If $ST=TS$, then is an eigenvector of one an eigenvector of the other? Is either diagonalisable?

Question

Let $S$ and $T$ be finite dimensional linear transformations that commute.

Which of the following, if any, are true?

1. If $x$ is an eigenvector of $S$, then $x$ is an eigenvector of $T$*

2. Converse of 1

3. $T$ is diagonalisable.

What if $S$ or $T$ is not finite dimensional?

In case none of them are true, I may be missing some assumptions. I think at least one is true.

What I tried:

Let $x_S$ and $\lambda_S$ be s.t.

$$Sx_S = \lambda_S x_S$$

We must check to see if $\exists \lambda_T$ s.t.

$$Tx_S = \lambda_T x_S$$

Try to prove it:

$$Sx_S = \lambda_S x_S$$

$$\to TSx_S = \lambda_S Tx_S$$

$$\to STx_S = \lambda_S Tx_S$$

I wasn't able to get further than this, whether or not $\lambda_S = 0$ so I'm guessing 1 and 2 are false. I can't think of any counterexamples. Please give hints.

Any particular reason a linear transformation that commutes with another might be diagonalisable?

*I am fairly certain the question said eigenvector and not eigenvalue.

Answer 1

Hint. Every linear transformation commutes with something, for example, with the identity. (Or for another example: with itself.) And every nonzero vector is an eigenvector of the identity.

So, in effect, your statements are

1. Every nonzero vector is an eigenvector of every transformation.
2. Same as 1.
3. Every transformation is diagonalisable.

I expect you can now decide whether these statements are true or false. Good luck!