-1

Suppose A is a square matrix such that $A^T A= AA^T$. Show that the nullspace of $A^T$ is equal to the nullspace of $A$.

Here is my attempt at a proof:

We must show that, given $A^T A= AA^T$: $$Ax=0 , A^T y=0 \iff x=y, \forall x ,y$$ Now, suppose $x$ is in the nullspace of $A$ and $y$ in the nullspace of $A^T$: $$Ax=0 , A^Ty=0$$ Multiplying the left most equation by $A^T$ and the rightmost equation by $A$: $$A^TAx=0, AA^Ty = 0$$ As $A^T A= AA^T$, we have: $$AA^Tx=0, A^TAy=0$$ At this point, I would like to claim that $x=y$, however I'm unsure if this is the case, as if we consider $AA^Tx=0$, it could be the case that $A^Tx$ is in the nullspace of $A$ rather $x$ being in the nullspace of $A^T$.

Thoughts?

David C. Ullrich
  • 92,839
William
  • 515
  • 1
    You have to show that $\ker A=\ker A^T. $ i.e. $Ax=0\ \iff A^Tx=0$ for any $x.$ Do you are dealing with one element $x$ at a time. The formula $\langle B^TBx,x\rangle =\langle Bx,Bx\rangle$ should be useful. – Ryszard Szwarc Aug 28 '22 at 08:21
  • No, $Ax=0,A^ty=0\iff x=y$ is absolutely not what you have to show. – David C. Ullrich Aug 28 '22 at 12:51

2 Answers2

0

We have $$ \ker(A)=\ker(A^TA)=\ker(AA^T)=\ker(A^T) $$ because of the assumption and this duplicate:

Prove that for a real matrix $A$, $\ker(A) = \ker(A^TA)$.

Dietrich Burde
  • 140,055
0

Proof: We begin by showing $Ax=0 \Rightarrow A^T x =0$ $$Ax=0$$ $$A^TAx=0$$ As $A^TA = AA^T$, $$AA^Tx=0$$ We now multiplt by $x^T$, noting that $x^T A = (A^Tx)^T$ $$x^T AA^Tx = (A^Tx)^T (A^Tx) = ||A^Tx||^2=0$$ Thus we have proven the initial statement, and proving its converse is done in a similar fashion. Q.E.D

William
  • 515