# Day 13:

Matrix representation of a linear map

### Matrix representation

Example 1. Let $$D:\mathbb{P}_{2}\to\mathbb{P}_{1}$$ be given by $D(f(x)) = f'(x).$

Fix the bases $$S = \{1,x,x^2\}$$ for $$\mathbb{P}_{2}$$ and $$T = \{1,x\}$$ for $$\mathbb{P}_{1}$$.

We compute $$D(v)$$ for each $$v\in S$$, and write the output as a linear combination of the vectors in $$T$$:

$D(1) = 0 = 0\cdot 1 + 0\cdot x$

$D(x) = 1 = 1\cdot 1 + 0\cdot x$

$D(x^{2}) = 2x = 0\cdot 1 + 2\cdot x$

Then, the matrix representation of $$D$$ with respect to $$S$$ and $$T$$ is

$\begin{bmatrix} 0 & 1 & 0\\ 0 & 0 & 2\end{bmatrix}$

### Matrix representation

Example 2. Let $$D:\mathbb{P}_{2}\to\mathbb{P}_{1}$$ be given by $D(f(x)) = f'(x).$

Fix the bases $$S = \{1,x,x^2\}$$ for $$\mathbb{P}_{2}$$ and $$T = \{1,2x\}$$ for $$\mathbb{P}_{1}$$.

We compute $$D(v)$$ for each $$v\in S$$, and write the output as a linear combination of the vectors in $$T$$:

$D(1) = 0 = 0\cdot 1 + 0\cdot (2x)$

$D(x) = 1 = 1\cdot 1 + 0\cdot (2x)$

$D(x^{2}) = 2x = 0\cdot 1 + 1\cdot (2x)$

Then, the matrix representation of $$D$$ with respect to $$S$$ and $$T$$ is

$\begin{bmatrix} 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}$

Note that this matrix representation is different from the previous one, but the linear map $$D$$ is the same.

### Matrix representation

Example 3. Let $$M:\mathbb{R}^{3}\to\mathbb{R}^{2}$$ be given by $M([x_{1}\ \ x_{2}\ \ x_{3}]^{\top}) = \begin{bmatrix} 1 & 2 & 0\\ 0 & 3 & 4\end{bmatrix}\begin{bmatrix}x_{1}\\ x_{2}\\ x_{3}\end{bmatrix}.$

Fix the bases $S = \left\{\begin{bmatrix}1\\ 0\\ 0\end{bmatrix},\begin{bmatrix}0\\ 1\\ 0\end{bmatrix},\begin{bmatrix}0\\ 0\\ 1\end{bmatrix}\right\}\quad\text{and}\quad T = \left\{\begin{bmatrix}1\\ 0\end{bmatrix},\begin{bmatrix}0\\ 1\end{bmatrix}\right\}$ for $$\mathbb{R}^3$$ and $$\mathbb{R}^{2}$$, respectively.

We compute $$M(x)$$ for each $$x\in S$$, and write the output as a linear combination of the vectors in $$T$$:

$M\left(\begin{bmatrix} 1\\ 0\\ 0\end{bmatrix}\right) = \begin{bmatrix} 1\\ 0\end{bmatrix} = 1\begin{bmatrix} 1\\ 0\end{bmatrix} + 0\begin{bmatrix} 0\\ 1\end{bmatrix}$

### Matrix representation

Example 3 continued.

$M\left(\begin{bmatrix} 0\\ 1\\ 0\end{bmatrix}\right) = \begin{bmatrix} 2\\ 3\end{bmatrix} = 2\begin{bmatrix} 1\\ 0\end{bmatrix} + 3\begin{bmatrix} 0\\ 1\end{bmatrix}$

$M\left(\begin{bmatrix} 0\\ 0\\ 1\end{bmatrix}\right) = \begin{bmatrix} 0\\ 4\end{bmatrix} = 0\begin{bmatrix} 1\\ 0\end{bmatrix} + 4\begin{bmatrix} 0\\ 1\end{bmatrix}$

Then, the matrix representation of $$M$$ with respect to $$S$$ and $$T$$ is

$\begin{bmatrix} 1 & 2 & 0\\ 0 & 3 & 4\end{bmatrix}$

### Matrix representation

Example 4. Let $$M:\mathbb{R}^{3}\to\mathbb{R}^{2}$$ be given by $M([x_{1}\ \ x_{2}\ \ x_{3}]^{\top}) = \begin{bmatrix} 1 & 2 & 0\\ 0 & 3 & 4\end{bmatrix}\begin{bmatrix}x_{1}\\ x_{2}\\ x_{3}\end{bmatrix}.$

Fix the bases $S = \left\{\begin{bmatrix}1\\ 1\\ 1\end{bmatrix},\begin{bmatrix}1\\ 1\\ 0\end{bmatrix},\begin{bmatrix}1\\ 0\\ 0\end{bmatrix}\right\}\quad\text{and}\quad T = \left\{\begin{bmatrix}1\\ 1\end{bmatrix},\begin{bmatrix}1\\ -1\end{bmatrix}\right\}$ for $$\mathbb{R}^3$$ and $$\mathbb{R}^{2}$$, respectively.

