= Solution
For a finite-dimensional linear operator $A$, the <Krylov subspace> is
$$
\mathcal K_m(A,v)=\operatorname{span}\{v,Av,\ldots,A^{m-1}v\}.
$$
These spaces are nested, so their dimensions are nondecreasing and can rise by at most one at each step. If $v\ne0$, let $k$ be the first index for which $A^kv$ is a <linear combination> of $v,\ldots,A^{k-1}v$. The preceding $k$ vectors are <independent>. Multiplying that dependence by $A$ and repeatedly substituting shows every higher power also lies in their span. Thus
$$
\boxed{d_m=m\ (1\leq m\leq k),\qquad d_m=k\ (m\geq k).}
$$
Such an index exists because the ambient space is finite-dimensional. If $v=0$, the separate answer is $d_m=0$ for every $m$, with $k=0$.
For <Krylov dimension from spectral components>, when $A$ is <diagonalizable>, group the <eigenvector> expansion by distinct <eigenvalues>:
$$
v=u_1+\cdots+u_k,\qquad Au_j=\lambda_ju_j,
$$
where each $u_j$ is its nonzero component in a distinct <eigenspace>. Then $A^rv=\sum_j\lambda_j^ru_j$. The vectors $u_j$ are <independent>, and the <Vandermonde matrix> $(\lambda_j^r)_{0\leq r<k}$ is invertible because the <eigenvalues> are distinct. Hence the first $k$ powers span exactly $\operatorname{span}\{u_j\}$, proving \b[the stabilized dimension equals the number of distinct <eigenspaces> with nonzero component in $v$]. These $u_j$ themselves are <eigenvectors>, so this is also the minimum number of <eigenvectors> needed to represent $v$ when the <eigenvectors> may be chosen freely.
If the printed phrase instead means the count of nonzero coefficients in an arbitrary fixed eigenbasis, it is false with repeated <eigenvalues>. For example $A=I$ and $v=e_1+e_2$ use two vectors of that <basis> but have $\dim\mathcal K_m=1$; the vector $v$ itself is an <eigenvector>. For a simple <spectrum> there is no distinction. This qualification resolves the claim without assuming an unstated simple <spectrum>.
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