Let $A$ be an $m\times n$ matrix of rank $m$ with $n\gt m$. If for some non-zero real number $a$, we have $x'AA'x=ax'x$, for all $x\in \Bbb R^m$ then $A'A$ has:

  1. exactly two distinct eigenvalues.
  2. $0$ is an eigenvalue with multiplicity n-m.
  3. $a$ is a non-zero eigenvalue.
  4. exactly two non-zero distinct eigenvalues.

I tried this problem. $A$ has rank $m$ and $AA'$ has rank $m$. Hence the nullity of $AA'$ is zero. But the kernel of $A$ has dimension $n-m$. The rank of $A'A$ is also $m$ hence its kernel has dimension $n-m$. Therefore 2 is the correct option.

$x'AA'x=ax'x$ is same as $\langle AA'x,x \rangle =\langle ax,x \rangle$. I am stuck here. How to tackle the other options?


Since, $x^t A A^t x = a x^tx$, for all $x \in \mathbb {R^m}$, applying transpose both sides will give you $x^t A^t A x = a x^tx$. Now, let $0 \neq \lambda$ be an eigenvalue of $A^tA$.
By definition, $(A^t A)x = \lambda x$, multiplying both sides with $x^t$ will give you $\lambda = a$. That is, every non-zero eigenvalue of $A^t A$ is $a$. This proves $0, a$ are the only eigenvalues, which proves 1 and 3.

  • $\begingroup$ I am impressed by your blog. Why you stopped writing? $\endgroup$ – StammeringMathematician May 22 at 5:57
  • $\begingroup$ Thank you, I am working on it. $\endgroup$ – Atul Anurag Sharma Jun 14 at 0:31

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