 Krylov dimension from spectral components

For a diagonalizable finite-dimensional operator, write $v={\sum }_j{u}_j$ with nonzero $u_j$ in distinct eigenspaces of eigenvalues ${\lambda }_j$. Then $A^{r}v={\sum }_j{\lambda }_j^r{u}_j$. The Vandermonde matrix of the first $k$ powers is invertible for $k$ distinct eigenvalues, so these powers span exactly the $k$ independent components $u_j$. Thus the stabilized Krylov subspace dimension is $k$. It counts distinct spectral components, not the number of nonzero coefficients in an arbitrary eigenbasis: $A=I$, $v=e_1+e_2$ has Krylov dimension one. Equivalently it is the minimum number of eigenvectors needed when those vectors may be chosen freely.