We compute $$M(x)$$ for each $$x\in S$$, and write the output as a linear combination of the vectors in $$T$$:

$M\left(\begin{bmatrix} 1\\ 1\\ 1\end{bmatrix}\right) = \begin{bmatrix} 3\\ 7\end{bmatrix} = 5\begin{bmatrix} 1\\ 1\end{bmatrix} + (-2)\begin{bmatrix} 1\\ -1\end{bmatrix}$

### Matrix representation

Example 4 continued.

$M\left(\begin{bmatrix} 1\\ 1\\ 0\end{bmatrix}\right) = \begin{bmatrix} 3\\ 3\end{bmatrix} = 3\begin{bmatrix} 1\\ 1\end{bmatrix} + 0\begin{bmatrix} 1\\ -1\end{bmatrix}$

$M\left(\begin{bmatrix} 1\\ 0\\ 0\end{bmatrix}\right) = \begin{bmatrix} 1\\ 0\end{bmatrix} = \tfrac{1}{2}\begin{bmatrix} 1\\ 1\end{bmatrix} + \tfrac{1}{2}\begin{bmatrix} 1\\ -1\end{bmatrix}$

Then, the matrix representation of $$M$$ with respect to $$S$$ and $$T$$ is

$\begin{bmatrix} 5 & 3 & \frac{1}{2}\\[1ex] -2 & 0 & \frac{1}{2}\end{bmatrix}$

### Matrix representation

Example. Let $$D:\mathbb{P}_{2}\to\mathbb{P}_{1}$$ be given by $D(f(x)) = f'(x).$

Fix the bases $$S = \{1,x,x^2\}$$ for $$\mathbb{P}_{2}$$ and $$T = \{1,x\}$$ for $$\mathbb{P}_{1}$$.

Then, the matrix representation of $$D$$ with respect to $$S$$ and $$T$$ is

$\begin{bmatrix} 0 & 1 & 0\\ 0 & 0 & 2\end{bmatrix}$

If we take an arbitrary vector $$v=ax^{2}+bx+c\in\mathbb{P}_{2}$$, then the coefficients of $$v$$ with respect to the basis $$S$$ are $$c,b,a$$ (NOTE THE ORDER). Arrange these as column vector, multiply by the matrix above, and we obtain

$\begin{bmatrix} 0 & 1 & 0\\ 0 & 0 & 2\end{bmatrix}\begin{bmatrix}c\\ b\\ a\end{bmatrix} = \begin{bmatrix}b\\ 2a\end{bmatrix}$ Use these as the coefficients on the vectors in $$T$$ (ORDER MATTERS) and we get $$b+2ax$$, which is $$D(v)$$.

### Matrix representation of a linear map

Let $$V$$ and $$W$$ be vector spaces with bases $$\{v_{i}\}_{i=1}^{n}$$ and $$\{w_{i}\}_{i=1}^{m},$$ respectively.

Let $$L:V\to W$$ be a linear map.

For each $$j\in\{1,2,\ldots,n\}$$ there are scalars $$a_{1j},a_{2j},\ldots,a_{mj}$$ such that

$L(v_{j}) = a_{1j}w_{1} + a_{2j}w_{2} + \cdots + a_{mj}w_{m} = \sum_{i=1}^{m}a_{ij}w_{i}$

The matrix

$\begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n}\\ a_{21} & a_{22} & \cdots & a_{2n}\\ \vdots & \vdots & \ddots & \vdots\\ a_{m1} & a_{m2} & \cdots & a_{mn}\end{bmatrix}$

is called the matrix representation of $$L$$ with respect to the bases $$\{v_{i}\}_{i=1}^{n}$$ and $$\{w_{i}\}_{i=1}^{m}$$.

### Coordinate vector

Let $$V$$ be vector space with basis $$\{v_{i}\}_{i=1}^{n}$$.

