Objective question linear algebra CSIR -2017 Let $A$ be a $m\times n$  real matrix of of rank $m$ such that $m<n $. If for some non zero real number $\alpha $ we have $$ x^{t} A^{t} Ax = \alpha x^{t}x $$, for all $x\in \mathbb{R}^n$, then $A^{t} A$ has 


*

*Exactly two distinct eigen values.

*$0$ as an eigen value of multiplicity $n-m$.

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

*Exactly two non zero distinct eigen values.
According to me as $A^{t}A $ is symmetric and so diagonalizable and rank of it is same as that of number of non zero eigen values. So I know only that second option is correct. Please suggest me. Thanks.

Since $A^tA$ is symmetric, it has only real eigenvalues and the eigenvectors form an eigenbasis.  Let $\lambda$ be an eigenvalue for $A^tA$ with corresponding eigenvector $v$.  Then you know that 
$$
v^tA^tAv=v^t(\lambda v)=\lambda v^tv=\alpha v^tv.
$$
Since $v$ is a nonzero vector, this implies that $\|v\|\not=0$, and since the given property applies to all vectors in $\mathbb{R}^n$,
$$
\lambda \|v\|^2=\alpha\|v\|^2.
$$
Therefore, $\lambda=\alpha$.  Since $\lambda$ was an arbitrary eigenvalue, this eliminates all options except for $3$.
On the other hand, since $A$ is $m\times n$, $A^tA$ is an $n\times n$ matrix of rank at most $m$.  Therefore, $A^tA$ has rank at most $m$, and so it must have at least $n-m$ $0$ eigenvalues.  This is a contradiction, so it appears that none of the options are correct. 
