# $2\times2$ matrix $A$ such that $A$ has one independent eigenvector while $A^{2}$ has two independent eigenvectors

Give an example of $2\times2$ matrix $A$ such that $A$ has one independent eigenvector while $A^{2}$ has two independent eigenvectors.

I would like to know a systematic answer of how to get this. My guess was that any matrix that has 1 independent eigenvector will have the same number of independent eigenvector for the squared matrix. I have tried random selections but obviously couldn't come up with one. What's more useful will be to get the intuition behind this problem.

## 1 Answer

A matrix $A$ with these requirements must be non-diagonalizable, that is, similar to a nontrivial $2\times 2$ Jordan block $$J=\pmatrix{\alpha&1\\0&\alpha}.$$ It is easy to see that $J^2$ is diagonalizable (and hence has 2 independent eigenvectors) iff $\alpha=0$ (giving $J^2=0$), that is, $$A=X\pmatrix{0&1\\0&0}X^{-1}$$ for some $2\times 2$ nonsingular $X$.

• This is also a good illustration of why "trying a random selection" often fails for problems like these. Note, for example, that "most matrices" are invertible. – Omnomnomnom Nov 10 '14 at 17:59