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?

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.

  • $\begingroup$ These properties are not preserved by a change to an arbitrary linear basis (i.e., a linearly independent spanning set), but they are preserved when changing to an orthonormal basis (however, the first two require that the basis be in $\mathbb{R}^n$). $\endgroup$ – J. Loreaux Nov 19 '14 at 16:44
  • $\begingroup$ @J.Loreaux OK. That makes sense. Because I've heard of "Hermitian transformations" or "unitary transformations" or "skew-symmetric" tranformations without respect to a defined basis, but it didn't make sense if one could just change the basis and get a non-Hermitian/unitary/skew-symmetric/etc tranformation. Thanks. $\endgroup$ – user193875 Nov 19 '14 at 16:54

If $A$ is a special matrix other than a scalar multiple of the identity matrix and $A$ and $B$ are similar, but $B$ has a basis of eigenvectors such that there are two non-orthogonal eigenvectors $u$ and $v$ for distinct eigenvalues $\lambda$ and $\mu$, respectively, then $B$ will not have the special property.

(Note, by the way, a self-adjoint matrix is the same thing as a Hermitian matrix.)


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).


$ \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.


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