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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 ?

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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.

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