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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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Hint: Recalculate the eigenvalues. They are much nicer than you think. Then find a basis of eigenvectors. The problem will fall apart.

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    @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.2012-06-28