Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The title says it.

I found out that there exist positive definite matrices which are non-symmetric, and I know that symmetric positive definite matrices have positive eigenvalues. Does this hold for non-symmetric matrices as well?

share|improve this question
Caution: there is no general agreement on what "positive definite" means for non-Hermitian matrices. Which definition are you using? –  Robert Israel 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 $$ \left|\matrix {1-\lambda &1\cr -1&1-\lambda } \right| =(1-\lambda)^2+1\ne 0. $$ –  David Mitra 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. –  Qiaochu Yuan Nov 17 '11 at 18:46
It's a nice sentiment, but the genie's out of the bottle. –  Michael Grant 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/… –  Fayyaz Ahmad Mar 22 at 19:43

2 Answers 2

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:


has eigenvalue $1 \pm i$. However, the real part of any eigenvalue $\lambda$ of $A$ is always positive.

Let $\mathbb{C} \in \lambda = \mu + \nu i$ 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.

share|improve this answer

If A is a real symmetric matrix, then all eigenvalues and eigenvectors of A are real. And we have that real positive definite matrices have positive eigenvalues.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.