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Let K be nonsingular symmetric matrix, prove that if K is a positive definite so is $K^{-1}$ .

My attempt:

I have that $K = K^T$ so $x^TKx = x^TK^Tx = (xK)^Tx = (xIK)^Tx$ and then I don't know what to do next.

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Well, somewhere you have to use the definition of, or some fact about, positive definite matrices --- so, what do you know about positive definite matrices? –  Gerry Myerson Oct 12 '12 at 3:56
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2 Answers 2

up vote 8 down vote accepted

If $K$ is positive definite then $K$ is invertible, so define $y = K x$. Then $y^T K^{-1} y = x^T K^{T} K^{-1} K x = x^T K x >0$ so is positive definite.

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Thank you very much! How did you know to define y = Kx ? –  diimension Oct 12 '12 at 4:10
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@diimension The thing you know is $K$ is PD. So you want to have a form of $x^T K x$ because we know it is positive. –  Patrick Li Oct 12 '12 at 4:34
    
So, essentially we are just being creative? It isn't an identity or axiom, just creativity? –  diimension Oct 12 '12 at 5:09
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Its just experience! But you must get used to that prooving things is not algorithmic, you must search for ideas. Comes with training! –  kjetil b halvorsen Oct 13 '12 at 1:16
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Here's one way: $K$ is positive definite if and only if all of its eigenvalues are positive. What do you know about the eigenvalues of $K^{-1}$?

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I haven't learned eigenvalues yet. So I can't really say anything about it. –  diimension Oct 12 '12 at 3:59
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