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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.

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    $\begingroup$ This is false unless you add some hypothesis to make $T$ nontrivial. As it stands, you can take $T$ to be the zero operator (which is self-adjoint) and take $\beta$ to be any basis that isn't orthonormal. The matrix of $T$ with respect to $\beta$ is the zero matrix, which is Hermitian. The same issue arises if $T$ is the identity operator; its matrix is the identity matrix with respect to any basis, even if the basis is not orthonormal. To avoid all such problems, you need to assume that $T$ has no multiple eigenvalues. $\endgroup$ – Andreas Blass Apr 7 '18 at 2:41
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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.

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