Finding a linear transformation with respect to basis

Question

Let $T\colon \mathbb{R}^3 \to \mathbb{R}^{2x2}$ a linear transform given by

\begin{equation} T(x,y,z) = \left({\begin{array}{cc} x + y + z & 2x -z \\ x-y+z & z-3y-2x \end{array}} \right), x, y, z \in \mathbb{R} \end{equation}

Calculate $[T]_{B_1,B_2}$, where bases $B_1$ and $B_2$ are $B_1 = \{(1,1,0), (0,1,1), (1,0,1)\}$,

$B_2 = \left\{ \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}, \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} \right\} $.

My reasoning is as follows...

I evaluate $T$ in each of the vectors of $B_1$ basis, and writing that output in basis $B_2$, obtaining three matrices:

$\frac{1}{5} \begin{pmatrix} 13 & 16 \\ 9 & -3 \end{pmatrix}, \frac{1}{5} \begin{pmatrix} 6 & -13 \\ -12 & 4 \end{pmatrix}, \frac{1}{5} \begin{pmatrix} 9 & -5 \\ 4 & 1 \end{pmatrix}$

If I write those matrices as the coefficient columns of a matrix $A$, I have:

$\frac{1}{5} \begin{pmatrix} 13 & 16 & 9 \\ 16 & -13 & -5 \\ 9 & -12 & 4 \\ -3 & 4 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix}$, I get the following transformation:

$[T]_{B_1,B_2} = \frac{1}{5} \begin{pmatrix} 13x+6y+9z & 16x-13y-5z \\ 9x-12y+4z & -3x+4y+z \end{pmatrix}$

It is correct?

Answer 1

$B_1$ has three elements and $B_2$ four, so $[T]_{B_1,B_2}$ should be a $4\times3$ matrix whose columns are the coordinates relative to $B_2$ of the images of the elements of $B_1$. That looks like what you tried to do when you constructed a matrix from the entries of the three matrices that you computed, but the entries of those matrices aren’t the correct coordinates.

Backing up a bit, the images of the elements of $B_1$ are $$Tv_1=\pmatrix{2&2\\0&-5},Tv_2=\pmatrix{2&-1\\0&-2},Tv_3=\pmatrix{2&1\\2&-1}.$$ Now you need the coordinate vectors of these three matrices relative to $B_2$. That is, for each $Tv_i$, you need scalars $a$, $b$, $c$ and $d$ such that $$Tv_i = a\pmatrix{1&0\\0&-1}+b\pmatrix{0&1\\0&0}+c\pmatrix{0&0\\1&0}+d\pmatrix{1&2\\3&4}.$$ You can solve each of these equations individually, or compute these coordinates all at once by flattening the matrices and applying a change of basis to them to obtain $$[T]_{B_1,B_2} = \pmatrix{1&0&0&1\\0&1&0&2\\0&0&1&3\\-1&0&0&4}^{-1}\pmatrix{2&2&2\\2&-1&1\\0&0&2\\-5&-2&-1}.$$

Answer 2

Let indicate with $M$ the matrix from the standard basis to $B_1$ (it has the vector of the new basis as columns) that is

$$v_{S_1}=M v_{B_1}$$

and let indicate with $N$ the matrix from the standard basis to $B_2$ (it has the vector of the new basis as columns) that is

$$w_{S_2}=N w_{B_2}$$

then

$$w_{S_2}=T_{S_1,S_2}v_{S_1}\implies N w_{B_2}= T_{S_1,S_2} M v_{B_1}\implies w_{B_2}= N^{-1}T_{S_1,S_2} M v_{B_1}$$

and therefore

$$T_{B_1,B_2}= N^{-1}T_{S_1,S_2} M$$