Do non-diagonalizable matrices with same eigenvectors commute?

Question

Let $A$ and $B$ be two matrices with the same set of eigenvectors, they are not supposed to be diagonalizable, how can I prove that $AB=BA$?

I guess this could be proved using some similitude argument, for example Jordan form. Thank you all for the help.

Answer 1

Hint: No, the set of upper triangular matrices is not commutative, for $n>1$.

Answer 2

This is not possible. Consider the two matrices $$ A=\pmatrix{ 0&1&0&0\\ 0&0&1&0\\ 0&0&0&0\\ 0&0&0&0\\ } , B=\pmatrix{ 0&1&0&0\\ 0&0&0&0\\ 0&0&0&0\\ 0&0&1&0\\ }. $$ Both have eigenvectors $e_1$ and $e_4$, but do not commute, as $$ ABe_3 = 0 \ne BAe_3=e_1. $$ Their Jordan forms are different: the blocks have size $(3,1)$ for $A$, but $(2,2)$ for $B$.

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If we assume that $A$, $B$ are $2\times 2$, upper triangular, and not diagonalizable, then $A$ and $B$ commute. This is no longer true for $3\times 3$.