# 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.