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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.

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2 Answers 2

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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 $M_{2\times 2}(\mathbb{R})$.


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

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

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  • $\begingroup$ Ok, and the resulting transition matrix will be 2x2? $\endgroup$
    – Aaron S
    Oct 1, 2016 at 0:48
  • $\begingroup$ I understand the strategy you're suvgesting but don't you need four 2x2 basis matrices in each set to compute the transition matrix? The transition matrix is 4x4 with four 2x2 basis matrices while the answer to the problem has to be a 2x2 matrix. $\endgroup$
    – Aaron S
    Oct 1, 2016 at 3:32
  • $\begingroup$ Should mention the vector space is lower triangular 2x2 matrices with trace zero... $\endgroup$
    – Aaron S
    Oct 1, 2016 at 3:35
  • $\begingroup$ YES, that's important. I add to my answer. $\endgroup$ Oct 1, 2016 at 13:24
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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]$

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  • $\begingroup$ Lets say I have the coordinates of M in the basis B. How would I find M? $\endgroup$
    – wheeler
    Aug 5, 2020 at 2:54

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