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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?

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    $\begingroup$ Caution: there is no general agreement on what "positive definite" means for non-Hermitian matrices. Which definition are you using? $\endgroup$ – Robert Israel Nov 17 '11 at 18:28
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    $\begingroup$ How about: $$[x\ y]\left[\matrix{1&1\cr -1&1}\right]\left[\matrix{x \cr y}\right]=[x\ y]\left[\matrix {x+y\cr-x+y}\right]=(x^2+xy)+(-xy+y^2)=x^2+y^2,$$ and $$ \left|\matrix {1-\lambda &1\cr -1&1-\lambda } \right| =(1-\lambda)^2+1\ne 0. $$ $\endgroup$ – David Mitra Nov 17 '11 at 18:42
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    $\begingroup$ I've said it before and I'll say it again: positive-definite should not be a term that applies to matrices. It should only apply to quadratic forms, which are naturally described by symmetric matrices only. $\endgroup$ – Qiaochu Yuan Nov 17 '11 at 18:46
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    $\begingroup$ It's a nice sentiment, but the genie's out of the bottle. $\endgroup$ – Michael Grant May 18 '13 at 21:08
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    $\begingroup$ There is a nice explanation about non-hermitian positive definite matrices. Please have a look into math.technion.ac.il/iic/ela/ela-articles/articles/… $\endgroup$ – user225624 Mar 22 '15 at 19:43
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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.

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    $\begingroup$ 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 $x \ne 0 \in \mathbb{R}^n$? What if we assume $A$ is diagonalizable? $\endgroup$ – nukeguy Feb 15 '17 at 18:24
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    $\begingroup$ @nukeguy The converse is false. example $A = \begin{bmatrix}3 & 7\\1 & 3\end{bmatrix}$ and $x = \begin{bmatrix}1\\-1\end{bmatrix}$. $\endgroup$ – achille hui Feb 16 '17 at 4:40
  • $\begingroup$ Can we put an extra condition (perhaps a bound on the off-diagonal entries in terms of the diagonal entries?) that in addition to a positive spectrum guarantees positivity of the matrix? $\endgroup$ – Oskar Limka Jan 23 at 19:32
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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 $x_1 = [-48 -10 -37]$ and $x_2 = [-48 -10 -37]$, then we have $𝑥_1𝐴𝑥_1^T = -1313$ and $𝑥_2𝐴𝑥_2^T = 37647.$

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