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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?

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Rotation matrix is not a good example since it is not symmetric. it is skew-symmetric – Maesumi Oct 1 '12 at 3:39
right i know, im just trying to understand if the eigenvalues of $A^2$ are fully characterized by the eigenvalues of $A$, squared. – Palace Chan Oct 1 '12 at 16:12
up vote 1 down vote accepted

The eigenvalues of $A^T A$ are always nonnegative. If $A$ is symmetric, the eigenvalues of $A^2$ are nonnegative because $A^2 = A^T A$. The eigenvalues of $A^2$ are the squares of the eigenvalues of $A$. A complex number whose square is nonnegative is a real number.

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