Help with Rank and Nullity of transpose matrices

Question

I'm stuck on this question;

Show that $Nul(A^tA) = Nul(A)$ for every matrix A.

I don't really know where to start on this. I know that rank is not changed by transposing, so nullity is also something I can figure out (rank nullity theorem).

But this is as far as I can get, and I have no idea how to tackle this in regards particularly to the matrix product, $A^tA$.

EDIT: I basically just added everything I know about the topic. Question doesn't change.

Answer 1

If $x \in \ker A$ then since $A^T(Ax) = A^T 0 = 0 $ we see that $x \in \ker A^T A$.

If $x \in \ker A^T A$, then $A^T A x = 0$ and so $x^T A^T A x = (Ax)^T (Ax) = \|Ax\|^2 = 0$ and so $x \in \ker A$.

Answer 2

In case you are working in linear algebra (as your tag specifies) you can think that the Null Space of a linear map (or application) represented by $A$ are all those vectors $x$ of the Domain such that:

$Ax = 0$

And $Ax = 0$ iff $x^TA^T = 0$. Then

$x^TA^TAx = 0$ will be true for exactly the same $x$ for which $Ax = 0$ is true...

Answer 3

You have to show that both sets are equal and equality of sets are shown through containment both ways.

It is clear that something in Null (A) is clearly in Null (A$^{t}$A).

Now the other way, if some vector x $\in$ Null (A$^{t}$A) then, A$^{t}$Ax = 0 which implies that x$^{t}$A$^{t}$A(x) = 0 which implies that,

(Ax)$^{t}$ Ax = 0 i.e. Ax.Ax=0 implying Ax = 0 implying x $\in$ Null (A).

So there is your two way, containment.