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?

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.

I am answering the first part of @nukeguy's comment, who asked

``Is the converse true? If all of the eigenvalues of a matrix $𝐴$ have positive real parts, does this mean that $𝑥^𝑇𝐴𝑥>0$ for any $𝑥≠0∈ℝ^𝑛$? What if we assume $𝐴$ 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 $𝑥_1𝐴𝑥_1^T = -1313$ and $𝑥_2𝐴𝑥_2^T = 37647.$