MATH 240

Mon. November 11th, 2019


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 Rn\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?


Relative Matrix for a Linear Transformation

Let V,WV,W be vector spaces.
Let B={b1,...,bn}B=\{b_1,...,b_n\} be an ordered basis for VV and C={c1,...,cm}C=\{c_1,...,c_m\} be an ordered basis for WW.
Let T:VWT: V \rightarrow W be a linear transformation.

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

Assume that x=r1b1+...+rnbnx=r_1b_1+...+r_nb_n.
[x]B=(r1...rn)T(x)=T(r1b1+...+rnbn)=r1T(b1)+...+rnT(bn)[T(x)]C=r1[T(b1)]C+...+rn[T(bn)]C=[[T(b1)]C...[T(bn)]C](r1...rn)=[[T(b1)]C...[T(bn)]C][x]B\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=[[T(b1)]C...[T(bn)]C].M=\begin{bmatrix} [T(b_1)]_C&...&[T(b_n)]_C \end{bmatrix}.

We call MM the matrix for TT relative to the bases BB and CC.


Ex. Suppose that B={b1,b2}B=\{b_1,b_2\} is an ordered basis for VV, C={c1,c2,c3}C=\{c_1,c_2,c_3\} is an ordered basis for WW, and T:VWT:V\rightarrow W is a linear transformation.
T(b1)=3c1+2c2+4c3T(b2)=c1+5c2+3c3\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 MM for TT relative to BB and CC.
[T(b1)]C=(324)[T(b2)]C=(153)M=(312543)\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 BB and CC are fixed, we can find a unique matrix for any linear transformation between VV and WW.


B-Matrix

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


Ex. We can represent taking the derivative as a linear transformation.
T:P3P3T(a0+a1t+a2t2+a3t3)=0a0+a1+2a2t+3a3t2\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 BB-matrix for TT where B={1,t,t2,t3}B=\{1,t,t^2,t^3\}.

[T(1)]B=(0000)[T(t)]B=(1000)[T(t2)]B=(0200)[T(t3)]B=(0030)\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}[T]B=[[T(1)]B[T(t)]B[T(t2)]B[T(t3)]B]=(0100002000030000)\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}


Theorem. Suppose A=PCP1A=PCP^{-1}. If BB is a basis for Rn\R^n formed from the columns of PP, then CC is the BB-matrix for xAxx\rightarrow Ax.

Recall that Rn\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 CC in that case to change coordinates between different basis systems.