Given a vector $$v\in V$$ and $$w\in W$$, there are scalars $$b_{1},b_{2},\ldots,b_{n}$$ such that

$v=\sum_{i=1}^{n}b_{i}v_{i}.$

The column vector

$\begin{bmatrix} b_{1}\\ b_{2}\\ \vdots\\ b_{n}\end{bmatrix}\in\mathbb{R}^{n}$

is the coordinate vector of $$v$$ with respect to the basis.

Theorem. Let $$V$$ and $$W$$ be vector spaces with bases $$S=\{v_{i}\}_{i=1}^{n}$$ and $$T=\{w_{i}\}_{i=1}^{m},$$ respectively. Let $$L:V\to W$$ be a linear map, and $$v\in V$$. If $$A\in\mathbb{R}^{m\times n}$$ is the matrix representation of $$L$$ with respect to $$S$$ and $$T$$, and $$x\in\mathbb{R}^{n}$$ is the coordinate vector of $$v$$ with respect to $$S$$, then $$Ax$$ is the coordinate vector of $$L(v)$$ with respect to $$T$$.

Proof. Let $$A = [a_{ij}]$$ and $$x=[b_{1}\ \ b_{2}\ \ \cdots\ \ b_{n}]^{\top}$$, then

$Ax=\begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n}\\ a_{21} & a_{22} & \cdots & a_{2n}\\ \vdots & \vdots & \ddots & \vdots\\ a_{m1} & a_{m2} & \cdots & a_{mn}\end{bmatrix}\begin{bmatrix} b_{1}\\ b_{2}\\ \vdots\\ b_{n}\end{bmatrix} = \begin{bmatrix} a_{11}b_{1} + a_{12}b_{2} + \cdots + a_{1n}b_{n}\\ a_{21}b_{1} + a_{22}b_{2} + \cdots + a_{2n}b_{n}\\ \vdots\\ a_{m1}b_{1} + a_{m2}b_{2} + \cdots + a_{mn}b_{n}\end{bmatrix}$

in particular, let $$c_{i}$$ denote that $$i$$th row of $$Ax$$, that is,

$c_{i} = \sum_{j=1}^{n}a_{ij}b_{j}.$

We wish to show that $$[c_{1}\ \ c_{2}\ \ \cdots c_{m}]^{\top}$$ is the coordinate vector of $$L(v)$$.

Proof continued.

$\sum_{i=1}^{m}c_{i}w_{i} = \sum_{i=1}^{m}\left(\sum_{j=1}^{n}a_{ij}b_{j}\right)w_{i} = \sum_{i=1}^{m}\sum_{j=1}^{n}a_{ij}b_{j}w_{i} = \sum_{j=1}^{n}\sum_{i=1}^{m}a_{ij}b_{j}w_{i}$

$= \sum_{j=1}^{n}b_{j}\sum_{i=1}^{m}a_{ij}w_{i} = \sum_{j=1}^{n}b_{j}L(v_{j}) = L\left(\sum_{j=1}^{n}b_{j}v_{j}\right) = L(v).$

$$\Box$$

Go down for that calculation without summation notation

Proof continued.



$\sum_{i=1}^{m}c_{i}w_{i} = c_{1}w_{1}+c_{2}w_{2}+c_{3}w_{3} + \cdots + c_{m}w_{m}$

$= \big(a_{11}b_{1} + a_{12}b_{2} + a_{13}b_{3} + \cdots + a_{1n}b_{n}\big)w_{1}$

$+\big(a_{21}b_{1} + a_{22}b_{2} + a_{23}b_{3} + \cdots + a_{2n}b_{n}\big)w_{2}$

$+\big(a_{31}b_{1} + a_{32}b_{2} + a_{33}b_{3} + \cdots + a_{3n}b_{n}\big)w_{3}$

$+ \big(a_{m1}b_{1} + a_{m2}b_{2} + a_{m3}b_{3} + \cdots + a_{mn}b_{n}\big)w_{m}$

$\vdots$

$= a_{11}b_{1}w_{1} + a_{12}b_{2}w_{1} + a_{13}b_{3}w_{1} + \cdots + a_{1n}b_{n}w_{1}$

$+ a_{21}b_{1}w_{2} + a_{22}b_{2}w_{2} + a_{23}b_{3}w_{2} + \cdots + a_{2n}b_{n}w_{2}$

$+ a_{31}b_{1}w_{3} + a_{32}b_{2}w_{3} + a_{33}b_{3}w_{3} + \cdots + a_{3n}b_{n}w_{3}$

$+ a_{m1}b_{1}w_{m} + a_{m2}b_{2}w_{m} + a_{m3}b_{3}w_{m} + \cdots + a_{mn}b_{n}w_{m}$

$\vdots$

Proof continued.



