Eigenvalues of $A^tA$?

Question

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

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 $n-m$. — A.Γ., Jun 16 '18 at 15:13

Devendra Singh Rana · Accepted Answer · 2018-06-16T15:20:35.063

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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.

edited Jun 16 '18 at 15:20

answered Jun 16 '18 at 15:15

Devendra Singh Rana

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How eigenvalues of $AA^t$ and $A^tA$ are the same?, as I thought of $(AA^t)^t = AA^t$? – BAYMAX Jun 16 '18 at 15:27
Haven't you checked the link above , in the comments, I can explain @BAYMAX – Devendra Singh Rana Jun 16 '18 at 15:39
Yes, i got that but how to think of the zero eigenvalues? – BAYMAX Jun 16 '18 at 15:40
perhaps you meant non-zero distinct eigenvalues? here -https://math.stackexchange.com/questions/146927/relation-between-rank-and-number-of-non-zero-eigenvalues-of-a-matrix – BAYMAX Jun 16 '18 at 15:44
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. – Devendra Singh Rana Jun 16 '18 at 15:46
well I am getting the point, thank you!! – BAYMAX Jun 16 '18 at 16:20

Eigenvalues of $A^tA$?

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