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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.

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    What is ELI5? As a hint, you can show it if you know about certain properties of dot (inner) products. – 2017-02-05
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    Explain Like I'm Five – 2017-02-05
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    @ArnoldFrenzy Oh ok, now that's CFC (clear from context) – 2017-02-05
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    Not a big friend of useless acronyms neither, just trying to help you – 2017-02-05
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    I decided to edit it out. Basically the point is, I'm only doing 1st year algebra, and I see a lot of explanations for things that simply go WAY over my head :) – 2017-02-05
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    I see. This does require bringing several ideas together which may seem unrelated (i.e. transposes and inner products). When first learning a topic in math, try to memorize the definitions and understand the first theorems/propositions that follow those definitions. This will give you an idea of how to work with the theory. – 2017-02-05
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    Yep, I think I see what you mean. In this case, I was overlooking that the null-space is a set of vectors where $Ax=0$ and not seeing how it related to the question. Now I can see how this is relevant, the question seems much less confronting :) – 2017-02-05
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    That would do it :) – 2017-02-05

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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 $x \in \ker A$.

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    Although it doesn't matter in the end, $(Ax)^T(Ax)=\|Ax\|^2$. – 2017-02-05
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    @Aweygan: Thanks for catching that, I am error prone... – 2017-02-05
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    No problem. Who isn't? – 2017-02-05
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    I am especially so :-). – 2017-02-05
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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.

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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...