W}^{B_V}=A^{-1}.

Equivalently, for every $w\in W$ ,

[T^{-1}(w)]_{B_V}=A^{-1}[w]_{B_W}.

Conversely, if the matrix representation $A$ is invertible, then $T$ is bijective.

The formula follows by writing $w=T(v)$ , so that $[w]_{B_W}=A[v]_{B_V}$ and hence $[v]_{B_V}=A^{-1}[w]_{B_W}$ . The two compositions are represented by $A^{-1}A$ and $AA^{-1}$ , which return the original coordinate vector. To check that $T^{-1}$ is linear, write $w_1=T(v_1)$ and $w_2=T(v_2)$ . Then $T(\alpha v_1+\beta v_2)=\alpha w_1+\beta w_2$ and uniqueness of preimages gives $T^{-1}(\alpha w_1+\beta w_2)=\alpha T^{-1}(w_1)+\beta T^{-1}(w_2)$ .

[EXAMPLE]

A scalar map and its inverse

Let $\alpha\in\mathbb R\setminus\{0\}$ and define

T:P_n(\mathbb R)\to P_n(\mathbb R),\qquad T(p)=\alpha p.

Relative to the standard polynomial basis $(1,x,\ldots,x^n)$ ,

A=[T]=\alpha I.

The candidate inverse is

S:P_n(\mathbb R)\to P_n(\mathbb R),\qquad S(p)=\frac{1}{\alpha}p,

whose matrix is

[S]=\frac{1}{\alpha}I=A^{-1}.

For every $p\in P_n(\mathbb R)$ ,

S(T(p))=S(\alpha p)=p, \qquad T(S(p))=T\left(\frac{1}{\alpha}p\right)=p.

Thus $T$ is bijective with inverse $S$ and its matrix $[S]=A^{-1}$ matches the theorem.

[routine]

[1]:Inverse maps in coordinates

The eigenvalues chapter asks when a choice of basis makes the matrix of an operator diagonal.

Matrix representations

Linear map

Coordinate vector

Basis

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