How to determine the eigen values of $A^tA$ where $x^tA A^tx=\alpha x^tx ~~~~ \forall x \in \Bbb{R}^m$ holds for some $\alpha$

Problem. A be an $$m\times n$$ matrix of rank $$m$$ with $$n>m$$. If for some nonzero real $$\alpha$$ we have $$x^tA A^tx=\alpha x^tx ~~~~ \forall x \in \Bbb{R}^m$$ then $$A^tA$$ has

1. exactly two distinct eigen values

2. $$0$$ as an eigen value with multiplicity $$n-m$$

3. $$\alpha$$ is a non zero eigen value.

4. exactly two non zero eigen values.

My attempt.

We know, $$rank(A^tA)=rank(A)=m$$. So, $$Nullity(A^tA)=n-m>0$$ So, $$0$$ is an eigen value of $$A^tA$$ with geometric multiplicity $$n-m$$. But $$A^tA$$ is symmetric so is diagonalizable i.e. $$0$$ is regular. Hence $$0$$ is an eigen value of (algebraic) multiplicity $$n-m$$ i.e. Option 2 is correct. (without using the given equation)

Now we have to conclude about the rest. I think for this we have to use the given equation.

I start by checking for the non zero eigen values of $$A^tA$$ (In fact, $$A^tA$$ should have $$m$$ non zero real eigen values)

Let $$\lambda$$ be a non zero eigen value of $$A^tA$$. Then there exists a non-null vector $$v \in \Bbb{R}^n$$ such that $$A^tAv=\lambda v$$. Now to use the given equation I think we need a vactor $$x \in \Bbb{R}^m$$. So, I try to choose $$x=Av$$ in that equation.....But I can't get any conclusion from there...!!

Any help is appreciated. Thank you.

N.B: This question has an answer here. But I cannot find out how to use $$AA^t=\alpha I$$ further.

EDIT: Well, I have found a way to proceed further:

Let $$\lambda$$ be a non-zero eigen value of $$A^tA$$. Then it can be proved it is also an eigen value of $$AA^t$$ (see here). Therefore, there exits a non-zero vector $$v \in \Bbb{R}^m$$ such that $$AA^tv=\lambda v$$. Again from the given equation we have, $$v^tAA^tv=\alpha v^tv$$, i.e., $$v^t(AA^tv)=\alpha v^tv$$, or, $$\lambda v^tv=\alpha v^tv$$ this implies, $$\lambda = \alpha$$.

Hence the only non-zero eigen value of $$A^tA$$ is $$\alpha$$.

So, finally Correct options are $$1,2$$ and $$3$$.

Your logic looks good to me, though it can be done more simply than you have it. You were on the right track with setting $$x = Av$$. Then $$A^tx = A^tAv = \lambda v\\AA^tx = \lambda Av = \lambda x\\x^tAA^tx = \lambda x^tx\\\alpha x^tx = \lambda x^tx$$ Since $$\|x\| \ne 0$$, we have $$\lambda = \alpha$$.