$A_{m \times n}, n>m, AA^t =\alpha I , \alpha> 0$, and rank A = m , what can we conclude about eigenvalues of $A^tA$ ?

Like we can conclude information about $AA^t$ of order $m \times m$ that since its a diagonal matrix so the eigenvalues will be the elements on the diagonal which is $\alpha$ with multiplicity $m$.But how to conclude that $A^t A$ has $\alpha$ as one eigenvalue of multiplicity $m$ and eigenvalue $0$ of multiplicity $n-m$?

  • $\begingroup$ Very related question if not duplicate. $\endgroup$ – A.Γ. Jun 16 '18 at 15:03
  • $\begingroup$ But how to think of the zero eigenvalue? $\endgroup$ – BAYMAX Jun 16 '18 at 15:05
  • $\begingroup$ One way: $A^tAx=0$ $\Rightarrow$ $x^tA^tAx=0$ $\Rightarrow$ $\|Ax\|^2=0$ $\Rightarrow$ $Ax=0$. Other way trivial, so any nonzero solution to $Ax=0$ is an eigenvector wrt zero eigenvalue. How "many" solutions can you get? Rang-nullity theorem: a subspace of dimension $n-m$. $\endgroup$ – A.Γ. Jun 16 '18 at 15:13

Well! It is clear from the link in the comments that if $\lambda$ is the eigenvalue of $A A^T $ then it is also the eigenvalue of $A^T A$.

Recall that rank($A A^T$)=rank($A^T A$) always holds and rank of a matrix is atleast the no. Of nonzero eigenvalues.

Hence, the only way you can manage the rank is to take all other eigenvalues to be zero i.e with n-m multiplicity.

  • $\begingroup$ How eigenvalues of $AA^t$ and $A^tA$ are the same?, as I thought of $(AA^t)^t = AA^t$? $\endgroup$ – BAYMAX Jun 16 '18 at 15:27
  • $\begingroup$ Haven't you checked the link above , in the comments, I can explain @BAYMAX $\endgroup$ – Devendra Singh Rana Jun 16 '18 at 15:39
  • $\begingroup$ Yes, i got that but how to think of the zero eigenvalues? $\endgroup$ – BAYMAX Jun 16 '18 at 15:40
  • $\begingroup$ perhaps you meant non-zero distinct eigenvalues? here -math.stackexchange.com/questions/146927/… $\endgroup$ – BAYMAX Jun 16 '18 at 15:44
  • $\begingroup$ Since it has m non-zero eigenvalues then if any of those remaining eigenvalues is non-zero then the rank ($A^T A$ ) will be m+1(atleast) which cannot be true as it should also be m, Hence all other eigenvalues must be zero. $\endgroup$ – Devendra Singh Rana Jun 16 '18 at 15:46

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.