# $\operatorname{rank} A = \operatorname{rank} A^2$ if and only if $\lim_{\lambda \to 0} (A+\lambda I)^{-1}A$ exists

Let $$A \in \mathbb C^{n \times n}$$. Prove that $$\operatorname{rank} A = \operatorname{rank} A^2$$ if and only if $$\displaystyle\lim_{\lambda \to 0} (A+\lambda I)^{-1}A$$ exists.

I am stuck on this problem, I don't understand what the limit is supposed to mean. I would guess that if the limit exists, it should be $$I$$ since the invertible matrices are dense. But how can I relate this to the rank?

• First step: show that $A$ and $A^2$ have the same rank iff the characteristic space for the eigenvalue $0$ is exactly the kernel of $A$. Second step: show that the limit exists iff for all $x$ in a characteristic space, $\lambda(A+\lambda I)^{-1}x$ has a limit as $\lambda$ goes to $0$. Third step: what happens for a characteristic space with a nonzero eigenvalue? Fourth step: what happens if, say, $A^2x=0$ but $Ax \neq 0$? Aug 25, 2019 at 22:10
• As $\operatorname{rank} A=\operatorname{rank} A^2$ and $\lim_{\lambda\to0}(A+\lambda I)^{-1}A=I$ whenever $A$ is invertible, it suffices to prove the problem statement when $A$ is a nilpotent Jordan block. Aug 25, 2019 at 22:29

Let $$V=\mathbb C^n$$. Since $$A^2V=A(AV)\subseteq A(V)=AV$$, if $$\operatorname{rank} A=\operatorname{rank} A^2$$, we must have $$A^2V=AV$$. Hence $$AV$$ is an invariant subspace on which $$A$$ is nonsingular. In turn, $$\lim_{\lambda\to0}(A+\lambda I)^{-1}A=I$$ on $$AV$$. Yet, we also have $$(A+\lambda I)^{-1}A=0$$ on $$\ker A$$ for every sufficiently small $$\lambda\ne0$$. Therefore $$\lim_{\lambda\to0}(A+\lambda I)^{-1}A$$ exists on $$AV+\ker A$$. This sum of subspaces must be equal to $$V$$, because $$AV\cap\ker A=0$$ (as $$A^2V=AV$$) and $$\dim(AV)+\dim(\ker A)=\dim(V)$$ (rank-nullity theorem). Thus $$\lim_{\lambda\to0}(A+\lambda I)^{-1}A$$ exists on $$V$$.

Conversely, observe that $$W=\ker A^2$$ is an invariant subspace of $$A$$ and \begin{aligned} &(A+\lambda I)^{-1}A=\frac{A}{\lambda}\left(I+\frac{A}{\lambda}\right)^{-1} =\frac{A}{\lambda}\left[I+\left(\frac{-A}{\lambda}\right)+\left(\frac{-A}{\lambda}\right)^2+\cdots\right] =\frac{A}{\lambda} \end{aligned} on $$W$$. So, if $$\lim_{\lambda\to0}(A+\lambda I)^{-1}A$$ exists, we must have $$A=0$$ on $$W$$, meaning that $$A^2x=0\Rightarrow Ax=0$$ for every vector $$x$$. But then $$A^2V=AV$$ and $$\operatorname{rank} A=\operatorname{rank} A^2$$.

Your limit exists iff for every $$x$$ in a characteristic space with eigenvalue $$\alpha$$, then $$(A+\lambda I)^{-1}Ax$$ converges as $$\lambda$$ goes to $$0$$.

Note that $$(A+\lambda I)^{-1}A=I-\lambda(A+\lambda I)^{-1}$$, so the limit exists iff for every $$x$$ in a characteristic space, $$\lambda(A+\lambda I)^{-1}x$$ converges as $$\lambda$$ goes to $$0$$.

Assume $$x$$ is in an characteristic space with eigenvalue $$\alpha \neq 0$$. Then $$(\lambda+\alpha)^nx=((A+\lambda I)-(A-\alpha I))^nx=(A+\lambda I)\sum_{k=1}^n{\binom{n}{k}(-1)^{n-k}(A+\lambda I)^k(A-\alpha I)^{n-k}}.$$

So if $$|\lambda|$$ is small enough, then $$(A+\lambda I)^{-1}x=\frac{1}{(\lambda+\alpha)^n}\sum_{k=1}^n\binom{n}{k}(-1)^{n-k}(A+\lambda I)^k(A-\alpha I)^{n-k}$$ is bounded as $$\lambda$$ goes to $$0$$, so $$\lambda(A+\lambda I)^{-1}x \rightarrow 0$$.

Let now $$x$$ be such that $$A^2x=0$$ and $$Ax=0$$. Then $$\lambda^2x=(A+\lambda I)(A-\lambda I)$$, so if $$\lambda$$ is small enough, $$\lambda(A+\lambda I)^{-1}x=\lambda\lambda^{-2}(\lambda I-A)x=x-\frac{Ax}{\lambda}$$ which does not converge.

Now, if $$\ker\,A^2 \subset \ker\,A$$, then the characteristic space for the eigenvalue $$0$$ is the kernel of $$A$$. Since $$\lambda(A+\lambda I)^{-1}x=x$$ if $$\lambda$$ is small enough and $$Ax=0$$, the limit exists iff the characteristic space of $$A$$ for the eigenvalue $$0$$ is the kernel of $$A$$, iff $$A$$ and $$A^2$$ have the same kernel.

It is standard linear algebra to show that for an endomorphism $$u$$ in finite dimension, $$u$$ and $$u^2$$ have the same kernel iff they have the same image iff they have the same rank.