Let A=$\left[ \begin {array}{cc} 0&-1\\ 1&0\end {array} \right] $
$L:Mat_2\mathbb{R}\rightarrow Mat_2\mathbb{R}$
$B\rightarrow AB-BA$
(1) Show that it is a linear transformation.
I really have no idea how to this, expect I know that:
$L(u+w)=L(u)+L(w)$
$\alpha*L(v)=L(\alpha*V)$
Can anyone help me prove this?
Let $B,C \in Mat_2\Bbb R$, $\alpha,\beta \in\Bbb R$.
$\begin{align}L(\alpha B + \beta C)& = A(\alpha B + \beta C)-(\alpha B + \beta C)A\\ & = \alpha AB + \beta AC - \alpha BA - \beta CA \\ & = (\alpha AB-\alpha BA) + (\beta AC - \beta CA) \\ & = \alpha(AB-BA) + \beta (AC-CA) \\ & = \alpha L(B) + \beta L(C) \end{align}$
This proves linearity.
We have: $$ L(B) = AB - BA $$ Example: Then for any $B \in \text{Mat}_2\mathbb{R}$, $\alpha \in \mathbb{R}$ we have: $$ L(\alpha B) = A(\alpha B) - (\alpha B) A = \alpha AB - \alpha BA = \alpha(AB - BA) = \alpha L(B) $$
The replies by Giulio and mvw are, of course, exactly correct! But, as a teacher, I believe that giving a formal answer to this question can sometimes make a simple concept seem complicated.
Here's a lesson where I try to explain linearity simply: http://lem.ma/Yz I think that if you understand how simple linearity is, you will be able to do the formal proof with great ease. Otherwise, the proof will continue to look like an a vexing stream of characters. (I've certainly been there.)
If you understand linearity, you will see $AB-BA$ as obviously linear in $B$. Maybe it'll help to see examples that are not linear in $B$: $$AB^2-B^2A$$ $$AB-BA+I$$ $$AB^{-1}-BA$$ $$A(B+I)-(B+I)A$$
Let me know if this, combined with the video, helps and, if not, we'll try to clarify the issue!