The stretch factors of $A^T A$ are the eigenvalues of $A^T A$

Question

Why are the stretch factors of $A^TA$ equal to the eigenvalues of $A^TA$? I understand that $A^TA$ is a symmetry matrix and have proved that the stretch factors of $A$ are square roots of the eigenvalues of $A^TA$. But why is it true that the stretch factors of $A^tA$ are the eigenvalues of $A^T A$?

Any verification would be appreciated.

Answer 1

The stretch factors of $A^TA$ would be the square root of the eigenvalues of $(A^TA)^T(A^TA)$.

If $A^TA = UDU^T$ where $D$ is a diagonal matrix and $U$ is orthogonal.

Then $$(A^TA)^T(A^TA)=UDU^TUDU^T=UD^2U^T$$

Hence eigenvalues of $(A^TA)(A^TA)$ are the squares of eigenvalues of $A^TA$.