Proving $x^TAx \geq 0$ for a positive definite $n \times n$ matrix A.

Question

I am trying to prove that $x^TAx \geq 0$ for a positive definite $n \times n$ matrix $\mathbf{A}$ and $x \in \mathbb{R}^n$.

I have already proven that for any eigenvalue $y$ of $\mathbf{A}$,$ y^TAy > 0$. I have also shown that that $\mathbf{A}$ is diagonalizable (using the fact that positive - definite matrices are symmetric) so we can find a basis of eigenvalues of $\mathbf{A}$ for the vector space $\mathbb{R}^n$, and therefore $x$ can be written as a linear combination of eigenvalues of $\mathbf{A}$, i.e: $$x = a_1y_1 + … + a_ny_n$$ where $y_i, 1 \leq I \leq n$ are eigenvalues and the $a_i$ are scalars. I am just not sure how to complete the final step of showing $x^TAx \geq 0$.

Thank you in advance!

Answer 1

Assuming you know $y_i^T A y_i \geq 0$ for all of $A$'s eigenvectors, which form an orthonormal basis of $\mathbb R^n$, then for $x$ a linear combination as you stated $$x^T A x=(\sum_{i=1}^n a_i y_i^T)A(\sum_{i=1}^n a_i y_i)=\sum_{i,j=1}^na_i a_j y_i^T A y_j = \sum_{i,j=1}^n a_i a_j \lambda_i y_i^T y_j = \sum_{i,j=1}^n a_i a_j \lambda_i\delta_{ij} = \sum_{i=1}^n a_i^2 \lambda_i \geq 0$$ So you were really close.

Answer 2

Look into principle component minors (PMk) and leading principle component minors (LPMk). Pretty sure its proof by induction using the PMk and LPMk.

For positive definiteness (> 0) you will need PMk and for semi-pos ($\ge$ 0) will need LPMk.

Answer 3

Since $A$ is symmetric, it can be diagonalized by an orthogonal matrix $U$: $\Lambda = \text{diag}(\lambda_1,\lambda_2,\dots,\lambda_n) = UAU^\top$, where $\lambda_i > 0$ is the $i$th eigenvalue of $A$. Therefore, let $x$ be any vector in $\mathbb{R}^n$. Then \begin{equation*} x^\top Ax = x^\top U^\top \Lambda Ux = \|\Lambda^{1/2}Ux\|_2^2 \ge 0, \end{equation*} where $\Lambda^{1/2}$ is the unique positive definite square root of $\Lambda$.