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.

$A$ is a nonzero $n \times n$ matrix such that $A^2=0$. If $n=2$, is it possible that I can show that there exists an invertible $2 \times 2$ matrix $S$ such that $S^{-1}AS=\begin{bmatrix} 0 & 0\\ 1 & 0 \end{bmatrix}$?

My first approach to this was to rearrange the equation so that it appears as $A=S\begin{bmatrix} 0 & 0\\ 1 & 0 \end{bmatrix}S^{-1}$ but it is kind of weird because this would mean $\begin{bmatrix} 0 & 0\\ 1 & 0 \end{bmatrix}$ is the eigenvalue matrix. And this isn't a diagonal matrix.

I got no more idea and so I looked at the hint and it says I could let $\vec{u} \in \mathbb{R}^n$ such that $A\vec{u}\neq0$ and then make $S=\begin{bmatrix} \vec{u} & A\vec{u} \end{bmatrix}$ to continue from here. But I don't understand how can I just anyhow throw in a vector in $\mathbb{R}^n$ into the eigenspace matrix $S$ without even confirming that $\vec{u}$ is a eigenvector of matrix $A$?

share|improve this question
    
The title doesn't make much sense to me. It says "Show that $S$ is invertible, if $A^2=0$ and $S^{-1}AS=...$. But if we do not know $S$ to be invertible, what is $S^{-1}AS$ supposed to mean? –  Oliver Braun Nov 5 '11 at 11:01
    
I kind of summarised the title from the actual question that I made in the post. So it may not be concise enough. That's how the book writes it. But could it be wrong? –  xenon Nov 5 '11 at 11:05
    
Who said that $u$ is an eigenvector of $A$? –  Phira Nov 5 '11 at 11:08
    
It didn't say that $\vec{u}$ is an eigenvector of $A$ but since $S$ is the eigenspace of $A$, and by throwing another vector into $A$, aren't we assuming that the vector is also a eigenvector? Otherwise, what is the meaning of putting the extra $\vec{u}$ into the eigenspace matrix $S$? Wouldn't it alter the whole equation? –  xenon Nov 5 '11 at 11:11

2 Answers 2

up vote 3 down vote accepted

This is basically the same answer as given by Listing, but formulated in "different language" - you can choose what suits you best.

Choose any non-zero vector $\vec x$ such that $\vec x A=\vec y$ is non-zero. (Note that I am using row vectors.)

Not that $\vec y A = \vec x A^2=0$. This also implies that the vectors $\vec x$ and $\vec y$ are linearly independent.

Now we have $$\begin{pmatrix} \vec y \\ \vec x \end{pmatrix} A = \begin{pmatrix} \vec 0 \\ \vec y\end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} \vec y \\ \vec x\end{pmatrix}.$$ If we denote $P=\begin{pmatrix} \vec y \\ \vec x\end{pmatrix}$, then the matrix $P$ is regular and $PA=\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}P$, i.e. $PAP^{-1}=\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}$.

(Basically, we just asked what is the matrix of the linear map $A$ in the basis $\vec x$, $\vec y$.)

share|improve this answer
    
Thanks! But since $P$ is an eigenspace matrix, how can we just anyhow put in another vector that isn't an eigenvector as one of its columns? –  xenon Nov 5 '11 at 13:13
    
@xEnOn In fact $y$ is an eigenvector, since $yA=0.y$ and $x$ is a generalized eigenvector, since $xA=y+0.x$. (Both for eigenvalue 0.) –  Martin Sleziak Nov 5 '11 at 13:56
    
Where does it say that $P$ is an eigenspace matrix? Since $x$ is not an eigenvector it clearly isn't. –  Sebastian Schoennenbeck Nov 5 '11 at 13:59

Let $A$ be such that $A^2=0$, then certainly $A$ can only have the eigenvalues $0$, and therefore, by the jordan normal-form, there exist $H$ invertible such that $A=H^{-1}\begin{bmatrix} 0 & 1\\ 0 & 0 \end{bmatrix}H$.

As $X=\begin{bmatrix} 0 & 0\\ 1 & 0 \end{bmatrix}$ has also only the eigenvalues $0$ there exist a $P$ invertible such that

$X=P^{-1}\begin{bmatrix} 0 & 1\\ 0 & 0 \end{bmatrix}P$

Now by transitivity of similar matrices your theorem is immediate, because $A$ is similar to $X$.

share|improve this answer
1  
If eigenvalues are zero, there are two possibilities for Jordan form of the matrix - the zero matrix and $\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$. This does not change your answer substantially, but you should included both possibilities. (In fact this case seems to be the important one - any matrix similar to zero matrix is zero. So perhaps I should have said that this is the only possibility for the Jordan form.) –  Martin Sleziak Nov 5 '11 at 11:39
    
Thank you, I changed it to use the jordan-decomposition as you suggested. –  Listing Nov 5 '11 at 11:43

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.