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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.

  • 1
    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?2012-10-12

3 Answers 3

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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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    Wouldn't this only work if the image of $K$ were all of $\mathbb{R}^n$? Is this generally true for positive definite matrices?2018-04-05
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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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    @GerryMyerson thanks for the advice, http://math.stackexchange.com/questions/2076912/inverse-of-a-positive-definite-automorphism-over-infinitely-generated-inner-prod2016-12-30
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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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    @AbhishekBhatia Because the inverse of a diagonal matrix with non-zero entries is the diagonal matrix of the reciprocals.2018-03-08