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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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    $\begingroup$ 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? $\endgroup$ – Gerry Myerson Oct 12 '12 at 3:56
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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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    $\begingroup$ Thank you very much! How did you know to define y = Kx ? $\endgroup$ – diimension Oct 12 '12 at 4:10
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    $\begingroup$ @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. $\endgroup$ – JACKY Li Oct 12 '12 at 4:34
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    $\begingroup$ Its just experience! But you must get used to that prooving things is not algorithmic, you must search for ideas. Comes with training! $\endgroup$ – kjetil b halvorsen Oct 13 '12 at 1:16
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    $\begingroup$ If we go in that direction, should we state that for any vector y in R we can some how express it as Kx? $\endgroup$ – Itay Jun 6 '16 at 4:25
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    $\begingroup$ @zxmkn if you're still here, it's true for invertible matrices, which is to say it's true for matrices which don't have zero as an eigenvalue, which means it's true for positive definite matrices, since they have only positive eigenvalues. $\endgroup$ – Gerry Myerson Oct 27 '19 at 0:49
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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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K is positive definite so all its eigenvalue are positive. The eigenvalues of $K^{-1}$ are inverse of eigenvalues of K, i.e., $\lambda_i (K^{-1}) = \frac{1}{\lambda_i (K)}$ which implies that it is a positive definite matrix.

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  • $\begingroup$ Why are the eigen-values of inverse of $K$ the reciprocal of those for $K$? $\endgroup$ – Abhishek Bhatia Jul 15 '16 at 6:25
  • $\begingroup$ @AbhishekBhatia Because the inverse of a diagonal matrix with non-zero entries is the diagonal matrix of the reciprocals. $\endgroup$ – Jonathan H Mar 8 '18 at 11:33

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