I came across the following problem that says:

Let $A$ and $B$ be $n \times n$ real matrices such that $AB=BA=0$ and $A+B$ is invertible. Then how can I prove the following:

  1. rank $A$+ rank $B$= $n$

  2. nullity $A$ + nullity $B$ =$n$

  3. $A-B$ is invertible.

Can someone point me in the right direction? Thanks in advance for your time.

  1. First $AB=0$ means $\mbox{Im}B\subseteq \mbox{Ker}A$, hence $\mbox{rank}B\leq \mbox{null} A=n-\mbox{rank} A$. Now $\mathbb{R}^n=\mbox{Im}(A+B)\subseteq \mbox{Im}A+\mbox{Im}B$, hence $n\leq \mbox{rank}A+\mbox{rank}B\leq \mbox{rank}A+n-\mbox{rank} A=n$. So $\mbox{rank}A+\mbox{rank}B=n$.

  2. Rank-nullity theorem on $A$ and on $B$.

  3. Observe that $(A-B)^2=A^2+B^2=(A+B)^2$.

Note: we only need $AB=0$ to obtain 1 and 2.

  • $\begingroup$ can you explain the point more clearly why $AB = 0$ means $Im B \subset Ker A$? $\endgroup$ – Germain Apr 17 '13 at 16:33
  • $\begingroup$ and why $\mathbb R^n =Im (A+B)$? $\endgroup$ – Germain Apr 17 '13 at 16:42
  • $\begingroup$ @analysis89 If $x$ belongs to $\mbox{Im} B$, we can find $y$ such that $x=By$. Then $Ax=ABy=0$. So $x$ belongs to $\mbox{Ker} A$. $\endgroup$ – Julien Apr 17 '13 at 16:42
  • $\begingroup$ $A+B$ invertible implies in particular $A+B$ surjective (and that's equivalent by the rank-nullity theorem). $\endgroup$ – Julien Apr 17 '13 at 16:44
  • 2
    $\begingroup$ The meaning of the Note is not clear to me. Do you mean that nothing but $AB=0$ is needed for 1. and 2. (contradicted by the fact that $A+B$ invertible is also used, as confirmed by the comments), or that $AB=0$ is not used to obtain 3. (but it would seem that $(A-B)^2=A^2+B^2$ requires $AB=BA=0$, or at least $AB+BA=0$). Maybe it just means that 1. and 2. can be obtained without the hypothesis that $BA=0$? It would be a rather cryptic way to say that. $\endgroup$ – Marc van Leeuwen Aug 28 '13 at 8:50

$BA = 0$ implies that the nullity of $B \geq$ the rank of $A$. $AB = 0$ implies that the nullity of $A \geq$ the rank of $B$.

$$ n(B) \geq r(A)$$ $$ n(A) \geq r(B)$$ $$ n(A) + r(A) + n(B) + r(B) = 2n$$

The final piece of the puzzle is that $A + B$ is invertible, this means that $r(A) + r(B) \geq n$ since we can't add two sets of vectors together and produce more linearly independent vectors than the sum of the rank of the two sets. The only conclusion is that $r(A) + r(B) = n$ and $n(A) + n(B) = n$.

To see that $A-B$ is invertible, $(A - B)^2 = A^2 + B^2 = (A + B)^2$


$\operatorname{rank}(AB) \geq \operatorname{rank} A+\operatorname{rank} B-n$ as $AB=0$ rank of $AB=0$ $$0 \geq \operatorname{rank} A+\operatorname{rank} B-n=0$$

  • 2
    $\begingroup$ You may want to add some explanations and format your answer to make it more readable. $\endgroup$ – mrf Aug 28 '13 at 8:52

(A)Show $\operatorname{rank}(A) + \operatorname{rank}(B) \ge \operatorname{rank}(A+B)$

$$Rank(A+B)\leq Rank (A)+Rank (B)$$ $$Rank(AB) \ge Rank(A)+Rank(b)-n$$ $$Rank(A+B)=n$$ Using these three results, $$Rank(A)+Rank(B)=n$$

(B) using the Rank-Nullity Theorem, we can easily prove.

(C)using this, $(A-B)^2=A^2+B^2=(A+B)^2$


Honestly, the asked conclusion is of no interest. A better writing would have been

Proposition. Let $A,B\in M_n(\mathbb{C})$ s.t. $AB=BA=0$ and $A+B$ is invertible.

Show that $A,B$ are simultaneously similar to

$A'=\begin{pmatrix}0_p&0\\0&U_{n-p}\end{pmatrix},B'=\begin{pmatrix}V_p&0\\0&0_{n-p}\end{pmatrix}$, where $U,V$ are invertible.

Proof. According to previous answers, we know that $rank(A)+rank(B)=n $. Then $Im(B)= \ker(A),Im(A)=\ker(B)$. Let $x\in \ker(A)\cap \ker(B)$; then $(A+B)x=0$ and therefore, $x=0$. According to the dimensions $p,n-p$ of the considered subspaces, $\mathbb{C}^n=\ker(A)\bigoplus\ker(B)$.

Considering an associated basis, we deduce that $A,B$ are simultaneously similar to

$A'=\begin{pmatrix}0_p&0\\0&U_{n-p}\end{pmatrix},B'=\begin{pmatrix}V_p&0\\0&0_{n-p}\end{pmatrix}$, where $U,V$ are invertible.


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