$\sum_{i=1}^{m}c_{i}w_{i} = a_{11}b_{1}w_{1} + a_{12}b_{2}w_{1} + a_{13}b_{3}w_{1} + \cdots + a_{1n}b_{n}w_{1}$

$+ a_{21}b_{1}w_{2} + a_{22}b_{2}w_{2} + a_{23}b_{3}w_{2} + \cdots + a_{2n}b_{n}w_{2}$

$+ a_{31}b_{1}w_{3} + a_{32}b_{2}w_{3} + a_{33}b_{3}w_{3} + \cdots + a_{3n}b_{n}w_{3}$

$+ a_{m1}b_{1}w_{m} + a_{m2}b_{2}w_{m} + a_{m3}b_{3}w_{m} + \cdots + a_{mn}b_{n}w_{m}$

$\vdots$

$=b_{1}\big(a_{11}w_{1} + a_{21}w_{2} + a_{31}w_{3} + \cdots + a_{m1}w_{m}\big)$

$+ b_{2}\big(a_{12}w_{1} + a_{22}w_{2} + a_{32}w_{3} + \cdots + a_{m2}w_{m}\big)$

$+b_{3}\big(a_{13}w_{1} + a_{23}w_{2} + a_{33}w_{3} + \cdots + a_{m3}w_{m}\big)$

$\vdots$

$+b_{n}\big(a_{1n}w_{1} + a_{2n}w_{2} + a_{3n}w_{3} + \cdots + a_{mn}w_{m}\big)$

Proof continued.



$\sum_{i=1}^{m}c_{i}w_{i} = b_{1}\big(a_{11}w_{1} + a_{21}w_{2} + a_{31}w_{3} + \cdots + a_{m1}w_{m}\big)$

$+ b_{2}\big(a_{12}w_{1} + a_{22}w_{2} + a_{32}w_{3} + \cdots + a_{m2}w_{m}\big)$

$+b_{3}\big(a_{13}w_{1} + a_{23}w_{2} + a_{33}w_{3} + \cdots + a_{m3}w_{m}\big)$

$\vdots$

$+b_{n}\big(a_{1n}w_{1} + a_{2n}w_{2} + a_{3n}w_{3} + \cdots + a_{mn}w_{m}\big)$

$= b_{1}L(v_{1}) + b_{2}L(v_{2}) + b_{3} L(v_{3}) + \cdots + b_{n}L(v_{n})$

$= L(b_{1}v_{1}) + L(b_{2}v_{2}) + L(b_{3} v_{3}) + \cdots + L(b_{n}v_{n})$

$= L(b_{1}v_{1}+b_{2}v_{2}) + L(b_{3} v_{3}) + \cdots + L(b_{n}v_{n})$

$= L(b_{1}v_{1}+b_{2}v_{2} + b_{3} v_{3}) + \cdots + L(b_{n}v_{n})$

$\vdots$

$= L(b_{1}v_{1}+b_{2}v_{2} + b_{3} v_{3} + \cdots + b_{n}v_{n}) = L(v)$

### Matrix representation of linear maps $$L:\mathbb{R}^{n}\to\mathbb{R}^{m}$$

Given a linear map $$L:\mathbb{R}^{n}\to\mathbb{R}^{m}$$, and bases $$S = \{v_{i}\}_{i=1}^{n}$$ and $$T = \{w_{i}\}_{i=1}^{m}$$ for $$\mathbb{R}^{n}$$ and $$\mathbb{R}^{m}$$, respectively, we can talk about the matrix representation of $$L$$ with respect to $$S$$ and $$T$$.

However, we will sometimes refer to the matrix representation of $$L$$ without referring to bases. In that case, we take the bases for both spaces to be the standard basis:

Definition. The standard basis for $$\mathbb{R}^{d}$$ is the set

$\left\{\begin{bmatrix}1\\ 0\\ 0\\ \vdots\\ 0\end{bmatrix},\begin{bmatrix}0\\ 1\\ 0\\ \vdots\\ 0\end{bmatrix},\begin{bmatrix}0\\ 0\\ 1\\ \vdots\\ 0\end{bmatrix},\ldots,\begin{bmatrix}0\\ 0\\ 0\\ \vdots\\ 1\end{bmatrix}\right\}\subset\mathbb{R}^{d}.$

The $$i$$th element in this set is called $$e_{i}$$.

### Matrix representation of linear maps $$L:\mathbb{R}^{n}\to\mathbb{R}^{m}$$

Let $$L:\mathbb{R}^{n}\to\mathbb{R}^{m}$$ be a linear map, and let $$A\in\mathbb{R}^{m\times n}$$ be the matrix representation of $$L$$.

