Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 16 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.

share|cite|improve this answer
Thank you very much! How did you know to define y = Kx ? – diimension Oct 12 '12 at 4:10
@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
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
If we go in that direction, should we state that for any vector y in R we can some how express it as Kx? – Itay Jun 6 at 4:25

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 $K^{-1}$?

share|cite|improve this answer
I haven't learned eigenvalues yet. So I can't really say anything about it. – diimension Oct 12 '12 at 3:59

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.

share|cite|improve this answer
Why are the eigen-values of inverse of $K$ the reciprocal of those for $K$? – Abhishek Bhatia Jul 15 at 6:25

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.