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.

  • $\begingroup$ A matrix with $m>n$ cannot have rank $m$ as $rank(A)\leq\min\{m,n\}$, so the maximum rank can be $n$. $\endgroup$ – Josu Etxezarreta Martinez Mar 21 '18 at 11:36
  • $\begingroup$ I edit it .....thanks .... $\endgroup$ – neelkanth Mar 21 '18 at 11:39
  • 1
    $\begingroup$ See the Rayleigh quotient $\endgroup$ – Michael Burr Mar 21 '18 at 11:41
  • $\begingroup$ Thanks......... $\endgroup$ – 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.


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