A is a matrix of positive defined quadratic form. How can I show, $A^{-1}$ is the same?

Question

Let a square matrix A is a matrix of positive defined quadratic form. How can I show, that $A^{-1}$ also a matrix of a positive defined quadratic form?

Positive defined quadratic form is A(x,y), that all it's corner minors are positive.

Inverse matrix is a matrix, contains all minors of the source matrix.

It must be linked somehow...

Answer 1

If $A$ is positive-definite, then $x^TAx >0$ for all $x$.

Since $A$ is invertible then for any $y$ we can find some $x$ such that $Ax=y$, but we have:

$Ax = y \implies x^TAx=x^Ty$

$x=A^{-1}y \implies y^TA^{-1}y=y^Tx$

$x^Ty$ is a real number, therefore $x^Ty=(x^Ty)^T=y^Tx$. This implies that:

$y^TA^{-1}y = x^TAx> 0$

Since $y$ was arbitrary, this proves that $A^{-1}$ is positive definite.

Answer 2

As $A \in S_{++}(\mathbb{R})$, then we have $x^T A x > 0$. Now recall the eigenvalues of $A^{-1}$ are precisely the reciprocals of $A$, and therefore $A^{-1}$ is also positive definite. In fact we get non-singularity as well.

Answer 3

Let $z=Ax$. $A$ is positive definite and hence also invertible. Thus, it follows that $ x=A^{-1}z. $ Premultiply by $z'$ to get $z'x=z'A^{-1}z$. But, $z'=x'A'$ so what you have is actually $$ x'A'x=z'A^{-1}z>0 $$ since $A=A'$ and $x'A'x>0$ by the nature of semi definite matrices.