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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?

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    $\begingroup$ 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 $Ax \neq 0$? $\endgroup$
    – Aphelli
    Aug 25, 2019 at 22:10
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    $\begingroup$ 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. $\endgroup$
    – user1551
    Aug 25, 2019 at 22:29

2 Answers 2

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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 $\operatorname{rank} A=\operatorname{rank} A^2$.

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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.

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