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.

Consider an $n\times n$ real matrix $K$ which satisfies $$ \det[K_{ij}]_{i,j=1}^k\geq 0,\qquad 1\leq k \leq n. $$ I know that if one assumes moreover that $K$ is symmetric, then $K$ is positive semi-definite, namely $$ c^tKc\geq0,\qquad c\in\mathbb R^n. $$ Does it still holds is the general case, that is when $K$ is not necessarily symmetric ?

share|improve this question

1 Answer 1

up vote 1 down vote accepted

The quick answer to your question is no. Indeed, you can see even when $n = 2$ that $$\left(\begin{array}{cc} 1 & 1\end{array}\right)\left(\begin{array}{cc}0 & -2\\ 1 & 0\end{array}\right)\left(\begin{array}{c}1 \\ 1\end{array}\right) = -1.$$

The longer answer to your question is that we usually don't talk about whether a general matrix $K$ is positive semidefinite. The reason we only do this with symmetric $K$ is that they define objects called quadratic forms, which show up very often in math. For a quadratic form, there is a notion of positive semidefiniteness, and this notion is equivalent to the defining matrix $K$ having nonnegative determinant.

share|improve this answer

Your Answer

 
discard

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.