Matrix rank linear algebra

Question

We consider two matrices $A, B \in M_4(\mathbb C)$, $a \in \mathbb{C}^*$ and $$AB + aA + aB = 0 .$$

How can I prove that $AB = BA$ and $\operatorname{rank}(A) = \operatorname{rank}(B)$?

Answer 1

Here is a proof of the second point.

$AB + aA + aB = 0$ implies $A(-a^{-1}B-I)=B$, which in turn implies $\text{image}(B)\subseteq \text{image}(A)$.

$AB + aA + aB = 0$ also implies that if $v \in \ker B$, then $v \in \ker A$. Thus, $\ker B \subseteq \ker A$.

Now $ n = \dim\ker(B)+\dim\text{im}(B) \le \dim\ker(A)+\dim\text{im}(A) = n $.

This implies that $\dim\ker(B)=\dim\ker(A)$ and $\dim\text{im}(B)=\dim\text{im}(A)$. In particular, $\text{rank} A = \text{rank} B$.

Answer 2

$AB+aA+aB=0$ is equivalent $$(aI+A)(aI+B)=a^2I$$ or $$(I+\frac{A}{a})(I+\frac{B}{a})=I.$$ That means $I+\frac{A}{a}$ is the inverse of $I+\frac{B}{a}$. So we have $$(I+\frac{B}{a})(I+\frac{A}{a})=I,$$ and $AB=BA=-(aA+aB)$ follows.

The other part follows from the fact that $$A(I+\frac{B}a)=-B$$ and that $I+\frac{B}{a}$ is invertible.