How is the P function applied in QSVT for the case of Hamiltonian simulation if it only modifies singular values?

Question

I am watching Andras Gilyen's talk on QSVT here.

On one slide he mentions the core of QSVT:

Given $U$--- a block encoding of matrix $A$ that has singular values $\lambda$, left singular vectors, $\left | w \right >$, and right singular vectors $\left < v \right |$

$ U = \begin{bmatrix} A & . \\ . & . \end{bmatrix} = \begin{bmatrix} \sum_i \lambda_i \left | w_i \right > \left < v_i \right | & . \\ . & . \end{bmatrix} $

We can construct $V_\vec{\phi}$ that is:

$ V = \begin{bmatrix} \sum_i P(\lambda_i) \left | w_i \right > \left < v_i \right | & . \\ . & . \end{bmatrix} $

However, on a following slide regarding Hamiltonian simulation, he brings this up:

What is $P(H)$ here? I thought $P$ could only act on singular values (and that is supported by $P$ having a domain of just $[-1, 1]$). Is this $P(H)$ correct? If so, is there another definition behind it that I am missing?

Answer 1

By definition, $P(A)$ is defined as acting in the singular basis of $A$, i.e. $$ \begin{align} P(A) = \sum_i P(\sigma_i) |w_i\rangle\langle v_i| = W P(\Sigma) V^\dagger. \end{align} $$ For Hermitian matrices, this just follows from the spectral decomposition of $A$. More generally, if $A$ is at least diagonalizable then $P$ will act over the eigenvalues of $A$ in the same way. Most generally, as the authors consider, any matrix has an SVD $A = W \Sigma V^\dagger$, in which case we can get $P$ to act as expected over the singular values (always real) if we make sure to alternate between $U$ and $U^\dagger$ (so that $W$ and $V$ are always hit by $W^\dagger$ and $V^\dagger$, respectively).