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$A_{m \times n}, n>m, AA^t =\alpha I , \alpha> 0$, 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 $n-m$?

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  • $\begingroup$ 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
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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.

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  • $\begingroup$ 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

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