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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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You must be more careful: A rectangular matrix cannot be symmetrix, so the two dimensions must be equal! so $m \times m$-matrix. – kjetil b halvorsen Oct 13 '12 at 1:04
You could also look at the eigenvalues of $S^2$, which are all going to be positive. – wj32 Oct 13 '12 at 1:32
@diimension You need to be very careful with your proofs. When you "suppose $q(x) >0$" you need to be able to justify that statement. Why do you suppose it's positive? What if it isn't? As you have it currently, that step is basically assuming the result you want to prove. – EuYu Oct 13 '12 at 1:34
@kjetilbhalvorsen you are right! Thank you for pointing that out! – diimension Oct 13 '12 at 1:45
@EuYu thank you very much for clarifying that! – diimension Oct 13 '12 at 1:46
up vote 2 down vote accepted

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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Thank you very much! – diimension Oct 13 '12 at 1:47

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.

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