Recall $A \in M_{m \times n}$ defines a linear map

$$ \begin{align*} L_A: \ &\mathbf{R}^n \rightarrow \mathbf{R}^m \\ &\bar{x} \rightarrow A\bar{x} \end{align*} $$

We will show that any linear map $T: V \rightarrow W$ between finite dimensional spaces can be represented by a matrix. For example the following linear transformations

$$ \begin{align*} T_d: \ &P_3 \rightarrow P_2 \\ &f \rightarrow f' \end{align*} $$

and

$$ \begin{align*} T_i: \ &P_2 \rightarrow P_3 \\ &\ f \rightarrow \int_0^x f(t)dt \end{align*} $$

are examples of linear transformations with finite dimensional vector spaces that can be represented by a matrix. But in order to show this, we will start with expressing vectors uniquely relative a basis in a vector space.

Coordinate Expression for a vector

Recall that if $\beta$ is a basis for $V$, then for any $v \in V$, $v$ can be expressed uniquely as an element of $Span(\beta)$. We proved this previously for any vector space whether infinite or finite dimensional. But if the basis is finite ($\beta = \{v_1, ... , v_n\}$), then for every $v \in V$, $v$ can be expressed uniquely in the form,

$$ \begin{align*} v = a_1v_1 + ... + a_nv_n \end{align*} $$

The coefficients $a_1,...a_n$ are unique for $v$. We want to think of this relationship as a map.

Definition

$$ \begin{align*} [v]_{\beta} = \begin{pmatrix} a_1 \\ . \\ . \\ . \\ a_n \end{pmatrix} \in \mathbf{R}^n \end{align*} $$ is the coordinate expression for $v$ with respect to $\beta$.

Example

Let $V = \mathbf{R}^2, v = (2,1), \beta = \{(1,0), (0,1)\}$ and $\beta' = \{(1,1), (1,-1)\}$.

$$ \begin{align*} [v]_{\beta} &= \begin{pmatrix} 2 \\ 1 \\ \end{pmatrix} \\ [v]_{\beta'} &= \begin{pmatrix} a_1 \\ a_2 \\ \end{pmatrix} \end{align*} $$

These $a_1$ and $a_2$ are the coefficients that appear in writing $v$ as a linear combinations of the vectors in the basis,

$$ \begin{align*} v = (2,1) = a_1(1,1) + a_2(1,-1). \end{align*} $$

From this, we get

$$ \begin{align*} a_1 + a_2 &= 2 \\ a_1 - a_2 &= 1 \\ \end{align*} $$

Of course we can use an augmented matrix and put the matrix in a reduced row echelon form to solve the system and find that the coefficients are

$$ \begin{align*} [v]_{\beta'} &= \begin{pmatrix} a_1 \\ a_2 \\ \end{pmatrix} \\ &= \begin{pmatrix} \frac{3}{2} \\ \frac{1}{2} \\ \end{pmatrix} \end{align*} $$

So we can think of $[\quad]_{\beta'}$ as a map:

$$ \begin{align*} [\quad]_{\beta'}: V \rightarrow \mathbf{R}^n \end{align*} $$

The idea is that each basis $\beta$ of $V$ with dimension $\dim V = n$ gives a map that identifies $V$ with $\mathbf{R}^n$. This map is linear, 1-1 and onto. “Bijective Correspondance”

Matrix Representation of linear transformations

Consider now $T: V \rightarrow W$ linear and both $V$ and $W$ are finite dimensional. We know we can identify $V$ with $\mathbf{R}^n$ and we can identify $W$ with $\mathbf{R}^m$ but now we want to represent $T$ as a matrix. Let $\beta = \{v_1, ..., v_n\}$ be the basis for $V$ and let $\gamma = \{w_1, ..., w_n\}$ be the basis for $W$. For $v \in V$, we have

$$ \begin{align*} T(v) = a_1w_1 + ... + a_mw_m. \end{align*} $$

(This is the image of $v$ which can be expressed uniquely in terms of the vectors of the basis $\gamma$ because we’re in the vector space $W$ now.) $v$ itself is a linear combination of the basis vectors of $\beta$. So for each $v_j$ we have,

$$ \begin{align*} T(v_j) &= a_{1j}w_1 + ... + a_{mj}w_m \\ &= \sum_i^m a_{ij}w_i \text{ for $j = 1,2,..,n.$} \end{align*} $$

