Linear algebra problem and What can we learn from tr(A) = 0? 
Does anyone else who can help? thanks.
According to lemma $tr(A) = tr(C^{-1} A C)$, $tr(A) = tr(C^{-1}AC) = tr(  \begin{matrix}
   a & 0 \\
   0 & b \\
  \end{matrix} \tag{1})$
$a + b = 0$ have two situation: 1. a = b = 0, 2. a and b are non-trivial.Situation 1 is impossible, so a and b are non-trivial and a+b=0.
$rank(A) = 2$
That's what I know, but I can not solve this problem solely with this conclusion.
 A: Hints:
For (1):

*

*If $\lambda\ne\mu$ are two distinct eigenvalues of $A$ with eigenvectors $v$ and $w$, then $v+w$ is not an eigenvector.

*If all vectors are eigenvectors, then $A=\lambda I$.

For (2): let $v$ and $Av$ be the columns of $P$, i.e. we write the matrix of $x\mapsto Ax$ in the basis $(v,Av)$.
(We even get a representation $P^{-1}AP=\pmatrix{0&a\\1&0}$.)
A: For $1:$
For contradiction assume it is not possible to find an independent set $(v, Av)$.
This says that $\forall v \in V$, $Av \in \text{span}(v)$, i.e., $Av = cv$ so that every non-zero vector in $V$ is an eigenvector. This implies that $A = \lambda I$ (you should prove this). Now we have $\text{tr} A = 2\lambda = 0 \implies \lambda = 0 \implies A = 0I = 0$, which is a contradiction.
For $2:$
Let $P = \begin{bmatrix} v & Av  \end{bmatrix}$, $D^* = \begin{bmatrix} 0 & a \\ 1 & 0  \end{bmatrix}$, and $d_i$ column $i$ of $D^*$. Then we have that
$AP = \begin{bmatrix} Av & A^2v  \end{bmatrix} = \begin{bmatrix} Av & av  \end{bmatrix} = \begin{bmatrix} Pd_1 & Pd_2  \end{bmatrix} = PD^*$.
Now the result follows since $P$ is invertible.
We know that every non-zero $v$ is an eigenvector of $A^2$ because $A^2$ is a scalar matrix. The easiest way to see this is to use Caley Hamilton which gives
$$A^2 - (\text{tr} A) A + (\det A)I = 0 \implies A^2 = -(\det A) I$$
It's possible to avoid Cayley Hamilton. Prove that $A^2$ is scalar (every non-zero vector is an eigenvector) instead by using the fact that the eigenspaces direct sum to $\mathbb{R}^2$ (exercise).
