 Solution

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.