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.


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.


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 $K^{-1}$?


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.