1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.