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I'm struggling to find the nullity $N(T)$ of the following linear transformation (in the canonical basis of $\mathbb{R^{2\times2}}$

$ M = \begin{bmatrix} 0 & 0 & 0 & 0 \\[0.3em] -1 & -3 & 3 & 0 \\[0.3em] 0 & 0 & 1 & 0 \\[0.3em] 0 & 0 & 0 & 1 \end{bmatrix} $

What's making me feel confused is that we are considering a basis of matrices. I knew how proceed in the case of for example the canonical basis of $\mathbb{R^2}$ (just multiply by a column vector (a,b,c,d) and the I would have a system of 4 linear equations (equal to zero)). But the presence of matrices 2x2 is driving me confused because I can't even try to multiply my matrix of the linear transformation to a matrix 2x2

Am I making myself clear? If you could help me, please...

Thanks

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2 Answers 2

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Hint:

A matrix $$ A= \begin{bmatrix} a&b\\c&d \end{bmatrix} $$ is represented in the canonical basis as: $$ A=a\begin{bmatrix} 1&0\\0&0 \end{bmatrix} +b\begin{bmatrix} 0&1\\0&0 \end{bmatrix} +c\begin{bmatrix} 0&0\\1&0 \end{bmatrix} +d\begin{bmatrix} 0&0\\0&1 \end{bmatrix} $$ so, in this basis, you can represent it as a vector: $$ \vec A= \begin{bmatrix} a\\b\\c\\d \end{bmatrix} $$ So your matrix $M$ acts as: $$ M\vec A= \begin{bmatrix} 0 & 0 & 0 & 0 \\ -1 & -3 & 3 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} a\\b\\c\\d \end{bmatrix} $$ and you have to solve : $$ \begin{bmatrix} 0 & 0 & 0 & 0 \\ -1 & -3 & 3 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} a\\b\\c\\d \end{bmatrix}=0 $$

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If $\mathbf{x}=(x,y,z,w)$, then right away you should see that $z=w=0$ and $x=-3y$. So the nullity has basis: $(-3,1,0,0)$ and is of dimension $1$.

If $\mathbf{x}$ is really a $2\times 2$ matrix, then you can think of it as $\left[\begin{matrix} x & y \\ z & w\end{matrix}\right]$. Converting between a matrix and its representation as a vector should be clear.

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