# How to show growth without bound in only certain cases and not in others?

I encountered the following problem. (We're working in a finite-dimensional real vector space, here.)

Suppose $$A=\frac{1}{2}\left(\begin{array}{cc}-2 & 4\\1 & 1\end{array}\right).$$ Find a vector, $y$, so that $\lVert A^nx\rVert\to\infty$ as $n\to\infty$, except when $x$ is perpendicular to $y$. Explain your reasoning.

My first thought was that we should try to express $A=zy^t$ for some vectors $y,z$, whence, if $x$ is perpendicular to $y$, we would have $$Ax=z(y^tx)=z\langle y,x\rangle=0,$$ and so $\lVert A^nx\rVert$ cannot grow without bound. Then, I'd hoped to show that for any other $x$, it did grow without bound. Unfortunately, when I assumed such $y,z$ existed and attempted to see what I could discern about them, I ended up with a $0\neq 0$ contradiction. It's possible I made a mistake, but I'm not finding one, and if there isn't a mistake, then that approach is not going to work.

I also found the eigenvalues of $A$--namely $\frac{-1}{8}\left(1\pm\sqrt{257}\right)$--both of which have modulus greater than $1$, so clearly $\lVert A^n\rVert$ grows without bound. Unfortunately, I'm not sure what else I can do from here. Any hints?

As a secondary question, for which matrices $A$ can we say that $a=zy^t$ for some $y,z$? (Of course, if it's all of them, then I goofed, and my first thought works after all.)

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But $y,z$ are $2\times 1$, so $y^t$ is $1\times 2$, and $zy^t$ is $2\times 2$. Why should all the entries be equal? –  Cameron Buie Jun 28 '12 at 22:29
If $$y=z=\left(\begin{array}{c}1\\2\end{array}\right),$$ then $$zy^t=\left(\begin{array}{cc}1 & 2\\2 & 4\end{array}\right).$$ –  Cameron Buie Jun 28 '12 at 22:40
@CameronBuie: You are right. In general $(a,b)$, $(c,d)$ yield matrix with entries first row $(ac,ad)$, second $(bc,bd)$. In particular the determinant is $0$. So we certainly do not get all. –  André Nicolas Jun 28 '12 at 22:45