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!

  • $\begingroup$ What is $\dim(A)$? Nullity of $A$? $\endgroup$ – Shubham Johri Nov 6 '20 at 8:07
  • $\begingroup$ I was intending to invoke the Dimension-Nullity Theorem... $\endgroup$ – a9302c Nov 6 '20 at 8:22
  • $\begingroup$ 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. $\endgroup$ – 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 $v=Au=0$.

  • $\begingroup$ May I know what the syntax of ∃u|Au=v means? I have not come across such expression before $\endgroup$ – a9302c Nov 6 '20 at 9:36
  • $\begingroup$ @a9302c It is read as: "there exists ($\exists$) $u$ such that (|) $Au=v$" $\endgroup$ – Shubham Johri Nov 6 '20 at 9:37
  • $\begingroup$ Thank you! I understand now. Would never have thought of this manipulation $\endgroup$ – a9302c Nov 6 '20 at 10:07
  • $\begingroup$ You are welcome $\endgroup$ – Shubham Johri Nov 6 '20 at 10:07

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