Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

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

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

share|improve this question
    
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? –  Bunuelian Trick 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 $a_{13}$ –  Bunuelian Trick 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
show 2 more comments

2 Answers 2

up vote 3 down vote accepted

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 $S_P$.

share|improve this answer
add comment

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 $\dim S_p=3$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.