# Characteristics of a Non-Hermitian Positive Definite Matrix

How does one determine if a non-hermitian matrix is positive definite, positive semi-definite, negative definite, etc...? Everything I find is for Hermitian matrices.

• What do you mean by "positive definite"? If you mean that $x^\ast Ax$ is real nonnegative for all complex vectors $x$, then $A$ is necessarily Hermitian. In fact, if $A$ is a matrix such that $x^\ast Ax\in\mathbb R$ for all complex vectors $x$, then $A$ is necessarily Hermitian. Commented Apr 1, 2020 at 23:21
• @user1551 That is where my confusion comes. My professor asked us to show the quadratic form of $$\begin{bmatrix} 8 & -6\\ 3 & -1 \end{bmatrix}$$ is psd, sd, nd, nsd, or indefinite, but it is my understanding this only holds for hermitian matrices
– nvm
Commented Apr 1, 2020 at 23:24
• @user1551 The only thing I can think of is to find the quadratic form and then find an equivalent symmetric matrix and use the eigenvalues of this matrix to show psd etc.
– nvm
Commented Apr 1, 2020 at 23:26
• What your professor concerns are probably quadratic forms over the real field. In this case, we can define the "positive definiteness" of a non-symmetric real matrix $A$ as $x^TAx>0$ for all real vectors $x$. Therefore $A$ is positive definite if and only if the symmetric matrix $A+A^T$ is positive definite in the usual sense. In other words, one may verify whether $A$ is PD, PSD, etc. by looking at the eigenvalues of $A+A^T$. Commented Apr 2, 2020 at 0:10
• To the OP. You obtained $2$ answers. Had you read them? The use in this website is, if one is satisfied with the answer, to upvote or to give the green ticket; otherwise one reports what is not suitable.
– user91684
Commented Apr 13, 2020 at 17:33

Assuming that you are looking at real matrices $$A$$ and real vectors $$v$$, use the facts that (i) $$v^TAv=v^T((A+A^T)/2)v$$ for all $$A,v$$ (ii) $$A+A^T$$ and $$(A+A^T)/2$$ are symmetric (iii) a matrix is positive-definite, negative-definite, positive-semi-definiteor negative semi-definite iff that matrix multiplied by a positive number is positive-definite, negative-definite, positive-semi-definite or negative semi-definite respectively, all that we have to do is apply a test for positive-definiteness, etc. for symmeric matrices to the symmetric matrix $$A+A^T.$$ $$*$$ Warning: We use the definitions that a matrix $$M$$ is positive definite if $$v^TAv>0$$ for all $$v \ne 0$$ and posive semi-definite if $$v^TMv \ge 0$$ for all $$v$$ with similar definitions for negative definite and negative semi definite. In the older literature "positive semi-definite" was often defined as " $$v^TMv \ge 0$$ for all $$v$$ and $$M$$ is not positive definite." We do NOT impose that additional equirement. For us,the only defining condition for being positive semi definite is $$v^TMv \ge 0$$ for all $$v$$. $$*$$ It is possible to give tests for these conditions in terms of eigenvalues of the matrix. We do not do so. Finding the eigenvalues of an $$n \times n$$ matrix requires solving an algebraic equation of degree $$n.$$ $$*$$ An $$n \times n$$ matrix $$A$$ is positive-definite iff in the matrix $$A+A^T$$ every upper left $$i \times i$$ block has positive determinant for $$1 \le i \le n.$$ $$*$$ An $$n \times n$$ matrix $$A$$ is negative -definite iff in the matrix $$A+A^T$$ every upper left $$i \times i$$ block has positive determinant if $$i$$ is even and negative determinant if $$i$$ is odd, for $$1 \le i \le n.$$ $$*$$ An $$n \times n$$ matrix $$A$$ is positive semi definite iff in the matrix $$A+A^T$$ every principal minor of order i is $$\ge 0$$ for $$1 \le i \le n.$$ $$*$$ An $$n \times n$$ matrix $$A$$ is negative semi definite iff in the matrix $$A+A^T$$ every principal minor of order i is $$\ge 0$$ if $$i$$ is even and $$\le 0$$ if $$i$$ is odd for $$1 \le i \le n.$$
Let $$A\in M_n(\mathbb{C})$$ (be eventually non-hermitian); as stated in other posts, the first thing to do is to calculate $$B=A+A^*$$. Now, to check if $$B$$ is $$>0,<0,\geq 0,\leq 0$$, then there are $$2$$ methods.
1. (The worst one) calculate the $$n$$ (upper-left) principal determinants of $$B$$. The complexity is $$\approx n^4/4$$. Unfortunately, this allows to conclude only if $$B>0,<0$$. Otherwise it is necessary to calculate all the principal determinants !!
2. Calculate an approximation of the (real) eigenvalues of $$B$$. The complexity is $$\approx$$ from $$10$$ to $$15n^3$$. It is necessary to check carefully the number of zero-eigenvalues.
3. Now, if you want to know if $$B$$ is $$>0$$ (resp. $$<0$$), then the best method is to calculate the Cholesky decomposition of $$B$$ (resp. $$-B$$). If the algorithm gives an output, then the answer is yes; if you obtain an error message, then the answer is no. The complexity is $$\approx n^3/2$$.