# Finding the matrix of an interesting transformation

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?

• Find the action on a basis for $V$. Mar 28, 2022 at 4:44
• 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. Mar 29, 2022 at 17:27
• He said the matrix would be a 4 by 4 but is that true? Because then the matrix multiplication would not make any sense Mar 30, 2022 at 20:12
• @user1551 won't we take standard basis if not stated? Apr 1, 2022 at 3:53
• @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. Apr 1, 2022 at 4:00

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 $$e_2,e_3,e_4$$.

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