Let $T: M_{2×2} \to \mathbb R^3$ have matrix $[T]_{B,A}=\begin{bmatrix}1&2&0&1\cr 0&1&1&0\cr 1&1&-1&-1\end{bmatrix}$ relative to $A=\{\begin{bmatrix}2&0\cr 0&0\end{bmatrix},\begin{bmatrix}0&3\cr 0&0\end{bmatrix},\begin{bmatrix}0&0\cr 5&0\end{bmatrix},\begin{bmatrix}0&0\cr 0&6\end{bmatrix} \}$ and $\beta=\{(1,1,1),(1,2,3),(1,4,9)\}$. Find the matrix of T relative to the bases $A'=\{\begin{bmatrix}1&0\cr 0&0\end{bmatrix},\begin{bmatrix}0&4\cr 0&0\end{bmatrix},\begin{bmatrix}0&0\cr 2&0\end{bmatrix},\begin{bmatrix}0&0\cr 0&7\end{bmatrix} \}$ and $\beta'=\{(1,1,1),(1,0,0),(1,1,0)\}$
To solve this, I've tried to find the transformation equation and got :
$$T(\begin{bmatrix}2&0\cr 0&0\end{bmatrix})=(2,5,10)$$
$$T(\begin{bmatrix}0&3\cr 0&0\end{bmatrix})=(4,8,9)$$
$$T(\begin{bmatrix}0&0\cr 5&0\end{bmatrix})=(2,5,12)$$
$$T(\begin{bmatrix}0&0\cr 0&6\end{bmatrix})=(0,-3,-8)$$
and we can let $X=(x,y,z)\in\mathbb R^3$ such that $X=(x,y,z)=x(1,0,0)+y(0,1,0)+z(0,0,1)$
where x,y,and z are constants
Now I'm supposed to find what the constants are to find the transformation equation and I assume I have to use the derived equation to work backwards on the second part of the problem to find the other transformation matrix. Both of which I don't have the slightest clue how to do and would like help to do so.
Best Answer
First of all, notice that $M_{2x2} \cong R^4 $ with $ \begin{bmatrix}a&b\cr c&d\end{bmatrix} \to (a,b,c,d)$.
Your first step is the right way to go, but I would be careful with the notation. Let $\alpha$ be the standard basis auf $\mathbb{R}^3$, then
$$T(\begin{bmatrix}2&0\cr 0&0\end{bmatrix})=(1,0,1)_B = (2,5,10)_\alpha =(10,-3,-5)_{B'}$$
$$T(\begin{bmatrix}0&3\cr 0&0\end{bmatrix})=(2,1,1)_B = (4,8,9)_\alpha =(9,-4,-1)_{B'}$$
$$T(\begin{bmatrix}0&0\cr 5&0\end{bmatrix})=(0,1,-1)_B =(2,5,12)_\alpha =(12,-3,-7)_{B'}$$
$$T(\begin{bmatrix}0&0\cr 0&6\end{bmatrix})=(1,0,-1)_B =(0,-3,-8)_\alpha =(-8,3,5)_{B'}$$
Thus, we have
$$[T]_{B',A} = \begin{bmatrix}10&9&12&-8\cr -3&-4&-3&3\cr -5&-1&-7&5\end{bmatrix}$$
Now we are almost done. The new basis $A'$ is good to handle because every new basis vector is a scalar multiple of a vector of $A$. For example the first basis vector of $A'$ is half of its counterpart of $A$. (So $(2,0,0,0)_{A'} = (1,0,0,0)_A$ ). So if you map $(1,0,0,0)_{A'}$ with our wanted matrix, it will have the same result as if we map $(\frac{1}{2},0,0,0)_A$ into $[T]_{B',A}$ (because they both map into the vector space generated by $B'$) Thus, the first column of $[T]_{B',A'}$ is $\begin{pmatrix}5\cr-\frac{3}{2}\cr-\frac{5}{2}\end{pmatrix}$. You can get the other columns with the same method. The solution is:
$$[T]_{B',A'} = = \begin{bmatrix}5&9\frac{4}{3}&12\frac{2}{5}&-8\frac{7}{6}\cr -\frac{3}{2}&-4\frac{4}{3}&-3\frac{2}{5}&3\frac{7}{6}\cr \frac{5}{2}&-1\frac{4}{3}&-7\frac{2}{5}&5\frac{7}{6}\end{bmatrix}$$