[proofplan] Part (1) follows from the containment $\mathrm{im}(\alpha^{k+1}) \subseteq \mathrm{im}(\alpha^k)$. Part (2) identifies $d_k = r_k - r_{k+1}$ as $\dim(\ker(\alpha) \cap \mathrm{im}(\alpha^k))$ via rank-nullity, then uses the nested containment of images. Part (3) shows that $r_k = r_{k+1}$ implies $\alpha|_{\mathrm{im}(\alpha^k)}$ is injective, forcing all subsequent images to equal $\mathrm{im}(\alpha^k)$. [/proofplan] [step:Show $r_k \geq r_{k+1}$ from the containment $\mathrm{im}(\alpha^{k+1}) \subseteq \mathrm{im}(\alpha^k)$] Since $\mathrm{im}(\alpha^{k+1}) = \alpha(\mathrm{im}(\alpha^k))$, and the image of a subspace under a linear map is contained in the original space's image: $\mathrm{im}(\alpha^{k+1}) \subseteq \mathrm{im}(\alpha^k)$. Therefore $r_{k+1} = \dim\mathrm{im}(\alpha^{k+1}) \leq \dim\mathrm{im}(\alpha^k) = r_k$. [/step] [step:Identify $d_k = \dim(\ker(\alpha) \cap \mathrm{im}(\alpha^k))$ and show $d_k$ is non-increasing] The map $\alpha: \mathrm{im}(\alpha^k) \to \mathrm{im}(\alpha^{k+1})$ is surjective (by definition of $\mathrm{im}(\alpha^{k+1})$). By the rank-nullity theorem applied to $\alpha|_{\mathrm{im}(\alpha^k)}$: \begin{align*} r_k = \dim\mathrm{im}(\alpha^k) = \dim\ker(\alpha|_{\mathrm{im}(\alpha^k)}) + \dim\mathrm{im}(\alpha^{k+1}) = d_k + r_{k+1}, \end{align*} where $d_k = \dim\ker(\alpha|_{\mathrm{im}(\alpha^k)}) = \dim(\ker(\alpha) \cap \mathrm{im}(\alpha^k))$. Since $\mathrm{im}(\alpha^{k+1}) \subseteq \mathrm{im}(\alpha^k)$: \begin{align*} \ker(\alpha) \cap \mathrm{im}(\alpha^{k+1}) \subseteq \ker(\alpha) \cap \mathrm{im}(\alpha^k). \end{align*} Therefore $d_{k+1} = \dim(\ker(\alpha) \cap \mathrm{im}(\alpha^{k+1})) \leq \dim(\ker(\alpha) \cap \mathrm{im}(\alpha^k)) = d_k$. [guided] The differences $d_k = r_k - r_{k+1}$ measure how much the rank drops at each step. The formula $d_k = \dim(\ker(\alpha) \cap \mathrm{im}(\alpha^k))$ has a clean interpretation: $d_k$ counts the dimension of the "newly killed" subspace — vectors in $\mathrm{im}(\alpha^k)$ that $\alpha$ sends to zero. The monotonicity $d_k \geq d_{k+1}$ follows from the nested containment $\mathrm{im}(\alpha^{k+1}) \subseteq \mathrm{im}(\alpha^k)$: intersecting a smaller set with $\ker(\alpha)$ can only decrease the dimension. Concretely, every vector in $\ker(\alpha) \cap \mathrm{im}(\alpha^{k+1})$ is also in $\ker(\alpha) \cap \mathrm{im}(\alpha^k)$, so $d_{k+1} \leq d_k$. This non-increasing property of the differences is stronger than the non-increasing property of the ranks themselves. It encodes the "concavity" of the rank sequence: the drops get smaller and smaller. [/guided] [/step] [step:Prove stabilisation: $r_k = r_{k+1}$ implies $r_k = r_{k+\ell}$ for all $\ell \geq 0$] If $r_k = r_{k+1}$, then $d_k = r_k - r_{k+1} = 0$, so $\ker(\alpha) \cap \mathrm{im}(\alpha^k) = \{\mathbf{0}\}$. This means $\alpha|_{\mathrm{im}(\alpha^k)}$ is injective. Since $\mathrm{im}(\alpha^{k+1}) = \alpha(\mathrm{im}(\alpha^k))$ and $\alpha|_{\mathrm{im}(\alpha^k)}$ is injective, $\dim\mathrm{im}(\alpha^{k+1}) = \dim\mathrm{im}(\alpha^k)$. Combined with $\mathrm{im}(\alpha^{k+1}) \subseteq \mathrm{im}(\alpha^k)$ and equal dimensions: $\mathrm{im}(\alpha^{k+1}) = \mathrm{im}(\alpha^k)$. By induction, $\mathrm{im}(\alpha^{k+\ell}) = \mathrm{im}(\alpha^k)$ for all $\ell \geq 0$: applying $\alpha$ to $\mathrm{im}(\alpha^k) = \mathrm{im}(\alpha^{k+1})$ gives $\mathrm{im}(\alpha^{k+2}) = \alpha(\mathrm{im}(\alpha^{k+1})) = \alpha(\mathrm{im}(\alpha^k)) = \mathrm{im}(\alpha^{k+1}) = \mathrm{im}(\alpha^k)$, and so on. Therefore $r_{k+\ell} = r_k$ for all $\ell \geq 0$. [/step]