# Eigenvalues of nonsymmetric positive definite matrices

I consider a non-symmetric matrix $$A\in\Re^{n\times n}$$ and try to estimate the region of the eigenvalues

1. Is it true, that a nonsymmetric matric $$A$$ is positive definite (in the sense of $$x^T\cdot A\cdot x>0$$), when the symmetric part of the matrix $$A_\text{sym}=\frac12(A^T+A)$$ is positive definite?
2. I know for a symmetric matric, that $$A_\text{sym} \text{ is pos. def} \Leftrightarrow \text{all Eigenvalues } \lambda_i > 0$$. What can I say about the eigenvalues of $$A$$? I read that the real part of these Eigenvalues must be $$\Re(\lambda_i)>0$$.

• You should use $x^TAx = x^T \frac{A^T+A}2 x$.
– daw
Commented Jun 30, 2021 at 9:05
• okay, so if i set $A=\underbrace{\frac12(A-A^T)}_{A_{usym}}+\underbrace{\frac12(A+A^T)}_{A_{sym}}$, the part $x^TA_{usym}x$ equals $0$?
– BigL
Commented Jun 30, 2021 at 9:15

Both claims are true, though be wary that your definition of a matrix being positive-definite in the sense that $$x^T A x > 0$$ for all non-zero real vectors $$x$$, is not common and usually restricted to Hermitian matrices. To be clear, let us denote $$A \succ 0$$ to mean that $$A$$ is positive-definite in the sense you have defined, for any real matrix.

1. We have $$x^T A x = \frac{1}{2} \left( x^T (A^T+A) x + x^T (A^T-A)x \right) = x^T A_{sym} x > 0$$ where $$x^T (A^T-A)x = 0$$ because $$x^T (A^T-A)x = (x^T (A^T-A)x)^T = x^T (A-A^T)x = -x^T (A^T-A)x \,.$$
2. Take any eigenvalue $$\lambda = a+ib$$ where $$a,b \in \mathbb{R}$$ of a positive-definite matrix $$A$$, with eigenvector $$x$$. Then $$Ax = \lambda x$$ implies $$x^* A x = a+ib$$ (where $$^*$$ denotes the conjugate transpose) and so $$a = Re(x^* A x) = \frac{1}{2} \left( x^* A x + x^* A^* x \right) = x^* A_{sym} x > 0$$ by using part 1 above and the fact that $$A^* = A^T$$ since $$A$$ is real.
• perfect, thanks for the answer!
– BigL
Commented Jun 30, 2021 at 9:17
• Part 2: Is the eigenvector $x$ necessarily real? If not, $x^\mathsf{T}x \ne 1$. From $Ax = \lambda x$, we can not get $x^\mathsf{T} Ax = \lambda$. We only get $x^\mathsf{H}A x = \lambda$ since $x^\mathsf{H} x = 1$. Commented Jun 30, 2021 at 9:45
• @RiverLi Thanks, I modified the proof using conjugate transpose instead of just transpose. Commented Jun 30, 2021 at 20:02
• $\frac{1}{2} \left( x^* A x + x^* A^* x \right) = x^* A_{sym} x$? Commented Jul 1, 2021 at 0:00
• (+1) It is fine now. Commented Jul 9, 2021 at 14:23

I think you misunderstood what you wrote in point 1. Any positive operator is Hermitian, so if your $$A$$ is nonsymmetric but positive definite, well... it fulfills $$A=A^\dagger$$. What I mean with this is that $$A$$ has at least two off-diagonal complex entries (otherwise it would be symmetric) such that these elements fulfill $$A_{ij}=A^*_{ji}$$.

The eigenvalues of $$H$$ positive definite, hence Hermitian, are positive (nonegative).