In a book on quantum mechanics, I encountered a statement equivalent to the following. Suppose $V$ is a finite dimensional inner product space over $\mathbb{C}$. Let $L : V \rightarrow V$ be a normal operator (or any operator that is diagonalizable). Suppose $\beta = \{ e_i \}$ is a corresponding orthonormal basis of eigenvectors, with eigenvalues $\{\lambda_i\}$. Let $f: \mathbb{C} \rightarrow \mathbb{C}$ be a function. Then $f(L)_\beta : V \rightarrow V$ defined by $f(L)_\beta(e_i) = f(\lambda_i)e_i$ is independent of the choice of the orthonormal basis of eigenvectors $\{e_i \}$, that is, if $\gamma$ is any other orthonormal basis of eigenvectors of $L$, then $f(L)_\beta = f(L)_\gamma$(So we can meaningfully speak of $f(L)$ if $L$ is diagonalizable).
The book does not provide a proof of this statement, and I seem to be only able to prove it if $L$ has no degenerate eigenvalues. I would be glad if someone could give me an approach to proving it in general.
For the distinct eigenvalues $\lambda_1,\dots,\lambda_k$, let $V_i$ be the space of eigenvectors for eigenvalue $\lambda_i$. Then $V=V_1+V_2+\cdots V_k$ and $f(L)(v_1+v_2+\cdots+v_k)=f(\lambda_i)v_1+\cdots +f(\lambda_k)v_k$ for $v_i\in V_i$, no matter which eigen-basis you chose. So it is less about a basis, and more about the eigenspaces $V_i$. And the $V_i$ are entirely determined by $L$.