Matrix representations of transformations

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A linear transformation can be represented in terms of multiplication by a matrix.

Suppose $V = \mathbb R^n, W = \mathbb R^m$ , and $L_A:V\to W$ is given by $L_A({\bf v}) = A*{\bf v}$ for some $m\times n$ real matrix $A$ . Then it follows immediately from the properties of matrix algebra that $L_A$ is a linear transformation: $L_A(\alpha {\bf v} + \beta {\bf w}) = A*(\alpha {\bf v} + \beta {\bf w}) = \alpha (A*{\bf v}) + \beta (A*{\bf w}) = \alpha L_A({\bf v}) + \beta L_A({\bf w})$

Conversely, suppose the linear transformation $L$ is given. Define the matrix $A_L$ by $A_L = [L({\bf e}_1)\ L({\bf e}_2)\ \dots L({\bf e}_n)]$ that is, the $m\times n$ matrix with $A(:,i) = L({\bf e}_i),\, 1\le i\le n$ . Then by construction $A_L*({\bf e}_i) = A(:,i) = L({\bf e}_i),\, 1\le i\le n$ so that ${\bf v}\mapsto L({\bf v})$ and ${\bf v}\mapsto A_L*{\bf v}$ are two linear transformations which agree on a basis for $\mathbb R^n$ , which by the previous corollary implies $L({\bf v}) = A_L*({\bf v})\qquad \forall {\bf v}\in \mathbb R^n$ Because of this, the matrix $A_L$ is referred to as a matrix representation of $L$ . Note that this representation is with respect to to the standard basis for $\mathbb R^n$ and $\mathbb R^m$ .

We see now that the same type of representation applies for arbitrary vector spaces once a basis has been fixed for both the domain and target. In other words, given

A vector space $V$ with basis $S = \{{\bf v}_1,\dots ,{\bf v}_n\}$ ,
a vector space $W$ with basis $T = \{{\bf w}_1,\dots ,{\bf w}_m\}$ , and
a linear transformation $L:V\to W$

we could ask if there is a similar representation of $L$ in terms of a matrix (which depends on these two choices of bases). The answer is “yes”.

For any ${\bf v}\in V$ ${}_TL({\bf v}) = {}_TL_S*{}_S{\bf v}$ where ${}_TL_S$ is the $m\times n$ matrix defined by ${}_TL_S = [{}_TL({\bf v}_1)\ {}_TL({\bf v}_2)\ \dots \ {}_TL({\bf v}_n)]$

Proof: Again by the above corollary it suffices to verify the equality for basis vectors. But ${}_S{\bf v}_i$ is the $n\times 1$ coordinate vector identical to the basis vector ${\bf e}_i$ for $\mathbb R^n$ . From this we get ${}_TL({\bf v}_i) = {}_TL_S(:,i) = {}_TL_S*{}_S{\bf v}_i,\qquad 1\le i\le n$ completing the proof. $\blacksquare$

Suppose $L:\mathbb R^3\to \mathbb R^2$ is the linear transformation given by $L({\bf x}) = L\left (\begin {bmatrix} x_1\\x_2\\x_3\end {bmatrix}\right ) = \begin {bmatrix} 2x_1 - 3x_2 + 7x_3\\ 4x_2 - 5x_3\end {bmatrix}$ To compute the matrix representation of $L$ with respect to the standard bases of $\mathbb R^3$ and $\mathbb R^2$ , we evaluate $L$ on the basis vectors ${\bf e}_i, 1\le i\le 3$ and then form the matrix $A_L = \begin {bmatrix} {L(\bf e}_1) & L({\bf e}_2) & L({\bf e}_3)\end {bmatrix}$ , yielding $A_L = \begin {bmatrix} 2 & -3 & 7\\ 0 & 4 & -5 \end {bmatrix}$

Recall that $P_n$ is the vector space of polynomials in one variable with real coefficients. Let $L:P_3\to P_3$ be given by $L(p(x)) = 2p'(x) - p(x)$ , where $p'(x)$ denotes the derivative of $p(x)$ . To find the matrix representation of $L$ with respect to the basis $\{1,x,x^2\}$ for $P_3$ , we first compute $L(1) = -1; L(x) = 2 - 2x; L(x^2) = 4x - x^2$ . Then $A_L = \begin {bmatrix} L(1) & L(x) & L(x^2)\end {bmatrix} = \begin {bmatrix} -1 & 2 & 0\\0 & -2 & 4\\0 & 0 & -1\end {bmatrix}$

Returning once more to the general case where $L:V\to W$ is linear, $S$ a basis for $V$ , $T$ a basis for $W$ , we note that the bases $S$ and $T$ can be used to identify the kernel and image of $L$ . Precisely, we have

Assume $Dim(V) = n$ , $Dim(W) = m$ . Let $A = {}_{T}L_{S}$ be the $m\times n$ matrix representation of $L$ with respect to the pair of bases $S,T$ . Then

${\bf v}\in ker(L)\subset V$ iff ${}_{S}{\bf v}\in N(A)\subset \mathbb R^n$ .
${\bf w}\in im(L)\subset W$ iff ${}_{T}{\bf w}\in C(A)\subset \mathbb R^m$ .

This theorem tells us that, once we have fixed a basis for $V$ and $W$ , the representation of $L$ by the matrix $A={}_{T}L_{S}$ further identifies i) the kernel $ker(L)$ of $L$ with the nullspace $N(A)$ of $A$ and ii) the image $im(L)$ with the column space $C(A)$ of $A$ .

Suppose $L: P_3\to P_3$ is the linear transformation represented with respect to the standard basis on $P_3$ by the matrix $A = \begin {bmatrix} 2 & 3 & 1\\ 3 & 9 & 6\\ 1 & 6 & 5\end {bmatrix}$ . Our objective is to find a minimal spanning set $ker(L)$ and $im(L)$ (with respect to the standard basis for $P^3$ ).

First we compute $rref(A) = \begin {bmatrix} 1 & 0 & -1\\0 & 1 & 1\\0 & 0 & 0\end {bmatrix}$ . We then see

The column space $C(A)$ of $A$ is $Span\{A(:,1), A(:,2)\} = Span\left \{\begin {bmatrix} 2\\3\\\1\end {bmatrix}, \begin {bmatrix} 3\\9\\6\end {bmatrix}\right \}$ ;
the nullspace $N(A)$ of $A$ is $Span\left \{\begin {bmatrix} 1\\-1\\1 \end {bmatrix} \right \}$ . Note that these are coordinate vectors.
The corresponding vectors, written as polynomials in $P_3$ , give $im(L) = Span\left \{(2 + 3x + x^2), (3 + 9x + 6x^2)\right \}, \quad ker(L) = Span\left \{(1 - x + x^2)\right \}$

Suppose $L:P_4\to P_4$ is the linear transformation whose representation in the standard basis for $P_4$ is given by $A =\begin {bmatrix} 10 & -16 & -2 & -2\\ -16 & 36 & 22 & 12\\ -2 & 22 & 53 & -4\\ -2 & 12 & -4 & 30\end {bmatrix}$ Following the method in the above example, write down a minimal spanning set for $ker(L)$ and $im(L)$ (the elements should be vectors in $P_4$ - i.e., polynomials, not their coordinate representations.)