The inverse of a positive definite matrix is also positive definite

Question

Let $K$ be nonsingular symmetric matrix, prove that if $K$ is positive definite so is $K^{-1}$ .

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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.

Answer 1

If $K$ is positive definite then $K$ is invertible, so define $y = K x$. Then $y^T K^{-1} y = x^T K^{T} K^{-1} K x = x^T K^{T} x >0$.

Since the transpose of a positive definite matrix is also positive definite, cf. here, this proves that $K^{-1}$ is positive definite.

Answer 2

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 $K^{-1}$?

Answer 3

inspired by the answer of kjetil b halvorsen

To recap, matrix $A \in \mathbb{C}^{n \times n}$ is HPD (hermitian positive definite), iff $\forall x \in \mathbb{C}^n, x \neq 0 : x^*Ax > 0$.

HPD matrices have full rank, therefore are invertible and $A^{-1}$ exists. Also full rank matrices represent a bijection, therefore $\forall x \in \mathbb{C}^n \enspace \exists y \in \mathbb{C}^n : x = Ay$.

We want to know if $A^{-1}$ is also HPD, that is, our goal is $\forall x \in \mathbb{C}^n, x \neq 0 : x^*A^{-1}x > 0$.

Let $x \in \mathbb{C}^n, x \neq 0$. Because $A$ is a bijection, there exists $y \in \mathbb{C}^n$ such that $x=Ay$. We can therefore write

$$x^*A^{-1}x = (Ay)^*A^{-1}(Ay) = y^*A^*A^{-1}Ay = y^*A^*y = y^*Ay > 0,$$

which is what we wanted to prove.