= Solution
The <Krylov subspace> is $K_m(A,v)=\operatorname{span}\{v,Av,\ldots,A^{m-1}v\}$. Let $s$ be the first index for which $A^s v$ depends linearly on $v,\ldots,A^{s-1}v$. Such an index exists with $1\leq s\leq n$, since $v\ne0$ and any $n+1$ vectors are dependent. The first $s$ vectors are independent. The relation for $A^sv$ 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. Therefore
$$
\boxed{\dim K_m=m\ (m\leq s),\qquad\dim K_m=s\ (m>s).}
$$
Once a <Krylov subspace> stops growing, it cannot resume growth.
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