Let $A$ and $B$ be square matrices of order $n$. Suppose that $\mathrm{rank}(A)=\mathrm{rank}(B)$ and $A^2B=A$.

Question

I am having difficulty trying to solve this problem during my practice:

Let $A$ and $B$ be square matrices of order $n$. Suppose that $\mathrm{rank}(A)=\mathrm{rank}(B)$ and $A^2B=A$. Prove the following:

1. $ B $ and $ A^2 $ have the same nullspace
   
2. $ B^2A = B $

I am given the following hints:

For 1)

Use the fact that:

(i) $\mathrm{rank}(AB) \le \min\{\mathrm{rank}(A),\mathrm{rank}(B)\}, A = (a_{ij})_{m \times n} $ and $ B = (b_{ij})_{n \times p} $
(ii) $\mathrm{rank}(A) + \mathrm{nullity}(A) = n$

to get a relation in $\mathrm{nullity}(B)$ and $\mathrm{nullity}(A^2)$. Then use the fact if $ U \subseteq V $ are vector spaces such that $\dim(U) = \dim(V) $, then $ U = V $.

For 2)

(i) Note that $ Bu = 0 \iff A^2u = 0 $
(ii) Prove that $(B^2A - B)x = 0 $ for all $ x \in \mathbb{R}^n $

For 1), my attempt:

$$ A^2B = A $$ $$ A^2Bx = 0 \iff Ax = 0, x \in \mathbb{R}^n $$ $$ Bx \in \mathrm{nullity}(A^2) $$

I am stuck here. My direction of thought seems wrong. But I really do not know what else to do.

For 2), my attempt:

$$ A^2x = 0, x \in \mathrm{nullity}(A^2), \mathrm{nullity}(B) $$ $$ Bx = 0, x \in \mathrm{nullity}(B), \mathrm{nullity}(A^2) $$ $$ B^2A = BBA = Bu \ \text{(Let }BA = u)$$ $$ \therefore Bu = 0 $$ $$ A^2u = 0 $$ $$ Bu - A^2u = 0 $$ $$ (B - A^2)u = 0 $$

I cannot seem how to get to my desired answer from here.

I spent hours trying to utilize the hints but to no avail. The final product of the proof seems too far from the given conditions. Could someone please advise me?

Answer 1

$\triangleright$ The relation $\mathrm{A}^2 \mathrm{B}=\mathrm{A}$ implies $\mathrm{ker}(\mathrm{B})\subset \mathrm{ker}(\mathrm{A})$. Since $\mathrm{rank}(\mathrm{A})=\mathrm{rank}(\mathrm{B})$, the rank formula applied to $\mathrm{A}$ and $\mathrm{B}$ gives $\dim(\mathrm{ker}(\mathrm{A}))=\dim(\mathrm{ker}(\mathrm{B}))$ and thus $\mathrm{ker}(\mathrm{A})=\mathrm{ker}(\mathrm{B})$.

$\triangleright$ As $\mathrm{A}^2 \mathrm{B}=\mathrm{A}$, we have $\mathrm{im}(\mathrm{A})\subset \mathrm{im}(\mathrm{A}^2)$. Universally, $\mathrm{im}(\mathrm{A}^2)\subset \mathrm{im}(\mathrm{A})$, resulting in $\mathrm{im}(\mathrm{A})=\mathrm{im}(\mathrm{A}^2)$ and thus $\mathrm{rank}(\mathrm{A})=\mathrm{rank}(\mathrm{A}^2)$. Applying the rank formula to $\mathrm{A}$ and $\mathrm{A}^2$ gives $\dim(\mathrm{ker}(\mathrm{A}))=\dim(\mathrm{ker}(\mathrm{A}^2))$. The universal inclusion $\mathrm{ker}(\mathrm{A})\subset \mathrm{ker}(\mathrm{A}^2)$ and the equality of dimensions result in $\mathrm{ker}(\mathrm{A})=\mathrm{ker}(\mathrm{A}^2)$.

Thus, we have $\mathrm{ker}(\mathrm{A})=\mathrm{ker}(\mathrm{B})=\mathrm{ker}(\mathrm{A}^2)$.

We have: \begin{align*} \mathrm{A}^2\mathrm{B}=\mathrm{A} & \quad \text{yields $\quad \mathrm{A}^2\mathrm{B}\mathrm{A}=\mathrm{A}^2$}\\ & \quad \text{yields $\quad \mathrm{A}^2(\mathrm{B}\mathrm{A}-I_n)=0$}\\ & \quad \text{yields $\quad \mathrm{im}(\mathrm{B}\mathrm{A}-I_n)\subset\mathrm{ker}(\mathrm{A}^2)$}\\ & \quad \text{yields $\quad \mathrm{im}(\mathrm{B}\mathrm{A}-I_n)\subset\mathrm{ker}(\mathrm{B})$}\\ \mathrm{A}^2\mathrm{B}=\mathrm{A} & \quad \text{yields $\quad \mathrm{B}^2\mathrm{A}=\mathrm{B}$.} \end{align*}

Have a nice day

Answer 2

A hint for 1):

Observe a) $\dim\ker A=\dim\ker B$ by the rank-nullity theorem.

b) Since $A^2B=A$, we have $\ker B\subseteq \ker A$. Further, $\ker A\subseteq\ker A^2$, so $\ker B\subseteq\ker A^2$, and they have the same dimension…