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Let $A$ and $B$ be $n$ $\times$ $n$ real matrices such that $AB$=$BA$=$0$ and $A+B$ is invertible. Which of the following are always true?

$1$. rank $(A)$=rank $(B)$.

$2$. rank $(A)$$+$rank $(B)$ =$n$

$3$. nullity $(A)$ $+$ nullity $(B)$ = $n$

$4$. $A-B$ is invertible.

I am not getting how to do. Atleast it is clear that Rank$(A+B)$= $n$. help me!

Topology
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2 Answers2

2

As has been stated by others, 1 is false taking $A=0$ and $B$ any invertible matrix.

Since $AB=0$, $A$ annihilates the image of $B$, so the nullity of $A$ must be at least as large as the rank of $B$, so $$\operatorname{nullity}(A)+\operatorname{nullity}(B)\geq \operatorname{rank}(B)+\operatorname{nullity}(B)=n$$ If $\operatorname{nullity}(A)+\operatorname{nullity}(B)>n$, then $\operatorname{nul}(A)$ and $\operatorname{nul}(B)$ are subspaces of $V$ with dimensions that sum to more than $n$. So the union of a basis for $\operatorname{nul}(A)$ with a basis for $\operatorname{nul}(B)$ has more than $n$ vectors and must be dependent. Therefore some nonzero vector $v$ is in both $\operatorname{nul}(A)$ and $\operatorname{nul}(B)$. But this contradicts the given condition that $A+B$ is invertible, since $(A+B)v=0$. Therefore $$\operatorname{nullity}(A)+\operatorname{nullity}(B)=n$$ proving 3.

2 is true by using 3 and the rank-nullity theorem: subtract each side of 3 from $2n$.

4 is true because in the given conditions, $(A+B)^2$ is invertible, and $$(A+B)^2=A^2+AB+BA+B^2=A^2-AB-BA+B^2=(A-B)^2$$

2'5 9'2
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Since $A$ and $B$ commute, each one stabilises the kernel and image of the other. Then $A+B$ stabilises each of these kernels and images, and since $A+B$ is invertible, so is each of its restrictions. Now the restriction of$~B$ to the image of$~A$ is zero (since $BA=0$), so the restriction of$~A$ to its own image coincides with the restriction of $A+B$, and therefore is invertible. This means the image if $A$ has zero-dimensional intersection with the kernel of $A$, and by the rank-nullity theorem the two subspaces are therefore complementary in $\mathbf R^n$. The same holds similarly for kernel and image of $B$. Since the kernel of $B$ contains the image of $A$ and vice versa, these two inclusions are equalities for dimension reasons: the kernel of $B$ equals the image of $A$ and vice versa.

Now for the points in the question:

  1. certainly there is no reason why the images of $A,B$ should have the same dimension; they even cannot do so if $n$ is odd;
  2. on the other hand the subspaces being complimentary, the dimensions of the images of $A,B$ do add up to$~n$;
  3. and so do the dimensions of their kernels, which are just the same subspaces but interchanged;
  4. the map $A-B$ coincides with $A+B$ on the kernel of $B$, and with $-(A+B)$ on the kernel of $A$; both restrictions to these complimentary subpaces being invertible, so is $A-B$ itself.

The final argument also shows that $\lambda A+\mu B$ is invertible whenever $\lambda\neq0\neq\mu$.

Marc van Leeuwen
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