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
exactly two distinct eigen values
$0$ as an eigen value with multiplicity $n-m$
$\alpha$ is a non zero eigen value.
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 $3$.
Please tell me whether I am right or made any mistake?
