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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. $\endgroup$ – Patrick Li Oct 12 '12 at 4:34
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    $\begingroup$ So, essentially we are just being creative? It isn't an identity or axiom, just creativity? $\endgroup$ – diimension Oct 12 '12 at 5:09
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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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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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    $\begingroup$ I haven't learned eigenvalues yet. So I can't really say anything about it. $\endgroup$ – diimension Oct 12 '12 at 3:59
  • $\begingroup$ @Par, if you have a question, post it as a new question, not as a comment that no one will see on an answer from four years ago. $\endgroup$ – Gerry Myerson Dec 30 '16 at 13:05
  • $\begingroup$ @GerryMyerson thanks for the advice, math.stackexchange.com/questions/2076912/… $\endgroup$ – Parisina Dec 30 '16 at 20:41
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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. $\endgroup$ – Sheljohn Mar 8 '18 at 11:33

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