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 
 A: 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."
A: Usually one speaks of positive definite matrices for the class of symmetric (or more generally Hermitian) matrices. The statement indeed follows easily if you assume that $A$ can be diagonalized by a orthogonal matrix $V$ ( I.e. such that $V^T V = V V^T = I$  ) which happens to be always possible if the matrix is symmetric. Then, assuming you have $A= V^T D V$ with $D$ diagonal with $D_{j,j}\geq 0$
$$
x^T A x = x^T V^T D V x
$$
And the last expression is the norm square of the vector $\sqrt{D} V x $ and so is positive for any $x$.
I denoted with $X^T$ the transpose of $X$.
A: Positive (negative) definite matrices are defined for symmetric ones, so assuming this your matrix is positive definite iff all its eigenvalues are positive. No need to mention diagonalization as any (complex or real) symmetric matrix is diagonalizable (even orthogonally so).
