# Symmetric Matrix with Positive Eigenvalues

Not all matrix with positive eigenvalues is positive definite, i.e. $\mathbf{x}^\mathsf{T}A\mathbf{x}>0$ for all non zero vector $\mathbf{x}$. For example consider matrix

$$A = \begin{bmatrix} 1 & -3 \\ 0 & 1 \end{bmatrix}.$$

How to prove that if we add symmetry into hypothesis then the assertion is true? That is, a symmetric matrix with positive eigenvalues is positive definite.

The fact builds upon the facts on eigenvalue and eigenvectors of symmetruc matrix. The one directly leads to the fact you asked is that: a symmetric matrix A can decomposed as $$A = Q^TDQ$$ where Q is an orthogonal matrix and D is diagonal matrix and entries of diagonal of D is eigenvalues $$\lambda_i$$ of A: $$d_{ii} = \lambda_i > 0$$
Now $$x^TAx = x^TQ^TDQx = (Qx)^TD(Qx) = y^TDy = \sum \lambda_i y_i^2 > 0$$ Here $$y =(y_1, y_2, ..., y_n) = Qx \ne 0$$ if $$x \ne 0$$. $$0$$ here is $$0$$ vector.
• Just to add information: $A$ can be decomposed as $A = Q^TDQ$ due to the Spectral Theorem, that can be obtained from the Schur Decomposition. Commented Aug 28, 2020 at 19:41