I'm wondering if special matrices are always similar only to other special matrices.
That is:
$\cdot$ Are all of the matrices similar to a symmetric matrix symmetric?
$\cdot$ Are all of the matrices similar to a skew-symmetric matrix skew-symmetric?
$\cdot$ Are all of the matrices similar to a normal matrix normal?
$\cdot$ Are all of the matrices similar to a self-adjoint matrix self-adjoint?
$\cdot$ Are all of the matrices similar to a Hermitian matrix Hermitian?
etc.
Let's try one:
A matrix $A$ is symmetric if $A=A^T$. Then if $A=P^{-1}BP$, we have $A^T=P^TB^T(P^T)^{-1}$. Therefore $A^T=A=P^TB^T(P^T)^{-1}=P^{-1}BP$. Therefore $B=PP^TB^T(P^T)^{-1}P^{-1}$. Because $P$ is not orthogonal in general, $B \ne B^T$?
NOTE: I could try this for all of them, but I'm pretty sure I'd reach a very similar conclusion in each of the above mentioned cases.
Assuming that this is the correct conclusion here, does this mean that in general the specialness (symmetry, Hermitianness, orthogonality, etc) of matrices is not preserved by a change of basis? If not, then there's no way to discuss a symmetric (or otherwise special) transformation without first defining our basis?
Just curious.
Best Answer
All of these properties fail to hold for similar matrices in general. However, they are true if we specify unitary similarity; that is, we only allow similarities such that $P^{-1} = P^*$ (or $P^{T}$, if $P$ is real).
Counterexamples:
$ \pmatrix{1\\&2} $ is normal, symmetric, self-adjoint, and Hermitian. But the similar matrix $ \pmatrix{1&1\\0&2} $ has none of those properties.
The matrix $ \pmatrix{0&-2\\2&0} $ is skew-symmetric, but the similar matrix $ \pmatrix{0&-1\\4&0} $ is not.