Let $V$ be the set of all $2\times2$ real matrices:
a) Show that $V$ is a $4$-dimensional vector space and find a basis.

I have the following:

$$\begin{bmatrix} 1 &0 \\ 0&0 \end{bmatrix}$$ $$\begin{bmatrix} 0 &0 \\ 0&1 \end{bmatrix}$$ $$\begin{bmatrix} 0&1 \\ 0&0 \end{bmatrix}$$ $$\begin{bmatrix} 0&0 \\ 1&0 \end{bmatrix}$$ $$\begin{pmatrix} 1 &0 &0 &1 \\ 0&0 &0 &0 \\ 0 &0 &0 &0 \\ 1&0 &0 &1 \end{pmatrix}$$

b) Let $C$ be a $2\times2$ matrix. Show that $L:V\rightarrow V$ given by $L(X)=CX$ and $R:V\rightarrow V$ given by $R(X)=XC$ are linear mappings

I figure I should show distributivity, but how?

  • 2
    $\begingroup$ I don't know what you mean by your fifth matrix. A base is simply made of the first four. For the question b), prove the defining properties of a linear mapping. Which are they? $\endgroup$ – Jean-Claude Arbaut Sep 26 '16 at 6:16

For b, you want to show $$L(X+Y)=L(X)+L(Y),\quad L(aX)=aL(X) $$ i.e., $$ C(X+Y)=CX+CY,\quad aCX=CaX$$ i..e, that matrix multiplication distributes over matrix addition, and that scalar $aI$ cmmutes with all matrices. But this is already a consequence of matrices representing linear maps: For each $v\in\Bbb R^2$, $C(X+Y)v=C(Xv+Yv)=CXv+CYv$ and $aCXv=CXav=CaXv$.

If this feels too abstract, just perform the explicit matrix multiplications required ...


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