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I am quite confused how the transformation matrix looks like in standard basis when the linear map is defined on square matrices.

Given a mapping $T: M_2(\mathbb{R}) \rightarrow M_2(\mathbb{R}) $ defined as follows:

\begin{equation} T \left( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \right) = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 3 & -1 \end{bmatrix}, \end{equation}

show that $T$ is linear. Find a transformation matrix in standard basis.

I have successfully proved that T is indeed linear. I know that the standard basis of $M_2(\mathbb{R})$ is:

\begin{equation} e_1=\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}, e_2=\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}, e_3=\begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}, e_4=\begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} \end{equation}

Applying T on each of the basis vectors gives me: \begin{equation} T(e_1)=\begin{bmatrix} 1 & 2 \\ 0 & 0 \end{bmatrix}, T(e_2)=\begin{bmatrix} 3 & -1 \\ 0 & 0 \end{bmatrix}, T(e_3)=\begin{bmatrix} 0 & 0 \\ 1 & 2 \end{bmatrix}, T(e_4)=\begin{bmatrix} 0 & 0 \\ 3 & -1 \end{bmatrix} \end{equation}

How to proceed from here?

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Hint: $T(e_1) = e_1 + 2 e_2 + 0 e_3 + 0e_4$. $T(e_2) = 3e_1 - e_2 + 0 e_3 + 0e_4$. etc. Write out each image of the standard basis in terms of the vector space basis. The coefficients of $T(e_i)$ will form a column of the matrix representation. Your matrix should have 16 elements.

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