From this, we now have this definition,

Definition

Using the above notation, let the matrix $A$ defined as $A_{ij} = a_ij$ be the matrix representation of T in the ordered $\beta$ and $\gamma$ and we write $$ \begin{align*} [T]_{\beta}^{\gamma} = A \end{align*} $$

Remark: The column vectors of $T$ are the images of the basis vectors $\beta$ written with respect to $\gamma$.

$$ \begin{align*} [T]_{\beta}^{\gamma} = \begin{pmatrix} | & & |\\ [T(v_1)]_{\gamma} & ... & [T(v_n)]_{\gamma} \\ | & & | \end{pmatrix} \end{align*} $$

Example

Let

$$ \begin{align*} T_d: \ &P_3 \rightarrow P_2 \\ &f \rightarrow f'. \end{align*} $$

Let $\beta = \{1, x, x^2, x^3\}$ be the standard basis of $P_3$. and $\gamma = \{1, x, x^2\}$ be the standard basis of $P_2$. We want to compute $[T_d]_{\beta}^{\gamma}$. The first thing that we want to do is to apply $T$ on the vectors of the basis $\beta$ so,

$$ \begin{align*} T_d(1) &= 0 \\ T_d(x) &= x \\ T_d(x^2) &= 2x \\ T_d(x^3) &= 3x^2 \end{align*} $$

Next, we want to write these with respect to the basis $\gamma = \{1,x,x^2\}$ meaning that we want to express each result as a linear combination of the vectors in the basis $\gamma$.

$$ \begin{align*} [T_d(1)]_{\gamma} &= [0]_{\gamma} = 0(1) + 0(x) + 0(x^2) = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \\ [T_d(x)]_{\gamma} &= [1]_{\gamma} = 1(1) + 0(x) + 0(x^2) = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \\ [T_d(x^2)]_{\gamma} &= [2x]_{\gamma} = 0(1) + 2(x) + 0(x^2) = \begin{pmatrix} 0 \\ 2 \\ 0 \end{pmatrix} \\ [T_d(x^3)]_{\gamma} &= [3x^2]_{\gamma} = 0(1) + 2(x) + 2(x^2) = \begin{pmatrix} 0 \\ 0 \\ 3 \end{pmatrix}. \end{align*} $$

So the final matrix is

$$ \begin{align*} [T_d]^{\gamma}_{\beta} &= \begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 3 \end{pmatrix}. \end{align*} $$

Theorem 2.15(a)

Let $A \in M_{m \times n}$ and let $$ \begin{align*} L_A: \ &\mathbf{R}^n \rightarrow \mathbf{R}^m \\ &\bar{x} \rightarrow A\bar{x}. \end{align*} $$ If $\beta$ and $\gamma$ are the standard bases for $\mathbf{R}^n$ and $\mathbf{R}^m$ respectively, then $$ \begin{align*} [L_A]^{\gamma}_{\beta} = A. \end{align*} $$

(Note: We earlier said that any linear transformation between finite dimensional vector spaces can be represented by a matrix. $L_A$ is a linear transformation. So what is the matrix representation of it? The claim is that as long as you use the standard basis for $\mathbf{R}^n$ and $\mathbf{R}^m$, then the representation is just $A$ itself!)

Proof: $\beta$ is the standard basis of $\mathbf{R}^n$ so $\beta = \{e_1,...,e_n\}$ where $e_1 = \begin{pmatrix} 1 & 0 & ... & 0 \end{pmatrix}^t$
We can then find $[L_A]_{\beta}^{\gamma}$ by applying $L_A$ on the vectors of $\beta$ and then re-writing them with respect to the basis $\gamma$,

$$ \begin{align*} [L_A]^{\gamma}_{\beta} = ([L_A(e_1)]_{\gamma},..., [L_A(e_n)]_{\gamma}) \end{align*} $$

