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My graduate student instructor for my linear algebra class provided this problem that piqued my interest:

Let $V$ be the vector space of ${2 \times 2}$ matrices. Let $T$ be an operator on $V$ such that ${T(A) = CA^T}$ , note that $C$ is given by ,\begin{bmatrix}2&-1\\0&1\end{bmatrix} Find $M(T)$, that is, the matrix of the transformation. I've looked at the problem for a while and I can't quite figure it out, anybody wanna give it a shot and talk me through it?

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  • $\begingroup$ Find the action on a basis for $V$. $\endgroup$
    – copper.hat
    Mar 28, 2022 at 4:44
  • $\begingroup$ Before talking about the matrix representation of a linear operator, one needs to choose a basis first. The matrix will depend on the chosen basis. There isn't a unique answer. $\endgroup$
    – user1551
    Mar 29, 2022 at 17:27
  • $\begingroup$ He said the matrix would be a 4 by 4 but is that true? Because then the matrix multiplication would not make any sense $\endgroup$ Mar 30, 2022 at 20:12
  • $\begingroup$ @user1551 won't we take standard basis if not stated? $\endgroup$ Apr 1, 2022 at 3:53
  • $\begingroup$ @CarsonNewman yes matrix will be $4 \times 4$ matrix. $T:V \to W $ is linear transformation with $ dim(V)=m , dim(W)=n$ than corresponding matrix will be $n \times m$. Here dimension of domain is 4 and that of co domain is also 4. $\endgroup$ Apr 1, 2022 at 4:00

1 Answer 1

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Let {$e_1,e_2,e_3,e_4$} is basis of $M_{2 \times2}(R)$, namely

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

now $T(e_1)=Ce_1^T=Ce_1$ since $e_1$ transpose is $e_1$ itself.

$$ \left[ \begin{matrix} 2&-1\\ 0&1\\ \end{matrix} \right] \left[ \begin{matrix} 1&0\\ 0&0\\ \end{matrix} \right] =\left[ \begin{matrix} 2&0\\ 0&0\\ \end{matrix} \right] $$

now write it in basis of $$M_{2 \times 2 }(R)$$ which will give you first column of Matrix which will be $4 \times 4 $ matrix.

Proceed similarly for $e_2,e_3,e_4$.

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  • $\begingroup$ so then the matrix would be: $$ \begin{matrix} 2 & -1 & 0 & 0 \\ 0 & 1 & 0 & 0\\ 0 & 0 & 2 & -1\\ 0 & 0 & 0 & 1\\ \end{matrix} $$ $\endgroup$ Mar 28, 2022 at 17:32

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