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Define for a fixed $A \in \mathbb{M}^{2 \times 2}(\mathbb{R})$ the mapping:

$$L_A : \mathbb{M}^{2 \times 2}(\mathbb{R}) \to \mathbb{M}^{2 \times 2}(\mathbb{R}) : X \mapsto AX-XA. $$ Write $M_A$ for the matrix such that for all $X \in \mathbb{M}^{2 \times 2}(\mathbb{R})$ it satisfies $[L_A (X)]_\xi=M_A [X]_\xi$. Here is $[\cdot ]_\xi$ the coordinate map that belongs to the standard basis

$$\xi = \{ E_1 = {\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}}, E_2={\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}}, E_3={\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}}, E_4={\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}} \}$$ of $\mathbb{M}^{2 \times 2}(\mathbb{R})$.

Calculate explicitly the matrices $M_{E_1}, M_{E_2}, M_{E_3}, M_{E_4}$.

I don't have a clue. All I get is the zero-matrix that satisfies the condition, for all $M_{E_i}$.

I wrote out $[L_{E_1} (X)]_\xi$ as follows:

$$\begin{align*} [L_{E_1} (X)]_\xi &= [E_1 X - X E_1]_\xi \\ &= \left[{\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}} {\begin{pmatrix} a & b \\ c & d \end{pmatrix}} - {\begin{pmatrix} a & b \\ c & d \end{pmatrix}} {\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}} \right]_\xi \\ &= \left[ {\begin{pmatrix} 0 & b \\ -c & 0 \end{pmatrix}} \right]_\xi \\ &= b {\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}} -c {\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}} \\ &= bE_3 - cE_2 \end{align*}$$

How do I continu to find $M_{E_1}$?

Thanks in advance.

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Remember that given your basis $\xi$, for $B \in M_2(\mathbb{R})$, $$ [B]_\xi = \begin{pmatrix}B_{11} \\ B_{21} \\ B_{12} \\ B_{22} \end{pmatrix}; $$ in particular, for $A \in M_2(\mathbb{R})$, $M_A := [L_A]_{\xi \xi}$ is the matrix of the linear transformation $L_A : M_2(\mathbb{R}) \to M_2(\mathbb{R})$ with respect to $\xi$, and hence a $4 \times 4$ matrix. Indeed, by introductory linear algebra, $M_A$ will be the $4 \times 4$ matrix whose $k$-th column is $[L_A(E_k)]_\xi$, as given by the equation above. So, given $A \in M_2(\mathbb{R})$, all you really need to do is compute $L_A(E_k)$ for each $k$, and express it as a linear combination of the $E_j$.

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Hint:do you know that the matrix you are looking for is $4\times4$?
Now you are looking for a $4\times4$ matrix such that $M_{E_1}[X]_\xi=[L_{E_1}(X)]_\xi$. In your case, as you calculated, $[L_{E_1}(X)]_\xi=(0,b,-c,0)$ column vector, while $[X]_\xi=(a,b,c,d)$ column. Do you see how to proceed?

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  • $\begingroup$ Thank you so much, this clarifies a bunch ! $\endgroup$ – Jeroen Apr 7 '13 at 20:56

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