Let \(A\)be an\(n \times n\)invertible symmetric matrix. Show that if the quadratic form \({x^T}Ax\) is positive definite, then so is the quadratic form\({x^T}{A^{ - 1}}x\).

When any Symmetric Matrix\(A\)is diagonalized orthogonally as \(PD{P^{ - 1}}\)we have:

\(\begin{aligned}{}{x^T}Ax = {y^T}Dy\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ {{\rm{as }}x = Py} \right\}\\{\rm{and}}\\\left\| x \right\| = \left\| {Py} \right\| = \left\| y \right\|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ {\forall y \in \mathbb{R}} \right\}\end{aligned}\)

As per the question, we have:

The positive \(n \times n\)matrix\(A\), which is an invertible symmetric matrix.

As the matrix \(A\) has positive eigenvalues.

The invertible matrix\({A^{ - 1}}\)will also have positive eigenvalues because, in the case of the invertible matrix, the eigenvalues just reciprocate without changing the sign.

\(\begin{aligned}{r}{x^T}Ax > 0\\ \Rightarrow {x^T}{A^{ - 1}}x > 0\end{aligned}\)

Hence proved, the quadratic form \({x^T}{A^{ - 1}}x\) is positive.