Show that positive eigenvalues determine that matrix is positive definite. I'm stuck in a problem on positive definite matrices. I should show that "if all eigenvalues of A are positive, then A >= 0." **EDIT:** assuming A is diagnosable. I know the definition for a positive definite matrix is that x'Ax >= 0 for all x (but does "x" herein refer to the eigenvectors?). I've spent many hours on trying to figure this out, but I don't really grasp the idea of how to get started here.. Would greatly appreciate help!! Best, Matthias

The statement is incorrect. For instance, consider the matrix $$A = \begin{bmatrix} 1 & -100\\\ 0 & 2\end{bmatrix}$$ The eigenvalues are $1$ and $2$. However, the matrix is not positive definite. For instance, $$\begin{bmatrix} 1 & 1\end{bmatrix} \begin{bmatrix} 1 & -100\\\ 0 & 2\end{bmatrix} \begin{bmatrix} 1 \\\ 1\end{bmatrix} = \begin{bmatrix} 1 & 1\end{bmatrix} \begin{bmatrix} -99 \\\ 2\end{bmatrix} = -97 < 0$$ The statement is true if the matrix $A$ is symmetric and diagonalizable. However, the following statement is always true: "If the matrix is positive definite and symmetric, then all its eigenvalues are positive."