# Multiplicity of Ritz eigenvalues

Consider a Krylov subspace $$K_m=\mathrm{span}\{v,Pv,...,P^{m-1}v\}$$, for $$P$$ a square matrix and a nonzero vector $$v$$. Let $$H_m$$ represent the projection of $$P$$ (seen as an application) restricted to $$K_m$$, onto $$K_m$$. That is, for $$V_m=[v_1,...,v_m]$$ a basis of $$K_m$$:

$$H_m = V^T_m P V_m$$

Assume, if necessary, that $$V_m$$ is orthonormal ($$V^T_m V_m =I$$), and that for no $$m' it holds that $$K_{m'}=K_m$$.

Question: is it true that all eigenvalues of $$H_m$$ (Ritz eigenvalues) have geometric multiplicity 1, and what is a (simple) proof of this fact? Any reference will be appreciated.

• This is true. An easy way of seeing this is to write down the matrix of $P$ wrt the basis $v,\ldots ,P^{m-1}v$ and look for eigenvectors for a given (putative) eigenvalue. (This property of the matrix corresponds to the fact that a constant coefficient ODE has solutions $e^{\lambda x}, x e^{\lambda x}, \ldots$ for a given characteristic value $\lambda$.) Feb 5, 2015 at 21:47
• After a little of thought, I think a very simple answer is the following: under the given conditions (orthonormality, etc.), $H_m$ is upper Hessenberg, and moreover all the elements of its first lower subdiagonal are $\neq 0$. This implies that $rk(\lambda I_m - H_m)=m-1$, for any $\lambda$, that is $ker(\lambda I_m - H_m)=1$. Feb 5, 2015 at 23:09
• Also, see here (the first equivalence in particular): en.wikipedia.org/wiki/Companion_matrix Feb 5, 2015 at 23:23