# Show that there exists an invertible $S$ for $A^2=0$ such that $S^{-1}AS=\begin{bmatrix} 0 & 0\\ 1 & 0 \end{bmatrix}$

$A$ 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 $A$?

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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

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 $\vec x$, $\vec y$.)

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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 $X$.

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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