# find the dimension of vector subspace $S_p$ of $M_3(\mathbb{R})$

I need to find the dimension of vector subspace $S_p$ of $M_3(\mathbb{R})$ we have fixed a matrix $P$ which is singular and $S_p=\{X: PX=0\}$ I defoned a linear map from $M_3(\mathbb{R})\to M_3(\mathbb{R}), X\to PX$ but not able to get the $\dim Image$, I mean I see the map is not onto as $P$ is singular, could any one tell me other way to find the dimension? actually $P$ was given a specific matrix which I am going to write. I wanted to apply rank nulity theorem.

$P=\begin{pmatrix}1&0&-1\\0&1&0\\1&1&-1\end{pmatrix}$

$PX=\begin{pmatrix}1&0&-1\\0&1&0\\1&1&-1\end{pmatrix}\times \begin{pmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{pmatrix}=\begin{pmatrix}a_{11}-a_{13}&a_{12}-a_{32}&a_{13}-a_{33}\\a_{21}&a_{22}&a_{23}\\a_{11}+a_{21}-a_{31}&a_{12}+a_{22}-a_{32}&a_{13}+a_{23}-a_{33}\end{pmatrix}$

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Why don't you just write down $PX=0$ explicitely in coordinates for an arbitrary $X$ and your specific $P$? – Julian Kuelshammer Jun 4 '13 at 19:35
@JulianKuelshammer I have done this! what can I conclude now? – Un Chien Andalou Jun 4 '13 at 19:48
Now you set that $=0$ and than you have a system of linear equations of which you can probably compute the dimension of the space of solutions. – Julian Kuelshammer Jun 4 '13 at 19:50
@JulianKuelshammer I am getting $X$ a atrix whose first row are same and has entry $a_{11}$, second row $a_{12}$ and third row $a_{13}$ – Un Chien Andalou Jun 4 '13 at 20:00
Sorry, didn't look into your matrix multiplication. It is incorrect, what did you compute? – Julian Kuelshammer Jun 4 '13 at 20:07

Writing the comments into an answer: We have $S_P=\{X|PX=0\}$. Write $X$ with entries $x_{ij}$. Then $PX=0$ leads to $x_{2i}=0$ and $x_{1j}=x_{3j}$. So you have that the dimension of the space of solutions is $3$, which is the same as the dimension of $S_P$.

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The only computation you need is that of the rank of $P$, or rather its nullity, which is the same modulo the rank-nullity theorem.

Note that $S_p$ is just the subspace of all matrices with range contained in $\ker P$. It is therefore isomorphic to $L(\mathbb{R}^3,\ker P)$, the vector space of all linear maps from $\mathbb{R}^3$ to $\ker P$.

So the dimension is $3\cdot \dim\ker P$, as $\dim L(E,F)=\dim E\cdot \dim F$ in general with finite dimensional vector spaces.

In this case, it is clear that $\dim\ker P=1$ whence $\dim S_p=3$.

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