Proof of The Eigenvalues of an Operator with an Orthonormal Eigenbasis

Let $\lVert\cdot\rVert$ denote the norm induced by the inner product.

Preliminary: every component of $e$ is nonzero. Let $j\in[n]$. Since $e$ is orthonormal, $e_{j}$ is a unit vector, so $\lVert e_{j}\rVert=1$ and hence $\langle e_{j},e_{j}\rangle=\lVert e_{j}\rVert^{2}=1$ by Norm Induced by a Complex Inner Product. Suppose $e_{j}=0_{V}$. By claim 4 of Elementary Identities in a Vector Space, $0_{V}=0\cdot 0_{V}$, so condition 3 of Complex Inner Product Space gives

contradicting $\langle e_{j},e_{j}\rangle=1$, since $1\ne0$ in a field. Hence $e_{j}\ne0_{V}$.

Sufficiency. Suppose $\mu=\lambda_{j}$ for some $j\in[n]$. Then $T(e_{j})=\lambda_{j}e_{j}=\mu e_{j}$ and $e_{j}\ne0_{V}$, so $e_{j}$ is an eigenvector of $T$ with eigenvalue $\mu$. Hence $\mu$ is an eigenvalue of $T$.

Necessity. Suppose $\mu$ is an eigenvalue of $T$, and let $x\in V$ be an eigenvector of $T$ with eigenvalue $\mu$, so that $x\ne0_{V}$ and $T(x)=\mu x$. Claim 1 of Orthonormal Expansion and Parseval's Identity in Finite Dimensions gives

and therefore, by claim 3 of Properties of Finite Sums of Vectors and condition 5 of Vector Space over a Field,

Claim 1 of Action of an Operator with an Orthonormal Eigenbasis gives

Fix $j\in[n]$. Applying claim 1 of Elementary Properties of an Orthonormal Family to the first of these two linear combinations gives $\langle e_{j},\mu x\rangle=\mu\langle e_{j},x\rangle$, and applying it to the second gives $\langle e_{j},T(x)\rangle=\lambda_{j}\langle e_{j},x\rangle$. Since $T(x)=\mu x$, the two left-hand sides coincide, so

Not every coefficient $\langle e_{k},x\rangle$ can be $0$: if $\langle e_{k},x\rangle=0$ for every $k\in[n]$, then $\langle e_{k},x\rangle e_{k}=0e_{k}=0_{V}$ for every $k\in[n]$ by claim 3 of Elementary Identities in a Vector Space, so claim 7 of Properties of Finite Sums of Vectors would give $x=0_{V}$, contrary to $x\ne0_{V}$. Choose $j\in[n]$ with $\langle e_{j},x\rangle\ne0$ and multiply the last display by the multiplicative inverse of $\langle e_{j},x\rangle$ in the field $\mathbb{C}$; this gives $\lambda_{j}=\mu$.