Action of an Operator with an Orthonormal Eigenbasis

Let $V$ together with $\langle\cdot,\cdot\rangle$ be a \reftext{def:complex-inner-product-space-2026a}{complex inner product space} with \reftext{lem:vector-space-basic-identities-2026a}{zero vector} $0_{V}$. Let $n$ be a \reftext{def:natural-numbers-2026a}{natural number}, let $[n]$ be the \reftext{def:initial-segment-natural-numbers-2026a}{initial segment} determined by $n$, and let $e\in V^{n}$ be an \reftext{def:finite-tuple-power-2026a}{$n$-tuple} in $V$ that is an \reftext{def:orthonormal-basis-2026b}{orthonormal basis} of $V$, with components $e_{k}$. Let $\lambda\in\mathbb{C}^{n}$ be an $n$-tuple with components $\lambda_{k}$ in the field $\mathbb{C}$ of \reftext{def:complex-numbers-2026a}{complex numbers}, and let $T$ be a \reftext{def:linear-operator-2026a}{linear operator} on $V$ satisfying

Such $e$ and $\lambda$ exist whenever $V$ is \reftext{def:finite-dimensional-vector-space-2026b}{finite-dimensional} with $V\ne\{0_{V}\}$ and $T$ is \reftext{def:self-adjoint-operator-2026b}{self-adjoint}, by \ref{thm:spectral-theorem-self-adjoint-2026a}.

Sums of vectors are \reftext{def:finite-sum-vector-space-2026a}{finite sums in $V$}, and $\lambda_{k}\langle e_{k},x\rangle e_{k}$ denotes the scalar multiple of $e_{k}$ by the complex number $\lambda_{k}\langle e_{k},x\rangle$. Then the following hold.

\textbf{1. (Diagonal action)}

\textbf{2. (Determination by the eigenbasis)} If $T'$ is a linear operator on $V$ with $T'(e_{k})=\lambda_{k}e_{k}$ for every $k\in[n]$, then $T'(x)=T(x)$ for every $x\in V$.