I just read this snippet in a textbook
"The eigenvalues of a symmetric real matrix are real (The proof follows by noting that if $A$ is symmetric, the eigenvalues of $A^TA$ are the eigenvalues of $A^2$, which are obviously nonnegative)"
How does it follow? I'm a bit confused about the above statement. First off, if $A$ were a rotation matrix in $R^2$ of $90$ degrees, then geometrically $-1$ is seen to the an eigenvalue of $A^2$ and it's negative...Though I'm not sure if the symmetric restriction is the necessary condition to take out all negative eigenvalues of the square. Also, I know that if $c$ is an eigenvalue of $A$ then clearly $c^2$ is an eigenvalue of $A^2$ - but does this characterize all the eigenvalues of $A^2?$ in other words, if $t$ is an eigenvalue of $A^2$ then is $\sqrt{t}$ an eigenvalue of $A$? And is this a one to one (or one to two I suppose) correspondence?