Proof of Operators Diagonal in an Orthonormal Basis

Write $0_{V}$ for the zero vector of $V$, let $\lVert\cdot\rVert$ be the norm induced by the inner product, and let $\overline{z}$ denote the complex conjugate of a complex number $z$.

Orthonormality relations. Since $e$ is orthonormal, each $e_{j}$ is a unit vector, so $\langle e_{j},e_{j}\rangle=\lVert e_{j}\rVert^{2}=1$ by Norm Induced by a Complex Inner Product; and for $j,k\in[n]$ with $k\ne j$ the vectors $e_{k}$ and $e_{j}$ are orthogonal, so $\langle e_{k},e_{j}\rangle=0$.

Claim 1: additivity. Let $x,y\in V$ and $k\in[n]$. Condition 2 of Complex Inner Product Space gives $\langle e_{k},x+y\rangle=\langle e_{k},x\rangle+\langle e_{k},y\rangle$, so by distributivity of multiplication over addition in the field $\mathbb{C}$ and condition 8 of Vector Space over a Field,

Claim 2 of Properties of Finite Sums of Vectors therefore gives $R(x+y)=R(x)+R(y)$.

Claim 1: homogeneity. Let $\nu\in\mathbb{C}$, $x\in V$ and $k\in[n]$. Condition 3 of Complex Inner Product Space gives $\langle e_{k},\nu x\rangle=\nu\langle e_{k},x\rangle$, so by associativity and commutativity of multiplication in $\mathbb{C}$ and condition 5 of Vector Space over a Field,

Claim 3 of Properties of Finite Sums of Vectors therefore gives $R(\nu x)=\nu R(x)$. Hence $R$ satisfies conditions 1 and 2 of Linear Map, so it is a linear map from $V$ to $V$, that is, a linear operator on $V$.

Claim 1: action on the basis. Fix $j\in[n]$. For $k\in[n]$ with $k\ne j$ the orthonormality relations give $\mu_{k}\langle e_{k},e_{j}\rangle=\mu_{k}\cdot0=0$, hence $\bigl(\mu_{k}\langle e_{k},e_{j}\rangle\bigr)e_{k}=0e_{k}=0_{V}$ by claim 3 of Elementary Identities in a Vector Space. Thus at most the summand with index $j$ is nonzero, and claim 7 of Properties of Finite Sums of Vectors gives

Claim 2. Assume every $\mu_{k}$ is a real number and let $x,y\in V$. The first identity of claim 6 of Properties of Finite Sums of Vectors, applied with the coefficients $\mu_{k}\langle e_{k},y\rangle$ and the vectors $e_{k}$, gives

and the second identity of that claim, applied with the coefficients $\mu_{k}\langle e_{k},x\rangle$, gives

both sums being finite sums in $\mathbb{C}$. By claim 1 of Properties of Complex Conjugation and Modulus, conjugation preserves products and fixes exactly the real numbers, so

Condition 1 of Complex Inner Product Space gives $\langle x,e_{k}\rangle=\overline{\langle e_{k},x\rangle}$. Hence, for every $k\in[n]$, the $k$-th summand of the first sum is $\mu_{k}\langle e_{k},y\rangle\overline{\langle e_{k},x\rangle}$ and the $k$-th summand of the second sum is $\mu_{k}\overline{\langle e_{k},x\rangle}\langle e_{k},y\rangle$; these are equal because multiplication in $\mathbb{C}$ is commutative. The two sums are therefore equal, so $\langle R(x),y\rangle=\langle x,R(y)\rangle$ for all $x,y\in V$, which is the condition of Self-Adjoint Operator.