Suppose that $A$ is a self-adjoint linear operator on a Euclidean finite-dimensional space $V$. Is it true that any orthogonal projector $P_\lambda$ onto an eigenspace of $A$ can be represented as a polynomial of $A$?
Comment: I was reading M. Taylor's book "Partial differential equations I" and here at p. 635 author says that if any polynomial of $A$ commutes with some unitary operator then the orthogonal projector as in my question also commutes with this operator.
What an interesting question! I think the answer is yes.
Consider the eigendecomposition of the self-adjoint matrix, $$A=Q D Q^*,$$ where $Q$ is unitary and $D$ diagonal. The orthogonal projector onto an eigenspace is $$P_\lambda = Q I_\lambda Q^*,$$ where $I_\lambda$ is the diagonal matrix with ones in locations corresponding to the eigenspace of interest and zeros elsewhere.
The question is then, can we find a polynomial $p$ such that $p(D) = I_\lambda$. For example, supposing we are interested in projecting onto the space associated with the first and third eigenvalue, then we need a polynomial such that \begin{align} p(\lambda_1) &= 1 \\ p(\lambda_2) &= 0 \\ p(\lambda_3) &= 1 \\ p(\lambda_4) &= 0 \\ \vdots \end{align}
But this is just a polynomial interpolation problem, where the interpolation points are the eigenvalues, and the values to be fit are either one or zero (depending on whether the eigenvector is projected onto or not). The solution to this 1D polynomial interpolation problem is the polynomial associated with the projector.