MHB Inverse Eigenvalues: A Puzzling Question?

[mathmari]

Hey!

Does it stand that the eigenvalues of $A^{-T}A^{-1}$ are the inverse of the eigenvalues of $A^TA$ ?? (Wondering)

 

[mathmari]

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)

 

[caffeinemachine]

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.