Suppose $n\geq 2$. Let $T:\mathbb{C}^n\to \mathbb{C}^n$ be linear. Prove that the following are equivalent:
(i) $T$ is diagonalizable.
(ii) For every two-dimensional subspace $W\subseteq \mathbb{C}^n$ invariant under $T$, the linear map $T|_W$ is diagonalizable.
(i) implies (ii) is clear to me. I wish to prove (ii) implies (i). It is not obvious to me why the result should hold. Suppose (ii) holds. Suppose $T$ is not diagonalizable. Then $(x-\lambda)^k$ divides the minimal polynomial of $T$, denoted $m_T(x)$, for some $k\geq 2$. Equivalently, letting $A$ be the JCF of the matrix of $T$, I know there is a Jordan block of size $k$ for eigenvalue $\lambda$. I guess that the idea is to create an invariant two-dimensional subspace of $T$ and then use our assumption (ii) to achieve a contradiction, but I don't see how this would go. In particular, how can I get a two-dimensional invariant subspace? It is clear to me that I am missing some part of the theory.
Any help at all is appreciated.
The first two basis vectors corresponding to any Jordan block of size greater than $1$ span an invariant subspace on which $T$ is given by a single Jordan block of size $2$. In particular, then, $T$ is not diagonalizable on that $2$-dimensional subspace.
Or, without referring to matrices at all, if $(x-\lambda)^2$ divides the minimal polynomial of $T$, this means there is a vector $v$ such that $(T-\lambda)v\neq 0$ but $(T-\lambda)^2v=0$. (To see this, let $p(x)=m_T(x)/(x-\lambda)^2$ and $q(x)=m_T(x)/(x-\lambda)$. Then $q(T)\neq 0$, so there is some $w$ such that $q(T)w\neq 0$. Now let $v=p(T)w$, so $(T-\lambda)v=w\neq 0$ but $(T-\lambda)^2v=m_T(T)w=0$.) The $T$-invariant subspace generated by $v$ is then $2$-dimensional, spanned by $v$ and $(T-\lambda)v$, and the restriction of $T$ is not diagonalizable since $(T-\lambda)^2v=0$ but $(T-\lambda)v\neq 0$.