Let $T$ be a self-adjoint operator on $V$ where $\beta$ is a basis for $V$. The matrix of $T$ is Hermitian if and only if $\beta$ is orthonormal. 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.

 A: I'm assuming you are working over complex numbers (I'll denote the complex conjugation by $*$) and let $(-,-)$ be the inner product. Let $M$ be the matrix of $T$ in basis $\beta=\{\beta_1, \cdots, \beta_n\}$, i.e. $T\beta_j = \sum_{i=1}^n M_{ij}\beta_i$. 
Suppose $\beta$ is orthonormal, then by taking the inner product of the above equation with $\beta_i$, you find
$$
(\beta_i, T\beta_j)=M_{ij}=(T\beta_j, \beta_i)^*
$$
But $T$ is self-adjoint, so $M_{ij}=(\beta_i, T\beta_j)=(T\beta_i, \beta_j)={M_{ji}}^*$, meaning $M$ is hermitian. 
Conversely, note that the whole argument is completely reversible, so you find that if $\beta$ is a basis such that $M$ is hermitian then, given any pair $i,j$ we have
$$
M_{ij} = M_{ji}^*\Longrightarrow (\beta_i, T\beta_j)=(T\beta_i,\beta_j)
$$
Therefore, by bilinearity of inner product and since $\beta$ is a basis, given any two  vectors $v,w$, one has $(v,Tw)=(Tv,w)$, i.e. $T$ is self-adjoint.
A: The claim is false, as the condition on the basis is sufficient but not necessary.
An easy counterexample, as noted by Andreas Blass in the comments to the question, is the identity operator, which has the same symmetric matrix ($I_n$) for every basis, even those which are not orthonormal. However, one may think that such a case can be avoided by imposing stricter conditions on $T$: this is usually not the case, as we will prove with a general counterexample. 
Let $T:\mathbb{C}^n\to \mathbb{C}^n$ be a self-adjoint operator of maximal rank which is not an orthonormal transformation on  of rank $n$, and let $M$ be its matrix in the canonical basis. If we define the basis $\beta=\{v_1,\dots,v_n\}$ as the basis composed by the column vectors of $M$, by the formula for the change of basis we obtain
$$T_{\beta}=(v_1|\dots |v_n)^{-1}M(v_1|\dots |v_n)=M^{-1}M M=M$$
Thus the matrix of $T$ in the basis $\beta$ (which is not orthonormal, for otherwise $T$ would be orthonormal) is symmetric, and we obtained a general counterexample
