Eigenvalues of $X$ from those of $X^TX$ Is it possible to compute  the eigenvalues of $X$ from those of $X^TX$? I found here that is it possible between $X$ and $X^2$ but I search for $X$ and $X^TX$. 
$X$ is a non symmetrical square matrix. 
 A: There are bounds relating the eigenvalues of $A$ and $A^TA$ (or, singular values of $A$). For example, it is well known that the eigenvalue of $A$ of the largest magnitude is bounded from above by the square root of that of $A^TA$. Then it becomes more complicated. See, e.g., this book. For a general square $A$ though, there is no relation between the spectra of $A$ and $A^TA$ which is independent of the matrix $A$ itself.
For example, consider
$$
A=\begin{bmatrix}1 & \alpha\\0 & 0\end{bmatrix}.
$$
Then
$$
\lambda(A)=\{0,1\}\quad\text{and}\quad\lambda(A^TA)=\{0,1+\alpha^2\}.
$$
The relation between the nonzero eigenvalues of the two matrices (assuming $\alpha\neq 0$) is rather arbitrary.
A: $A$ doesn't need to have any (real) eigenvalues. But all eigenvalues of $A^T A$ are real and non negative. So there is no mapping between them. 
A: No, it's not possible to compute the eigenvalues of $A$ from the eigenvalues of $A^2$. The eigenvalues of $A^2$ give you a lot of information about the eigenvalues of $A$ but do not determine them: Let $A=I$ and $B=-I$. Then $A^2=B^2$, so $A^2$ and $B^2$ have the same eigenvalues, but $A$ and $B$ have different eigenvalues. (The eigenvalues of $A$ do determine the eigenvalues of $A^2$.)
