If $\det(A)$ is $< 0$, then $A$ is not a positive semidefinite matrix.
How can I prove that statement?
Since $\det A$ is the product of the eigenvalues of $A$, $A$ has an eigenvalue $\lambda<0$. Let $v$ be an eigenvector with eigenvalue $\lambda$. But then $v^t.A.v=\lambda\lVert v\rVert^2<0$.
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Since $\det A$ is the product of the eigenvalues of $A$, $A$ has an eigenvalue $\lambda<0$. Let $v$ be an eigenvector with eigenvalue $\lambda$. But then $v^t.A.v=\lambda\lVert v\rVert^2<0$.