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

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?

## 2 Answers

$$\triangleright$$ 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

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…

• Why does $ker B \subseteq ker A$ if $A^2B = A$? Commented Mar 11, 2017 at 17:36
• Take any element in $\ker B$ . What's its image by $A^2B$? Commented Mar 11, 2017 at 17:42
• I am not sure. If $A^2B = A$, that means that both $A^2B$ and $A$ have the same column space. But I don't quite get what you are trying to suggest. Commented Mar 11, 2017 at 18:25
• I think in terms of linear maps between vector spaces. However I see I'm missing an argument, I'll complete my answer in a moment. Commented Mar 11, 2017 at 19:16
• You didn't show that $\ker B$ and $\ker A^2$ have the same dimension. Commented Aug 25, 2021 at 10:15