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?

  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.