Action of an Operator with an Orthonormal Eigenbasis

Let $V$ together with $\langle\cdot,\cdot\rangle$ be a complex inner product space with zero vector $0_{V}$. Let $n$ be a natural number, let $[n]$ be the initial segment determined by $n$, and let $e\in V^{n}$ be an $n$-tuple in $V$ that is an 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 complex numbers, and let $T$ be a linear operator on $V$ satisfying

$$T(e_{k})=\lambda_{k}e_{k}\qquad\text{for every }k\in[n].$$

Such $e$ and $\lambda$ exist whenever $V$ is finite-dimensional with $V\ne\{0_{V}\}$ and $T$ is self-adjoint, by Spectral Theorem for a Self-Adjoint Operator in Finite Dimensions.

Sums of vectors are 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.

1. (Diagonal action)

$$T(x)=\sum_{k=1}^{n}\lambda_{k}\langle e_{k},x\rangle e_{k}\qquad\text{for every }x\in V.$$

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 $x\in V$.