# 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?

• Caution: there is no general agreement on what "positive definite" means for non-Hermitian matrices. Which definition are you using? Nov 17 '11 at 18:28
• 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 $$\left|\matrix {1-\lambda &1\cr -1&1-\lambda } \right| =(1-\lambda)^2+1\ne 0.$$ Nov 17 '11 at 18:42
• 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. Nov 17 '11 at 18:46
• It's a nice sentiment, but the genie's out of the bottle. May 18 '13 at 21:08
• 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/…
– user225624
Mar 22 '15 at 19:43

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.

• 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? Feb 15 '17 at 18:24
• @nukeguy The converse is false. example $A = \begin{bmatrix}3 & 7\\1 & 3\end{bmatrix}$ and $x = \begin{bmatrix}1\\-1\end{bmatrix}$. Feb 16 '17 at 4:40
• 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? Jan 23 '19 at 19:32
• @achillehui Nice answer, and upvoted: Does the counter example matrix you gave above have a square root? Nov 19 '19 at 14:29
• @Mathmath, yes, $A$ do have square roots, e.g. $\frac{\sqrt{2}I_2+A}{\sqrt{6 + 2\sqrt{2}}}$ is one such square root. Nov 19 '19 at 14:54

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 $$𝑥_2𝐴𝑥_2^T = 37647.$$