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Prove that if $S = S^T$ is symmetric and non-singular, then $S^2$ is positive definite.

My attempt:

Suppose $S$ is an $m\times n$ symmetric matrix with linearly independent columns, and suppose $q(x) > 0$, then the matrix $q(x) = \mathbf{x}^\mathrm{T}S\mathbf{x}$ is a positive definite $n\times n$ matrix thus $S^2$ is also positive definite. Am I right, or completely off?

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    @EuYu thank you very much for clarifying that!2012-10-13

2 Answers 2

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It's easy: Since $S$ is symmetric and nonsingular, $x^T S^T S x= (Sx)^T (Sx) \ge 0$ since it is a sum of squares. Now, the above can be zero only if $Sx=0$, and since $S$ is non-singular, $Sx=0$ is possible only if $x=0$. That's it.

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Excellent answer from @kjetil. I would like to add another one. $S$ is symmetric and non-singular. So it's eigen decomposition is $S=U\Lambda U^{T}$ with non-zero eigen values. So $S^2=U\Lambda^{2}U^{T}$. Clearly $S^2$ has positive eigen values, so it should be positive definite.