Does non-symmetric positive definite matrix have positive eigenvalues? I found out that there exist positive definite matrices that are non-symmetric, and I know that symmetric positive definite matrices have positive eigenvalues.
Does this hold for non-symmetric matrices as well?
 A: I am answering the first part of @nukeguy's comment, who asked:

Is the converse true? If all of the eigenvalues of a matrix $$ have
positive real parts, does this mean that $^>0$ for any
$≠0∈ℝ^$? What if we assume $$ is diagonalizable?

I have a counterexample, where $A$ has positive eigenvalues, but it is not positive definite:
$    A=
    \begin{bmatrix}
        7   & -2 & -4  \\
        -17 & 40 & -19 \\
        -21 & -9 & 31
    \end{bmatrix}
$. Eigenvalues of this matrix are  $1.2253$, $27.4483$, and $49.3263$, but it indefinite because if $x_1 = \begin{bmatrix}-48 & -10& -37\end{bmatrix}$ and $x_2 = \begin{bmatrix}-48 &10 &-37\end{bmatrix}$, then we have $_1_1^T = -1313$ and $_2_2^T = 37647.$
A: Let $A \in M_{n}(\mathbb{R})$ be any non-symmetric $n\times n$ matrix but "positive definite" in the sense that: 
$$\forall x \in \mathbb{R}^n, x \ne 0 \implies x^T A x > 0$$
The eigenvalues of $A$ need not be positive. For an example, the matrix in David's comment:
$$\begin{pmatrix}1&1\\-1&1\end{pmatrix}$$
has eigenvalue $1 \pm i$. However, the real part of any eigenvalue $\lambda$ of $A$ is always positive.
Let $\lambda = \mu + i\nu\in\mathbb C $ where $\mu, \nu \in \mathbb{R}$ be an eigenvalue of $A$. Let $z \in \mathbb{C}^n$ be a right eigenvector associated with $\lambda$. Decompose $z$ as $x + iy$ where $x, y \in \mathbb{R}^n$.
$$(A - \lambda) z = 0 \implies \left((A - \mu) - i\nu\right)(x + iy) = 0
\implies \begin{cases}(A-\mu) x + \nu y = 0\\(A - \mu) y - \nu x = 0\end{cases}$$
This implies
$$x^T(A-\mu)x + y^T(A-\mu)y = \nu (y^T x - x^T y) = 0$$
and hence
$$\mu = \frac{x^TA x + y^TAy}{x^Tx + y^Ty} > 0$$
In particular, this means any real eigenvalue $\lambda$ of $A$ is positive.
