$\mathrm{rank}$ inequality about $A,B$ such that $AB=BA$

Question

Question: Let $A,B\in Mat_n(F)$ such that $AB=BA.$ Show that

1. $r(A)+r(B)\geq r(A+B)+r(AB);$
2. $r(A^2)+r(B^2)\geq 2r(AB).$

---

My work:

1. we show that $$\dim\ker A+\dim\ker B\leq \dim\ker (A+B)+\dim\ker AB.$$ By dimension formula, it's equivalent to $$\dim(\ker A+\ker B)+\dim \ker A\cap\ker B\leq\dim\ker(A+B)+\dim\ker AB.$$
2. By 1, we have $$\dim\ker(A ^2)+\dim\ker(B^2)\leq \dim\ker(A^2+B^2)+\dim\ker(ABAB).$$ Since $$\dim\mathrm{Im}\ ABAB=\dim\mathrm{Im}\ AB-\dim \ker AB\cap\mathrm{Im}\ AB,$$ then $$\dim\ker ABAB=\dim\ker AB+\dim\ker AB\cap\mathrm{Im}\ AB.$$ If we show that $$\dim\ker (A ^2+B^2)+\dim\ker AB\cap\mathrm{Im}\ AB\leq \dim\ker AB,$$ it's done. But i have no idea about this.

Answer 1

$\newcommand{\b}[1]{\begin{bmatrix}#1\end{bmatrix}}$

When $AB=BA$, it is not necessarily that $r(A^2)+r(B^2)\geq 2r(AB).$

Here is a counterexample.

$$A=\b{0&1&0&0\\0&0&0&0\\0&0&0&1\\0&0&0&0}$$ $$B=\b{0&1&1&0\\0&0&0&1\\0&0&0&1\\0&0&0&0}$$

Then $$A^2=0$$ $$B^2=\b{0&0&0&2\\0&0&0&0\\0&0&0&0\\0&0&0&0}$$ $$AB=BA=\b{0&0&0&1\\0&0&0&0\\0&0&0&0\\0&0&0&0}$$ We have $$r(A^2)+r(B^2)=0+1=1\lt 2= 2r(AB).$$

Answer 2

The first from the standard observation $\text{Ima}(A+B)\subset\text{Ima}(A)+\text{Ima}(B)$,

noting that $\text{Ima}(A)\cap\text{Ima}(B)\supset\text{Ima}(AB)$.

Where $\text{Ima}(A)$ is the image of the linear map $x\mapsto Ax$, the subspace spanned by the columns of $A$, the set of all vectors that can be written as $Ax$ for some $x$.