Note that the coordinate vector of $$x$$ with respect to the standard basis is $$x$$. Hence, by the previous theorem we have $$L(x) = Ax$$ for all $$x\in\mathbb{R}^{n}$$. In particular,

$\operatorname{im}(L) = \{L(x) : x\in\mathbb{R}^{n}\} = \{Ax : x\in\mathbb{R}^{n}\} = C(A)$

$\operatorname{ker}(L) = \{x : L(x) = 0\} = \{x : Ax=0\} = N(A)$

Hence, $$\operatorname{im}(L)$$ is the column space of the matrix representation of $$L$$, and $$\operatorname{ker}(L)$$ is the nullspace of the matrix representation of $$L$$.

### Matrix representation of linear maps $$L:\mathbb{R}^{n}\to\mathbb{R}^{m}$$

Again, note that the coordinate vector of $$x$$ with respect to the standard basis is $$x$$. Hence, if $$B\in\mathbb{R}^{m\times n}$$ is the matrix representation of $$L$$, then $$L(x) = Bx$$.

Note that $$Be_{i}$$ is the $$i$$th column of $$B$$, and $$Ae_{i}$$ is the $$i$$th column of $$A$$. Hence, for each $$i\in\{1,2,\ldots,n\}$$ we have

$Ae_{i} = L(e_{i}) = Be_{i},$

that is, the $$i$$th columns of $$A$$ and $$B$$ are equal for each $$i$$. Therefore, $$A=B$$, that is, the matrix representation of $$L$$ is $$A$$.

Since a matrix $$A$$ gives us a linear map $$L$$, we often use the notation $$\operatorname{im}(A)$$ and $$\operatorname{ker}(A)$$ for the image and kernel of $$L$$.

Let $$A$$ be an $$m\times n$$ matrix, and define the linear map $$L:\mathbb{R}^{n}\to\mathbb{R}^{m}$$ by $$L(x) = Ax.$$

### Matrix of an isomorphism

Proposition. Let $$V$$ and $$W$$ be vector spaces with bases $$S = \{v_{i}\}_{i=1}^{n}$$ and $$T = \{w_{i}\}_{i=1}^{m}$$, respectively. A linear map $$L$$ is an isomorphism if and only if the matrix representation of $$A$$ with respect to $$S$$ and $$T$$ is invertible.

Proof. Suppose $$L$$ is an isomorphism. This implies that $$\operatorname{im}(L) = W$$ and $$\operatorname{ker}(L) = \{0\}$$.

Suppose that $$x\in\mathbb{R}^{n}$$ is in the nullspace of $$A$$. Let $$v\in V$$ be the vector whose coordinate vector is $$x$$. By the previous theorem $$Ax$$ is the coordinate vector of $$L(v)$$. However, $$Ax=0$$, and hence $$L(v) = 0$$, that is, $$v\in\operatorname{ker}(L) = \{0\}$$. We conclude that $$v=0$$, and hence $$x=0$$. This shows $$N(A) = \{0\}$$.

A similar argument (which you should write out) shows that $$C(A) = \mathbb{R}^{m}.$$ By the Rank-Nullity Theorem we have

$n = \operatorname{rank}(A) + \operatorname{nullity}(A) = m+0 = m.$

### Matrix of an isomorphism

Proof. Thus, $$m=n$$ and $$A$$ is a square matrix.

Since $\operatorname{rank}(A) = m = (\# \text{ rows of }A) = n = (\#\text{ columns of }A) ,$ we see that $$\operatorname{rref}(A)$$ has a pivot in each column and each row. Therefore, $$\operatorname{rref}(A) = I$$, and hence by a previous theorem $$A$$ is invertible. $$\Box$$

Corollary. If $$V$$ and $$W$$ are isomorphic finite-dimensional vector spaces, then $$\operatorname{dim}(V) = \operatorname{dim}(W)$$.

Proof. The assumption $$V$$ and $$W$$ are isomorphic means that there is a linear bijection $$L:V\to W$$. The assumption that the spaces are finite dimensional means that there are finite bases $$S = \{v_{i}\}_{i=1}^{n}$$ for $$V$$ and $$T = \{w_{i}\}_{i=1}^{m}$$ for $$W$$. By the previous theorem, the matrix representation of $$L$$ with respect to $$S$$ and $$T$$ is an invertible matrix, in particular, it is a square matrix. Therefore $$n=m$$. $$\Box$$

By John Jasper

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