# What can you say about these matrices.

Let $A$ and $B$ be two $n\times n$ matrices over $\mathbb{R}$ such that $AB=BA=0$ and $A+B$ is invertible then:

(1) Is there any relation between rank$(A)$ and rank$(B)$?

(2) Is $A-B$ invertible?

(3) Is there any relation between their nullity?

Thanks!

-
What do you know? What have you tried? Is this homework? –  Julian Kuelshammer Dec 29 '12 at 11:04
I cannot guess anything, I got it in a competitive exam. –  rajkamalmath Dec 29 '12 at 11:10
@rajkamalmath: The reason you're in a competitive exam is presumably to show that you know something, and/or have the wits to try something new to find a result. Thus Julian's questions still apply. –  Henning Makholm Dec 29 '12 at 12:22

(1) On one hand, as $AB=0$, we have $\operatorname{range}B\subseteq\ker A$. Therefore $\operatorname{rank}B \le \operatorname{nullity}A$ and in turn $$(\dagger):\ \operatorname{rank}(A+B) \le \operatorname{rank}A+\operatorname{rank}B \le \operatorname{rank}A+\operatorname{nullity}A = n.$$ On the other hand, as $A+B$ is invertible, we have $\operatorname{rank}(A+B)=n$. Therefore, equality must hold in $(\dagger)$ and hence $\operatorname{rank}A+\operatorname{rank}B=n$.

(2) Yes, $A-B$ must be invertible. Suppose the contrary. Then there exists some nonzero vector $u$ such that $(A-B)u = 0$. Let $v=Au=Bu$. If $v\neq0$, then $(A+B)v=ABu+BAu=0$, which is a contradiction because $A+B$ is invertible. If $v=0$, then $(A+B)u = 2v = 0$, which is also a contradiction.

(3) Using the rank-nullity theorem and the result of (1), we have $\operatorname{nullity} A+\operatorname{nullity} B=n$.

-
thank you very much, for your help. –  rajkamalmath Dec 29 '12 at 17:18

1) AB = 0 implies that rank(A) + rank (B) $\leq n$ (since range B is in null A). A+B is invertible gives $n =$rank(A+B), which is $\leq$ rank(A) + rank(B) (by dimension of basis). Hence rank(A) + rank(B) = n.

2) [I can only do this over $\mathbb{C}$, need algebraically closed. The result might still hold over $\mathbb{R}$, but I'm not certain due to complex eigenvalues/eigenvectors.] Consider A, B as matrices in $M_n (\mathbb{C})$. AB=BA=0 means that they are simultaneously triangularisable. With this basis, AB=0 means that if a diagonal entry of A is non-zero, then the corresponding entry of B is 0. Likewise, if a diagonal entry of B is non-zero, then the corresponding entry of A is 0. A+B is invertible (in this basis) means that an entry on a diagonal cannot be simultaneously 0. Hence, looking the diagonal entries of A-B are all non-zero, so A-B is invertible. In fact, Det(A-B) = (-1)^k Det(A+B), where k = rank (B).

3) Since rank(M)+null(M) =Dim(M), it follows from 1) that Null(A) + Null(B) = n.

-
Very nice Clive! Good, Good... +1 –  B. S. Jan 18 '13 at 8:15