# 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

1. Exactly two distinct eigen values.

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

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

4. 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.

• A matrix with $m>n$ cannot have rank $m$ as $rank(A)\leq\min\{m,n\}$, so the maximum rank can be $n$. – Josu Etxezarreta Martinez Mar 21 '18 at 11:36
• I edit it .....thanks .... – neelkanth Mar 21 '18 at 11:39
• See the Rayleigh quotient – Michael Burr Mar 21 '18 at 11:41
• Thanks......... – neelkanth Mar 21 '18 at 12:56

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.