Let $W$ be a finate-dimensional inner product space over $\mathbb{R}$ (or $\mathbb{C}$) and let $M: W\to W$ be a self-adjoint linear operator. Then there exists an orthonormal basis $B=\{\phi_i\}$ consisting of the eigenvectors of $M$. Denote the eigenvalues by $\lambda_i$ respectively. Let there be $v\in W$. Since $B$ is a basis of $W$, there exists a unique combination of $\alpha_i$ such that $v=\sum_i \alpha_i \phi_i$. Denote by \begin{equation*} \lambda_B(v) := \max_i\{\lvert\lambda_i\rvert \mid \alpha_i\neq 0\} \end{equation*} the largest eigenvalue (in absolute value) that matches an eigenvector that spans $v$. Equivalently, since $B$ is an orthogonal basis, \begin{equation*} \lambda_B(v) := \max_i\{\lvert\lambda_i\rvert \mid v\not\perp\phi_i \}. \end{equation*} where $v\not\perp\phi$ is defined as $\langle v, \phi \rangle \neq 0$.
Note that $B$ is not unique: there can be another orthonormal basis consisting of eigenvectors of $M$. My question is whether $\lambda_B(v)$ is independent of $B$. What is the simplest way to prove it? By simplest I mean easiest to understand for someone who has a math or CS degree.
Thank you!
The vector space $W$ can be written as $$ W=\bigoplus_i E_{\lambda_i}, $$ where $E_{\lambda_i}$ is the eigenspace associated to the eigenvalue $\lambda_i$. Then if $\lambda_B(v)=\lambda$ for some basis $B$, it has the same value for any other basis as the the vector $\phi_\lambda$ associated to $\lambda$ cannot be written as a sum of the vectors in the other eigenspaces.
Notice that if the eigenvalues are all different, the orthonormal basis consisting of normalized eigenvectors of $M$ is unique, except for a reordering.