Do commutative matrices share the same eigenvectors? Let two square matrices $A$ and $B$ represent linear operators on a vector space $V$ over $\mathbb{C}$. Suppose they are commutative. Then

$$ABx=BAx,\,\,\forall x\in V$$

Then let $\tilde{x}$ be an eigenvector of $B$. Setting, $x=\tilde{x}$, we see that

$$\begin{align}
AB\tilde{x}&amp;=BA\tilde{x}\\
A\lambda \tilde{x} &amp;= BA\tilde{x},\,\,\exists \lambda\in\mathbb{C}\\
\lambda (A\tilde{x}) &amp;= B(A\tilde{x})\\
&amp;\\
\implies A\tilde{x}&amp;=\alpha\tilde{x},\,\,\exists \alpha\in\mathbb{C}
\end{align}$$

So every eigenvector of $B$ is also an eigenvector of $A$. $\blacksquare$

MY QUESTIONS: Is this valid? If it isn't, what have I done wrong? If it is, does this mean that the two matrices are similar by the matrix formed by making each column an eigenvector of the two matrices?
It's obviously false. The identity matrix commutes with every other matrix, but any nonzero vector is an eigenvector for the identity, which can be not true for a non identity matrix.
Yes, it is true. May be $P$ the Matrix of all eigenvectors and $diag([\lambda])$ the diagonal Matrix of all Eigenvalues $\lambda$. Then $A=P^Tdiag([\alpha])P$ and $B=P^Tdiag([\lambda])P$. These matrices are similar; only the Eigenvalue spectrum can differ.