Is it true that if $A$ is unitary diagonalizable (i.e. $A$ is normal) and $B$ is similar to $A$ then $B$ is also unitary diagonalizable (i.e. $B$ is normal) ?
Here are my thoughts:
If $A$ is unitary diagonalizable matrix then there exist unitary matrix $P$ and diagonal matrix $D$ such that $P^*AP=D$. $(*)$
We also know from the Spectral theorem that $A$ is normal ($\because$ $A$ is unitary diagonalizable $ \iff A$ is normal).
From the fact that $B$ is similar to $A$ we can derive that there exists invertible matrix $Q$ such that $Q^{-1}BQ=A$.
Finally, by replacing $A$ in $(*)$ with $Q^{-1}BQ$ we get:
$$ P^*Q^{-1}BQP=D \ \ (**)$$
The only thing I can derive from this equation $(**)$ is that $B$ is diagonalizable, because :
$$ P^*Q^{-1}BQP=D \implies P^{-1}Q^{-1}BQP=D \implies (QP)^{-1}B(QP)=D \implies C^{-1}BC=D$$
Am I missing something? Is it possible that I can also derive that $B$ is unitary diagonalizable ?
Best Answer
For a unitarily diagonalizable matrix, eigenvectors for distinct eigenvalues are orthogonal.
Thus let $A$ be a diagonal matrix with distinct eigenvalues and $B = S A S^{-1}$. Then $B$ is similar to $A$, and the columns of $S$ are eigenvectors. If $S$ has columns that are not orthogonal, $B$ is not unitarily diagonalizable.