Does $V= \ker T \oplus \operatorname{im} T$ imply diagonalizable?

Question

(FYI: I posted this same question very recently but think I accidentally deleted it. Sorry for the repost.)

For $V$ finite dimensional, if $V= \ker T \oplus \operatorname{im} T$, is $T$ necessarily diagonalizable? I'm guessing that it's not true and am looking for a counterexample.

By the way, I saw that when I posted this question earlier, someone answered by offering the counterexample $T(x,y)=(y,x)$ where $(x,y)\in \mathbb{R}^2$. But I think this wouldn't work, since $T$ has eigenvalues $\pm 1$.

Answer 1

The usual counterexample works: $$ \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} $$ has one eigenvalue (namely $1$), so it has trivial kernel, and its columns form a basis, so it has image the whole of $\mathbb{R}^2$, and indeed $\mathbb{R}^2 = \{0\} \oplus \mathbb{R}^2$, so it satisfies the conditions in the question. It is also famously not diagonalisable.

Answer 2

No, take $$T=\left(\begin{array}{cc} 1&1\\ 0&1 \end{array}\right)$$ This matrix is not diagonalizable, but it is surjective, so it satisfies the direct sum decomposition in your statement trivially since $V$ is the image.