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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.

Rodrigo de Azevedo
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diimension
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4 Answers4

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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.

Hermi
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kjetil b halvorsen
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    Thank you very much! How did you know to define y = Kx ? – diimension Oct 12 '12 at 04:10
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    @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. – JACKY88 Oct 12 '12 at 04:34
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    So, essentially we are just being creative? It isn't an identity or axiom, just creativity? – diimension Oct 12 '12 at 05:09
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    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 01:16
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    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 06 '16 at 04:25
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    Wouldn't this only work if the image of $K$ were all of $\mathbb{R}^n$? Is this generally true for positive definite matrices? – zxmkn Apr 05 '18 at 14:25
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    @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. – Gerry Myerson Oct 27 '19 at 00:49
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    hi @kjetilbhalvorsen, I have a question about your proof. As you know if we want to say a matrix is PD, we should say it is symmetric and should satisfy the inequality $x^tKx > 0$ for all $x$. The point about your proof is that in your proof $y$ belongs to the range space of $K$. It can not take all possible values. Can you help me? What Am I doing wrong? – Green Falcon Apr 01 '20 at 22:42
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    @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. – kjetil b halvorsen Apr 01 '20 at 23:04
  • @kjetilbhalvorsen thank you! – Green Falcon Apr 01 '20 at 23:35
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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 $K^{-1}$?

Gerry Myerson
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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.

Sholeh Yasini
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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.

Jakub Homola
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