Inverse Eigenvalues: A Puzzling Question?

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SUMMARY

The eigenvalues of the matrix $A^{-T}A^{-1}$ are definitively the inverses of the eigenvalues of $A^{T}A$, provided that the matrix $A$ is invertible. This conclusion is supported by the relationship between the eigenvalues of $A^{-T}A^{-1}$ and $(A^{-T}A^{-1})^{-1}=AA^{T}$, where the eigenvalues of $AA^{T}$ are equivalent to those of $A^{T}A$. Thus, the proposition holds true for all invertible matrices.

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mathmari
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Hey! :o

Does it stand that the eigenvalues of $A^{-T}A^{-1}$ are the inverse of the eigenvalues of $A^TA$ ?? (Wondering)
 
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It is as followed, right??

The eigenvalues of $A^{-T}A^{-1}$ are the inverse of the eigenvalues of $(A^{-T}A^{-1})^{-1}=AA^{T}$ and the eigenvalues of $AA^{T}$ are the same as the eigenvalues of $A^{T}A$. So the eigenvalues of $A^{-T}A^{-1}$ are the inverse of the eigenvalues of $A^{T}A$.

(Wondering)

But for which matrices $A$ does this stand?? (Wondering)
 
mathmari said:
It is as followed, right??

The eigenvalues of $A^{-T}A^{-1}$ are the inverse of the eigenvalues of $(A^{-T}A^{-1})^{-1}=AA^{T}$ and the eigenvalues of $AA^{T}$ are the same as the eigenvalues of $A^{T}A$. So the eigenvalues of $A^{-T}A^{-1}$ are the inverse of the eigenvalues of $A^{T}A$.

(Wondering)

But for which matrices $A$ does this stand?? (Wondering)
As long as $A$ is invertible, the proposition is valid as you have shown.
 

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