# Relationship between nullspaces of A when $rank(A) = rank(A^2)$

I have a square matrix A such that $$rank(A) = rank(A^2)$$. I am first tasked to prove that nullspace of $$A$$ is equal to the nullspace of $$A^2$$. For this, my reasoning is:

Let $$v$$ be a solution to the nullspace of $$A$$. As such, $$v$$ will also be a solution to the nullspace of $$A^2 = A*A$$. Hence nullspace $$A \subseteq \$$subset of $$A^2$$.Then, because $$rank(A) = rank(A^2)$$, $$dim(A) = dim(A^2)$$. Hence nullspace of $$A$$ = nullspace of $$A^2$$. How can I make this more airtight as a proof?

In addition, how can I possibly extend this concept to show that (nullspace of $$A$$) $$\cap$$ (column space of $$A$$) = $$\{0\}$$? I don't quite understand the relationship between the nullspace and column space.

Thank you!

• What is $\dim(A)$? Nullity of $A$? – Shubham Johri Nov 6 '20 at 8:07
• I was intending to invoke the Dimension-Nullity Theorem... – a9302c Nov 6 '20 at 8:22
• Rank of $A$ is the dimension of range space of $A$. But $\dim A$ does not make sense, you could say $\dim\text{Im(A)}$ (dimension of image of $A$) instead. – Shubham Johri Nov 6 '20 at 8:23

I don't understand what $$\dim A$$ means. Perhaps you want to say nullity of $$A$$. Then your proof is "air-tight" because $$\text{rank}(A)=\text{rank}(A^2)\implies \text{nullity}(A)=\text{nullity}(A^2)=n-\text{rank}(A)$$ by the rank-nullity theorem, where $$n$$ is the size of $$A$$.
The column space of $$A$$ is the range space of $$A$$. Let $$v$$ belong to both the range space and null space of $$A$$. Then $$\exists u|Au=v$$ and $$Av=0$$. This gives $$A(Au)=A^2u=0$$, thus $$u$$ belongs to the null space of $$A^2$$. But since the null space of $$A$$ is the same as the null space of $$A^2$$, $$u$$ belongs to the null space of $$A$$. Thus $$v=Au=0$$.
• @a9302c It is read as: "there exists ($\exists$) $u$ such that (|) $Au=v$" – Shubham Johri Nov 6 '20 at 9:37