But then consider $[L_A(e_1)]_{\gamma}$. We know $L_A$ by definition takes a vector $\bar{x}$ and produces $A\bar{x}$. So $[L_A(e_1)]_{\gamma}$ = $[Ae_1]_{\gamma}$. Moreover, $Ae_1 = \bar{a}_1e_{11}+\bar{a}_2e_{12}+...+\bar{a}_ne_{1n}$ where $\bar{a}_1,...,\bar{a}_n$ are the columns of $A$. But all the coefficients of $e_1$ are zero except for the first one. So $Ae_1 = 1\bar{a}_1$ which is just the first column of $A$. Finally, the coordinate representatives of the first column of $A$ with respect to basis $\gamma$ is just the first column of $A$ since it’s the standard basis. Similarly, we can use the same argument to find out the remaining columns and so

$$ \begin{align*} [L_A]^{\gamma}_{\beta} = A. \blacksquare \end{align*} $$

So far, we’ve taken a linear transformation from a vector space $V$ to another $W$ and represented it with a matrix $[T]_{\beta}^{\gamma}$. Above, we can also take a matrix $A$ and turn it into a linear map $L_A$ and if we do so with the standard bases $\mathbf{R}^n$ and $\mathbf{R}^m$, we can recover the matrix $A$.

Linear Transformation as a Matrix Multiplication

Theorem 2.14

Let $T: V \rightarrow W$ be a linear transformation, Let $\beta, \gamma$ be fixed bases.
For all $v \in V$, $$ \begin{align*} [T(v)]_{\gamma} = [T]_{\beta}^{\gamma} [v]_{\beta} \end{align*} $$ $T$ acts like matrix multiplication.

Proof: Let $\beta=\{v_1,...,v_n\}$, $\gamma=\{w_1,...,w_m\}$. We can write $v$ below as a linear combination of the basis vectors in $\beta$. We can then apply $T$ which is linear.

$$ \begin{align*} [T(v)]_{\gamma} &= [T(a_1v_1 + ... + a_nv_n)]_{\gamma} \\ &= [T(a_1v_1) + ... + T(a_nv_n)]_{\gamma} \\ &= [a_1T(v_1) + ... + a_nT(v_n)]_{\gamma} \end{align*} $$

But this map $[ ]_{\gamma}$ is also linear. Therefore, we can distribute it in

$$ \begin{align*} [T(v)]_{\gamma} &= [a_1T(v_1) + ... + a_nT(v_n)]_{\gamma} \\ &= [a_1T(v_1)]_{\gamma} + ... + [a_nT(v_n)]_{\gamma} \\ &= a_1[T(v_1)]_{\gamma} + ... + a_n[T(v_n)]_{\gamma} \end{align*} $$

The expression above is exactly matrix multiplication so we can re-write this as

$$ \begin{align*} a_1[T(v_1)]_{\gamma} + ... + a_n[T(v_n)]_{\gamma} &= \begin{pmatrix} [T(v_1)]_{\gamma} & ... & [T(v_n)]_{\gamma} \end{pmatrix} \begin{pmatrix} a_1 \\ . \\ . \\ . \\ a_n \\ \end{pmatrix} \\ &= [T]_{\beta}^{\gamma} [v]_{\beta}. \end{align*} $$

This is exactly what we wanted to show. $\blacksquare$

Example

Let’s continue the same example from earlier where we found the matrix representative for the linear transformation

$$ \begin{align*} T_d: \ &P_3 \rightarrow P_2 \\ &f \rightarrow f'. \end{align*} $$

Suppose we have the function $f(x) = 3x^3 + 4x + 5$ in $P_3$. We can express this function in terms of the basis vectors in $\beta = \{1, x, x^2, x^3\}$ as the coefficients vector $(5, 4, 0, 3)$. So now, let’s multiply this vector by the linear transformation matrix

$$ \begin{align*} \begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 3 \end{pmatrix} \begin{pmatrix} 5 \\ 4 \\ 0 \\ 3 \end{pmatrix} = \begin{pmatrix} 4 \\ 0 \\ 9 \end{pmatrix}. \end{align*} $$

This gives us the coefficients vector with respect to the basis $\gamma$. This means that the function will be $4 + 9x^2$ which is exactly what we would get if manually applied the transformation on $f(x)$ to get $f'(x)$ but now instead we have a matrix to do this.

References

Math416 by Ely Kerman