Question about a linear operator and its matrix being normal and self-adjoint So I have a linear operator $T$, an orthonormal basis $\epsilon$, and I have calculated its matrix, $A=[T]_\epsilon$. I know that the matrix is normal, but not self adjoint. I found this by calculating $A^*$, and then showing that $A^*A=AA^*$, and that $A\neq A^*$. However, using I am not quite sure how I would show if $T$ is normal or self-adjoint, without calculating $T^*$. Is it possible, using the information I have about $A$?
Since $\epsilon$ is an orthonormal basis, you have

$$ [T \circ T^{*}]_{\epsilon} = [T]_{\epsilon} [T^{*}]_{\epsilon} = [T]_{\epsilon} ([T]_{\epsilon})^{*} = AA^{*} = A^{*}A = [T^{*} \circ T]_{\epsilon} $$

which implies that $T \circ T^{*} = T^{*} \circ T$ and so $T$ is normal. More generally, this argument shows that an operator $T$ is normal / unitary / Hermitian / etc if and only if the matrix representing $T$ with respect to some (or any) orthonormal basis is normal / unitary / Hermitian / etc.
An operator on a finite-dimensional vector space is normal iff it can be diagonalized with orthogonal eigenvectors.