Past exam of the mathematics course of the University of Cambridge / 2015 / ii / Paper 4 / 37E / a

The Krylov subspace is $K_m(A,v)=\mathrm{span}\{v,Av,\dots ,A^{m-1}v\}$. Let $s$ be the first index for which $A^{s}v$ depends linearly on $v,\dots ,A^{s-1}v$. Such an index exists with $1\le s\le n$, since $v\ne 0$ and any $n+1$ vectors are dependent. The first $s$ vectors are independent. The relation for $A^{s}v$ shows that their span is invariant under $A$: multiplying any generator by $A$ stays in that span. Repeated multiplication makes every higher power stay there too. ThereforeOnce a Krylov subspace stops growing, it cannot resume growth.

Let the distinct eigenvalues be ${\lambda }_1,\dots ,{\lambda }_{s^{\prime }}$. Distinct-eigenvalue eigenvectors are independent. A relation ${\sum }_{k=0}^{m-1}{a}_k{A}^{k}v=0$ becomes ${\sum }_i{c}_{i}p({\lambda }_i)w_i=0$, with $p(z)={\sum }_{k<m}{a}_k{z}^k$. Since every $c_i\ne 0$, it requires $p({\lambda }_i)=0$ for every $i$. No nonzero polynomial of degree below $s^{\prime }$ has that many distinct roots. Thus the first $s^{\prime }$ powers are independent, while ${\prod }_i(A-{\lambda }_{i}I)v=0$ gives the next dependence. Hence $\boxed{s=s^{\prime }}$. Equivalently the coefficients of these powers form a full-rank Vandermonde matrix through this index.