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Let $M_2(\mathbb {F})$ denote the set of all $2 \times 2$ matrices over the field $\mathbb {F}$. Observe that $M_2(\mathbb {F})$ becomes a vector space if addition and scalar multiplication are defined by $ \begin{bmatrix} x_1 & x_2 \\ x_3 & x_4 \end{bmatrix} + \begin{bmatrix} y_1 & y_2 \\ y_3 & y_4 \end{bmatrix} = \begin{bmatrix} x_1+y_1 & x_2+y_2 \\ x_3+y_3 & x_4+y_4 \end{bmatrix} $

$ \lambda * \begin{bmatrix} x_1 & x_2 \\ x_3 & x_4 \end{bmatrix} = \begin{bmatrix} \lambda*x_1 & \lambda*x_2 \\ \lambda*x_3 & \lambda*x_4 \end{bmatrix} $

Let $\mathcal{E} = \{E_1, E_2, E_3, E_4\}$ where $E_1 = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$, $E_2 = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$, $E_3 = \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}$, $E_4 = \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}$, and let $A = \begin{bmatrix} 1 & 0 \\ -1 & 1 \end{bmatrix}$ and let $S: M_2(\mathbb {F})\rightarrow M_2(\mathbb {F})$ be the linear mapping $S(X) = AX - XA$. Find $M(S) _\mathcal{E} ^ {\mathcal{E}} $.

My attemp to solve this was:

I first tried to investigate where the given mapping sends $\mathcal{E}$:

$S(E_1) = \begin{bmatrix} 0 & 0 \\ -1 & 0 \end{bmatrix} = -E_3$

$S(E_2) = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} = E_1-E_4$

$S(E_3) = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = E_n - E_n$ where $1 \leq n \leq 4$

$S(E_4) = \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} = E_3$

but then I am not sure as to how present $M(S) _\mathcal{E} ^ {\mathcal{E}} $. I initially thought it would be: $M(S) _\mathcal{E} ^ {\mathcal{E}} = $\begin{bmatrix} 0 & 1& 0 & 0 \\ 0 & 0 & 0 & 0 \\ -1 & 0 & 0 & 1 \\ 0 & -1 & 0 & 0 \end{bmatrix} (i.e. I wrote out coefficients in front of $E_n$ in columns as this is how we have been taught)

but then this cannot be true since as fas as I understand $M(S) _\mathcal{E} ^ {\mathcal{E}} $ has to be a $2 \times 2$ matrix.

Thank you in advance!

  • 4
    You are correct until you say $M(S)$ is a $2x2$ matrix. It is a $4x4$ matrix as it represents a linear map from a $4$-dimensional vector space to itself. – 2017-01-02
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    But should not we be able to obtain, for example $S(E_1)$ by finding the product of $M(S)$ and $E_1$? If yes, then how can we multiply 4x4 matrix by 2x2? – 2017-01-02
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    Given the basis $(E_1,E_2,E_3,E_4)$, $E_1$ corresponds to the vector $(1,0,0,0)^T$. Applying $M(S)$ gives the vector $(0,0,-1,0)^T$ which corresponds to $-E_3$ under the basis $(E_1,E_2,E_3,E_4)$. – 2017-01-02
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    Oh right, thank you! However, why are we not representing $E_1$ as a column vector? When thinking about it I actually did the following: $M(S)*E_1$ where I represented $E_1$ as a column vector $(1,0,0,0)$. But then it did not works as I got all 0s when multiplying the matrices. Could you please explain why in multiplication we first write the vector and not the transformation? – 2017-01-05

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