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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 $m'<m$ 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.

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  • $\begingroup$ 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 $\lambda$.) $\endgroup$
    – Christian Remling
    Commented Feb 5, 2015 at 21:47
  • $\begingroup$ 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 $ker(\lambda I_m - H_m)=1$. $\endgroup$
    – Michele
    Commented Feb 5, 2015 at 23:09
  • $\begingroup$ Also, see here (the first equivalence in particular): en.wikipedia.org/wiki/Companion_matrix $\endgroup$
    – Christian Remling
    Commented Feb 5, 2015 at 23:23

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