Suppose $A$ and $B$ are diagonalizable matrices. Prove or disprove that $A$ is similar to $B$ iff $A$ and $B$ are unitarily equivalent. Alt-Solution?

Question

Suppose $A$ and $B$ are diagonalizable matrices. Prove or disprove that $A$ is similar to $B$ iff $A$ and $B$ are unitarily equivalent.

Is there a way to disprove this without using the idea of showing: $\Sigma_{i,j}^n|A_{i,j}|^2 = \Sigma_{i,j}^n|B_{i,j}|^2$? I ask because the question that establishes using the summation criterion doesn't come up until after this question I'm working on.

Answer 1

Take $A=\left[\begin{smallmatrix}1&0\\0&0\end{smallmatrix}\right]$ and $B=\left[\begin{smallmatrix}1&1\\0&0\end{smallmatrix}\right]$. They are similar, since, if you take $P=\left[\begin{smallmatrix}1&1\\0&1\end{smallmatrix}\right]$, then $B=P^{-1}AP$. However, $A$ and $B$ are not unitarily equivalent. That would mean that there would be an unitary matrix $U$ such that $B=U^{-1}AU$. Then we would have $A=UBU^{-1}$ and therefore, if $v_1$ and $v_2$ are the first and the second column of $U^{-1}$ respectively, then $v_1$ would be an eigenvector of $B$ with eigenvalue $1$ and $v_2$ would be an eigenvalue of $B$ with eigenvalue $0$. But (since $U^{-1}$ is unitary) $v_1$ and $v_2$ would be orthogonal. But the eigenvalues of $B$ with eigenvalue $1$ are those of the form $(x,0)$ and the eigenvalues of $B$ with eigenvalue $0$ are those of the form $(x,x)$ (with $x\ne0$). They are never orthogonal.

Answer 2

$\begin{pmatrix}1&0\\0&0\end{pmatrix}$ and $\begin{pmatrix}1&-1\\0&0\end{pmatrix}$ are similar (projections of rank $1$) but not unitarily equivalent (the second one is not hermitian).