How do I perform matrix rotation in 3-space? I have been asked the following question in a tutorial:

Let A be a 3x3 matrix which is invertible. Show that you can always perform a rotation of 3-space to make the last row of A be [0 0 A33]

I haven't the faintest idea how to do this and have been deeply confused during lectures. I have tried speaking to my lecturer in person when I've previously had problems, but I never understand him. (He's new and has difficulty pitching at first-year level.)
Could anyone here show me how to do this and explain, as simply and clearly as possible, how they derived the answer? (I need to be able to apply this in a test this Thursday.) You guys are my last resource.

You need to show that A has at least one eigenvector. That is, that the equation $det(A-\lambda I)=0$ has at least one real solution (note that this is a third-degree polynomial in $\lambda$). Showing this, you have to calculate one eigenvector $v$ corresponding to this $\lambda$ (solving $A\cdot v=\lambda v$). Since $det(A-\lambda I)=0$, the equation has nontrivial solutions. Once you have this vector, just make a rotation that takes one of the canonical vectors to $\frac{v}{\mid v \mid}$.
You will obtain a matrix that has the last column $[ 0 0 \lambda ]$. To obtain the desired result, you have to apply the method above to $A^t$ (note that $A$ and $A^t$ have the same eigenvalues.)
