73

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?

Rodrigo de Azevedo
  • 23,223
  • 26
    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
  • 18
    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
  • 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
  • 10
    It's a nice sentiment, but the genie's out of the bottle. – Michael Grant May 18 '13 at 21:08
  • 3
    There is a nice explanation about non-hermitian positive definite matrices. Please have a look into http://www.math.technion.ac.il/iic/ela/ela-articles/articles/vol20_pp621-639.pdf –  Mar 22 '15 at 19:43
  • 1
    For reference, Golub and Van Loan have an entire subsection dedicated to “Unsymmetric Positive Definite Systems”. It is subsection 4.2.2 of “Matrix Computations”, 4th edition. This lends credence and some canonicity to the idea. – Just Some Old Man Aug 23 '24 at 19:27

3 Answers3

76

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.

achille hui
  • 125,323
  • 11
    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? – nukeguy Feb 15 '17 at 18:24
  • 19
    @nukeguy The converse is false. example $A = \begin{bmatrix}3 & 7\1 & 3\end{bmatrix}$ and $x = \begin{bmatrix}1\-1\end{bmatrix}$. – achille hui Feb 16 '17 at 04:40
  • 2
    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? – Oskar Limka Jan 23 '19 at 19:32
  • @achillehui Nice answer, and upvoted: Does the counter example matrix you gave above have a square root? – Mathguest Nov 19 '19 at 14:29
  • 1
    @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. – achille hui Nov 19 '19 at 14:54
  • @achillehui Thanks a lot, let me check the calculation on my own. – Mathguest Nov 19 '19 at 14:56
  • The =0 part in the equation after "This implies" just comes from the fact that the dot product of x and y (which is a scalar) is equal to its transpose... – Golden_Ratio Aug 18 '22 at 08:11
15

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

Is the converse true? If all of the eigenvalues of a matrix $$ have positive real parts, does this mean that $^>0$ for any $≠0∈ℝ^$? What if we assume $$ 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 $_1_1^T = -1313$ and $_2_2^T = 37647.$

Daniel Weissman
  • 600
Salivan
  • 447
  • 4
  • 13
0

If you use the defintion that $A$ is positive definite if and only if $\langle Ax, x \rangle > 0 $ for all $x \in \mathbb{C}^n \setminus \{0\}$, then we see that the following are equivalent:

  1. $A$ is positive definite, i.e., $\langle Ax, x \rangle > 0 $ for all $x \in \mathbb{C}^n \setminus \{0\}$
  2. $A = A^*$ (symmetric) and spec$(A) >0 $
  3. $A$ is unitarily diagonalizable with positive diagonals ($\exists \ U$ unitary such tat $UAU^* = $diag$(\lambda_i)$ with $ \lambda_i > 0.$ Note that the example @DavidMitra posted is not positive over $\mathbb{C}^2$ since $v = [1, i]^t $ gives $v^*Mv = 2+2i$. See: https://en.wikipedia.org/wiki/Definite_matrix#Consistency_between_real_and_complex_definitions

In short and to answer your question, when working over $\mathbb{C}^n$, $A$ is positive definite is if and only if symmetric with positive eigenvalues.

travelingbones
  • 376