Why Are Symmetric & Non-Symmetric Matrices Not Unitarily Equivalent?

[namegoeshere]

I heard this assertion during a discussion:

If two matrices are similar, but one is symmetric and the other is not, then the two matrices are not unitarily equivalent.

Why is this true? This was only mentioned in passing, and I definitely don't understand why.

 

[jbunniii]

I assume that when you say "symmetric" you really mean conjugate symmetric (Hermitian). A matrix $A$ is Hermitian if it equals the complex conjugate of its transpose: $A = A^*$.

If $A$ and $B$ are unitarily equivalent, then there exists a unitary matrix $U$ (i.e., $U$ satisfies $U^* U = U U^* = I$) such that $A = U^* B U$.

If $A$ is Hermitian, then $A = A^* = U^* B^* U$. Therefore,
$$U^* B U = U^* B^* U$$
Multiplying this equation on the left by $U$ and on the right by $U^*$, we conclude that $B = B^*$.

In exactly the same way, we can show that if $B^* = B$, then $A^* = A$.

Therefore $A$ is Hermitian if and only if $B$ is Hermitian.

If all the matrices involved are real-valued, then you can replace "unitary" with "orthogonal" and "Hermitian" with "symmetric."

 

[namegoeshere]

I assume that when you say "symmetric" you really mean conjugate symmetric (Hermitian). A matrix $A$ is Hermitian if it equals the complex conjugate of its transpose: $A = A^*$.

If $A$ and $B$ are unitarily equivalent, then there exists a unitary matrix $U$ (i.e., $U$ satisfies $U^* U = U U^* = I$) such that $A = U^* B U$.

If $A$ is Hermitian, then $A = A^* = U^* B^* U$. Therefore,
$$U^* B U = U^* B^* U$$
Multiplying this equation on the left by $U$ and on the right by $U^*$, we conclude that $B = B^*$.

In exactly the same way, we can show that if $B^* = B$, then $A^* = A$.

Therefore $A$ is Hermitian if and only if $B$ is Hermitian.

If all the matrices involved are real-valued, then you can replace "unitary" with "orthogonal" and "Hermitian" with "symmetric."

D'oh. Yes, I meant conjugate symmetric. Thanks. :)