If an operator $T$ is self-adjoint, why is the matrix of $T$ symmetric if and only if the basis is orthonormal?
I'm seeing that this is what is being said in Artin's Algebra, Treil's Linear Algebra Done Wrong, and Hoffman and Kunze's Linear Algebra, but no justification is given.
How do I show that:
Let $T$ be a self-adjoint operator on a complex inner product space $V$ and let $\beta$ be a basis for $V$ . The matrix of $T$ is Hermitian if and only if $\beta$ is orthonormal.