# Does QSVT/QSP require singular/eigen decomposition?

Basically the title says it all. Every construction I've seen for QSP and QSVT involves finding 1 and 2-dimensional invariant subspaces of the iterate, $$O \equiv U_A \exp(i \pi/2 Z_\Pi)$$, where $$U_A$$ is an $$(\alpha = 1, m, \epsilon=0)$$ block encoding of $$A$$ and $$Z_\Pi$$ is a reflection operator formed from the projection $$\Pi$$ as $$Z_\Pi = 2I_2-\Pi$$. Without loss of generality we can always dilate (at the cost of an extra ancilla) $$A\mapsto\pmatrix{0 & A^\dagger\\A & 0}$$ and therefore we may assume that $$U_A^\dagger=U_A$$ and that $$U_A$$ is a Hermitian block-encoding. For each eigenvalue $$\lambda_i \neq \pm 1$$ of $$A$$ there exists a (2D) invariant subspace spanned by $$H_i = \text{span}(\{|0^m\rangle|v_i\rangle,|\perp_i\rangle\})$$, where $$|v_i\rangle$$ is an eigenvector of the block encoded matrix $$A$$. In the case where $$\lambda_i = \pm 1$$, we get a 1D invariant subspace $$H_i = \text{span}(|0^m\rangle|v_i\rangle)$$. The whole QSP procedure becomes interleaving a controlled phase difference between these invariant subspaces, the phase angles determining the matrix polynomial implemented by the QSP.

Do we need to know the spectrum of $$A$$ to find these invariant subspaces? If so, why even go through this process to begin with? If not, how does this algorithm work without knowing the spectrum of $$A$$ $$\textit{a priori}$$?