MATH 240

Mon. November 11th, 2019

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Recall that we can answer questions about vectors in an abstract vector space (like whether they’re linearly independent) by finding a basis for the space, moving the vectors into $\R^n$ using the basis, answering the question there, and then moving back into the abstract space.

Can we do the same for linear transformations?

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Relative Matrix for a Linear Transformation

Let $V,W$ be vector spaces.
Let $B=\{b_1,...,b_n\}$ be an ordered basis for $V$ and $C=\{c_1,...,c_m\}$ be an ordered basis for $W$.
Let $T: V \rightarrow W$ be a linear transformation.

We want to find some $m\times n$ matrix $M$ such that $[T(x)]_C=M[x]_B$. If we do this, then we can completely represent $T(x)$ as a transformation from $\R^n$ to $\R^m$ rather than as an abstract transformation.

Assume that $x=r_1b_1+...+r_nb_n$.
$\begin{aligned} [x]_B&=\begin{pmatrix} r_1\\...\\r_n \end{pmatrix}\\\\ T(x)&=T(r_1b_1+...+r_nb_n)\\ &=r_1T(b_1)+...+r_nT(b_n)\\\\ [T(x)]_C&=r_1[T(b_1)]_C+...+r_n[T(b_n)]_C\\ &=\begin{bmatrix} [T(b_1)]_C&...&[T(b_n)]_C \end{bmatrix}\begin{pmatrix} r_1\\...\\r_n \end{pmatrix}\\ &=\begin{bmatrix} [T(b_1)]_C&...&[T(b_n)]_C \end{bmatrix}[x]_B \end{aligned}$
$M=\begin{bmatrix} [T(b_1)]_C&...&[T(b_n)]_C \end{bmatrix}.$

We call $M$ the matrix for $T$ relative to the bases $B$ and $C$.

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Ex. Suppose that $B=\{b_1,b_2\}$ is an ordered basis for $V$, $C=\{c_1,c_2,c_3\}$ is an ordered basis for $W$, and $T:V\rightarrow W$ is a linear transformation.
$\begin{aligned} T(b_1)&=3c_1+2c_2+4c_3\\ T(b_2)&=c_1+5c_2+3c_3 \end{aligned}$ Find a matrix $M$ for $T$ relative to $B$ and $C$.
$\begin{aligned} [T(b_1)]_C&=\begin{pmatrix} 3\\2\\4 \end{pmatrix}\\ [T(b_2)]_C&=\begin{pmatrix} 1\\5\\3 \end{pmatrix}\\ M&=\begin{pmatrix} 3&1\\2&5\\4&3 \end{pmatrix} \end{aligned}$

You can see that once the bases $B$ and $C$ are fixed, we can find a unique matrix for any linear transformation between $V$ and $W$.

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B-Matrix

Now we’re going to look at a special case: $V=W,\ B=C$.
In this case, we call $M$ the matrix for $T$ relative to $B$, or simply the $B$-matrix for $T$, and denote it $[T]_B$ (i.e. $[T(x)]_B=[T]_B[x]_B$).

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Ex. We can represent taking the derivative as a linear transformation.
$\begin{gathered} T:P_3\rightarrow P_3\\T(a_0+a_1t+a_2t^2+a_3t^3)=0a_0+a_1+2a_2t+3a_3t^2 \end{gathered}$
Find the $B$-matrix for $T$ where $B=\{1,t,t^2,t^3\}$.

$\begin{aligned} [T(1)]_B&=\begin{pmatrix} 0\\0\\0\\0 \end{pmatrix}\\ [T(t)]_B&=\begin{pmatrix} 1\\0\\0\\0 \end{pmatrix}\\ [T(t^2)]_B&=\begin{pmatrix} 0\\2\\0\\0 \end{pmatrix}\\ [T(t^3)]_B&=\begin{pmatrix} 0\\0\\3\\0 \end{pmatrix} \end{aligned}$$\begin{aligned} [T]_B&=\begin{bmatrix} [T(1)]_B&[T(t)]_B&[T(t^2)]_B&[T(t^3)]_B \end{bmatrix}\\ &=\begin{pmatrix} 0&1&0&0\\0&0&2&0\\0&0&0&3\\0&0&0&0 \end{pmatrix} \end{aligned}$

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Theorem. Suppose $A=PCP^{-1}$. If $B$ is a basis for $\R^n$ formed from the columns of $P$, then $C$ is the $B$-matrix for $x\rightarrow Ax$.

Recall that $\R^n$ has infinitely many bases – we just tend to use the standard basis because it’s computationally convenient. However, you may encounter data represented in a different basis. We can use $C$ in that case to change coordinates between different basis systems.