Let $T\colon \mathbb{R}^3 \to \mathbb{R}^{2×2}$ 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?
Best Answer
$B_1$ 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 $$[T]_{B_1,B_2} = \pmatrix{1&0&0&1\\0&1&0&2\\0&0&1&3\\-1&0&0&4}^{-1}\pmatrix{2&2&2\\2&-1&1\\0&0&2\\-5&-2&-1}.$$