Normal Operators In Different Basis If we have normal operators, for example symmetric, can we "see" that it symmetric in one basis (orthonormal) but not in another basis?
 A: Let $V$ be a finite-dimensional complex inner product space; for $T: V \to V$ an operator and $\beta$ a basis for $V$, let $[T]_\beta$ denote its matrix representation with respect to $\beta$.
Suppose that $\beta$ is an orthonormal basis for $V$. Then for any operator $T : V \to V$,
$$
 [T^\ast]_\beta = [T]_\beta^\ast.
$$
Thus, the operator $T$ is normal [unitary, self-adjoint, positive] if and only if $[T]_\beta$ is normal [unitary, self-adjoint, positive].
Now, suppose that $\gamma$ is a basis for $V$, which need not be orthonormal. Let $C_\beta^\gamma$ denote the change of coordinates matrix from $\beta$ to $\gamma$. Then
$$
 [T^\ast]_\gamma = C_\beta^\gamma[T^\ast]_\beta(C_\beta^\gamma)^{-1} = C_\beta^\gamma [T]_\beta^\ast (C_\beta^\gamma)^{-1},
$$
whilst
$$
 [T]_\gamma^\ast = (C_\beta^\gamma [T]_\beta (C_\beta^\gamma)^{-1})^\ast = ((C_\beta^\gamma)^{-1})^\ast [T]_\beta^\ast (C_\beta^\gamma)^\ast,
$$
so that
$$
[T^\ast]_\gamma = \left(C_\beta^\gamma (C_\beta^\gamma)^\ast\right)[T]_\gamma\left(C_\beta^\gamma (C_\beta^\gamma)^\ast\right)^{-1}.
$$
The moment that $\gamma$ isn't orthonormal (actually, orthogonal with vectors all of the same norm), there's no guarantee that $C_\beta^\gamma (C_\beta^\gamma)^\ast$ will commute with $[T]_\gamma$, in which case
$$
 [T^\ast]_\gamma \neq [T]_\gamma^\ast,
$$
and hence a normal operator $T$ could give rise to a non-normal matrix representation $[T]_\gamma$. Note, though, that the zero operator $O_V$ and the identity operator $I_V$ won't have this issue, since $[O_V]_\gamma = 0$ and $[I_V]_\gamma = I$ will both commute with any matrix.
A: No.
For example, the identity operator $I$ is normal and symmetric, but it is represented with the identity matrix in any basis, not only in an orthonormal basis.
Same thing with the zero operator $0$.
