Prove that $A$ is diagonalizable and find similar matrices

Question

Let $A$ be a matrix $(3x3)$ such that:

$A(1,1,1)^t=(2,2,2)^t$ and $rank(2I + A) \lt rank(2I-A)$

I need to prove that $A$ is an diagonalizable matrix and find all the matrices that are similar to it.

I'm having difficulties solving this, can anyone please tell me what I should do?

Thanks

Answer 1

From the first equality you get that $2$ is an eigenvalue of $A$, so $\operatorname{rank}(2I-A)<3$. Also $\operatorname{rank}(2I+A)<3$, and therefore $-2$ is an eigenvalue as well. The geometric multiplicity of $2$ is thus $1$ or $2$, meaning, respectively, $\operatorname{rank}(2I-A)=2$ and $\operatorname{rank}(2I-A)=1$. The latter is impossible, because it would imply $\operatorname{rank}(2I+A)=0$ and so the geometric multiplicity of $-2$ would be $3$.

Hence $\operatorname{rank}(2I-A)=2$ and $\operatorname{rank}(2I+A)=1$.

Can you go on?

Answer 2

A isn't necessarily orthogonal.

$$ A = \begin{bmatrix}-2&0&4\\0&-2&4\\0&0&2\end{bmatrix} $$

satisfies your conditions and isn't orthogonal, i.e. its columns aren't orthogonal.