How to find the transition matrix for ordered basis of 2x2 diagonal matrices

Question

The problem:
For the vector space of lower triangular matrices with zero trace, given ordered basis: $B=${$$ \begin{bmatrix} -5 & 0 \\ 4 & 5 \\ \end{bmatrix}, $$ \begin{bmatrix} -1 & 0 \\ 1 & 1 \\ \end{bmatrix}}

and $C=${$$ \begin{bmatrix} -5 & 0 \\ -4 & 5 \\ \end{bmatrix}, $$ \begin{bmatrix} -1 & 0 \\ 5 & 1 \\ \end{bmatrix}}

find the transition matrix $C$ to $B$.

I know how to find a transition matrix when the basis consists of $n \times 1 $ vectors, but my textbook doesn't address this scenario where the basis consists of a set of $2 \times 2$ matrices and haven't found applicable guidance online.

Answer 1

Hint:

If you know how to solve the problem for $n\times 1$ vectors than consider that your matrices can be considered as vectors of a vector space with standard basis the $2\times 2$ matrices that have only one element $=1$ and the other elements $=0$. In this basis, the matrix: $$ \begin{bmatrix} -5 & 0 \\ 4 & 5 \\ \end{bmatrix} $$ is the vector $$ \begin{bmatrix} -5 \\ 0 \\ 4\\ 5 \\ \end{bmatrix} $$

You can do the same for the other matrices and solve the problem as for usual vectors, but note that yours sets $B$ and $C$ are not basis for $M_{2\times 2}(\mathbb{R})$.

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If we work in the space of lower triangular and null trace matrix in $M_{2\times 2}(\mathbb{R})$, than this subspace has dimension $2$ and any matrix in it can be expressed as $$ \begin{bmatrix} a&0\\ b&-a \end{bmatrix} =a \begin{bmatrix} 1&0\\ 0&-1 \end{bmatrix} +b \begin{bmatrix} 0&0\\ 1&0 \end{bmatrix} =a\hat i +b \hat j $$ so $a$ $b$ can be seen as the componets of a vector $(a,b)^T$ in the basis $\{\hat i, \hat j\}$.

In this notation your basis are: $$ B=\{(-5,4)^T,(-1,1)^T\} \qquad B=\{(-5,-4)^T,(-1,5)^T\} $$

Now you can find the $2\times2$ matrix that represents the transformation (in the basis$\{\hat i, \hat j\}$) solving:

$$ \begin{pmatrix} x&y\\z&t \end{pmatrix} \begin{pmatrix} -5\\4 \end{pmatrix}= \begin{pmatrix} -5\\-4 \end{pmatrix} $$ and

$$ \begin{pmatrix} x&y\\z&t \end{pmatrix} \begin{pmatrix} -1\\1 \end{pmatrix}= \begin{pmatrix} -1\\5 \end{pmatrix} $$

Answer 2

Thank you for your direction. I was able to use your ideas to find the correct solution to the problem. First I expressed B and C in terms of the basis

$\left[ \begin{matrix} -1 & 0 \\ 0 & -1 \end{matrix} \right]$,$\left[ \begin{matrix} 0 & 0 \\ 1 & 0 \end{matrix} \right]$

The basis B is thus equivalent to $\left[ \begin{matrix} -5\\ 4 \end{matrix} \right]$,$\left[ \begin{matrix} -1 \\ 1\end{matrix} \right]$

And the basis C to $\left[ \begin{matrix} -5\\ -4 \end{matrix} \right]$,$\left[ \begin{matrix} -1 \\ 5\end{matrix} \right]$

Then I computed the transition matrix from C to B by multiplying the inverse of B times C:

$\left[ \begin{matrix} -5 & -1 \\ 4 & 1 \end{matrix} \right]^{-1}$$\left[ \begin{matrix} -5 & -1 \\ -4 & 5 \end{matrix} \right]$=$\left[ \begin{matrix} 9 & -4 \\ -40 & 21 \end{matrix} \right]$