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Let $K$ be nonsingular symmetric matrix, prove that if $K$ is 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? $\endgroup$ Commented Oct 12, 2012 at 3:56

4 Answers 4

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

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

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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. $\endgroup$
    – JACKY88
    Commented Oct 12, 2012 at 4:34
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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! $\endgroup$ Commented Oct 13, 2012 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
    Commented Jun 6, 2016 at 4:25
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    $\begingroup$ @zxmkn if you're still here, it's true for invertible matrices, which is to say it's true for matrices which don't have zero as an eigenvalue, which means it's true for positive definite matrices, since they have only positive eigenvalues. $\endgroup$ Commented Oct 27, 2019 at 0:49
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    $\begingroup$ @Media: But the result only makes sense for invertible matrices ... as it is a claim about a property of $K^{-1}$. So the range space is the full space. $\endgroup$ Commented Apr 1, 2020 at 23:04
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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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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 $K$? $\endgroup$ Commented Jul 15, 2016 at 6:25
  • $\begingroup$ @AbhishekBhatia Because the inverse of a diagonal matrix with non-zero entries is the diagonal matrix of the reciprocals. $\endgroup$
    – Jonathan H
    Commented Mar 8, 2018 at 11:33
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    $\begingroup$ @AbhishekBhatia $Av = \lambda v \implies \frac{1}{\lambda}v = A^{-1}v$ $\endgroup$ Commented Oct 10, 2021 at 5:12
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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.

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