For an endomorphism of $rank=1$, prove that $Im(u)\subset Ker(u)$ iff $u$ is NOT diagonilsable.

Question

Well, I got this question that I couldn't solve:

Problem: Let $E$ be a vector space over $R$ of finite dimension, and $u$ and an endomorphism of $E$ of rank 1. Prove that $$Im(u)\subset Ker(u)$$ if and only if $u$ is NOT diagonalisable.

My approach:
$\Rightarrow$: (I don't know how to prove it general, but say $n=3$): For $n=3$, we have $$dim(E)=3$$ $$dim(Im(u))=rank(u)=1$$ so $$dim(ker(u))=2$$ Now, $u\circ u=u^{2}$, so for all $x\in E$, $u(x)\in Im(u)\subset Ker(u)$. Hence $$u\circ u=u(u(x))=u^{2}=0$$ So $P(x)=X^{2}$ annihilates $u$. The only possible eigenvalue of $u$ is $0$. If $u$ was diagonalisable, $u=0$, which is impossible since $rank(u)=1$.

$\Leftarrow$: $...$

Answer 1

Since the rank of $u$ is $1$ then the matrix of $u$ is

$$A=(\lambda_1 c\cdots \lambda_n c)=cv^T$$ where $c$ is a non zero column and the $\lambda_i$ aren't all zero and $v=(\lambda_1,\ldots,\lambda_n)^T$. We see that

$$A^2=cv^Tcv^T=(v^Tc)cv^T=\operatorname{trace}(A)A$$ hence the polynomial $P=x^2-\operatorname{trace}(A)x=x(x-\operatorname{trace}(A))$ annihilates $A$ so we see that $A$ isn't diagonalizable iff $P$ has not simple roots iff $\operatorname{trace}(A)=0$ iff $A^2=0$ iff $Im(A)\subset \ker A$.

Answer 2

The rank of $u$ has very little to see here. The simple fact that $\operatorname{im} u\subset \ker u$ implies $u^2=0$. Hence if $u\neq 0$, the minimal polynomial of $u$ is $x^2$, which has a double root. The condition for an endomorphism to be diagonisable (over $\mathbf C$) is all roots of its minimal polynomial are simple.

More generally, no nilpotent endomorphism is diagonisable, except the null endomorphism.