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

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?

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).