Prove that $A$ is diagonalizable if and only if $\mbox{tr} A\neq 0$.
$A$ is an $n\times n$ matrix over $\mathbb{C}$, and $\mbox{rk} A=1$.
If $p(t)$ is the characteristic polynomial of $A$, I know that $a(n-1)\neq0$ because $\mbox{tr} A = (-1)^{n+1}a(n-1).$
I also know that $\dim\ker(A-0\cdot I)=\dim\ker A=n-\mbox{rk} A=n-1$ (so the geometric multiplicity of $t=0$ as an eigenvalue is $n-1$).
Though I don't know how to continue from here (on both directions).
Any suggestions?
Thanks
Best Answer
$A$ is diagonalizable $\implies$ $trA\neq 0$
$A=PDP^{-1}$ where $$ D=\begin{pmatrix} \alpha & 0 & \cdots & 0 \\ 0 & 0 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 0 \end{pmatrix}$$ with $\alpha\in\mathbb{C}^*$
so $$tr(A)= tr(PDP^{-1}) = tr(P^{-1}PD) = tr(D) = \alpha \neq 0$$
$trA\neq 0$ $\implies$ $A$ is diagonalizable
Since $rk(A)=1$, the first $n-1$ eigenvalues of A are $0$. The last one is different from $0$ since $trA\neq 0$ and fulfill the eigenspace since $dim(E_\alpha)=rk(A)=1$. Therefore, A is diagonalizable.