I want to prove, that the matrix $ \Gamma:=\Big(\min\{t_i,t_j\}-t_i\cdot t_j\Big)_{1\leq i,j\leq k}\in\mathbb{R}^{k\times k} $ is positive definite where $0\leq t_1<\ldots<t_k\leq 1$.
I tried several things, but nothing worked. Thanks for help!
Edit: I already know, that $\Gamma$ is positive semi-definite, because it is the covariance function of the brownian bridge. So it would be enough to show, that 0 is'nt an eigenvalue of $\Gamma$, which has a lot of equvivalences, like $\text{rg}\,\Gamma=k$ for example. But even this I cannot proof.
I am not sure that this claim is correct. For instance, consider $\{t_{1}, t_{2} \} = \{1/2, 1 \}$ Then. $\Gamma = \left( \begin{matrix} 1/4 & 0 \\ 0 & 0 \end{matrix} \right)$. Then $\Gamma$ has eigenvalues $0,1/4$, which implies that $\Gamma$ is not positive definite because $0$ is an eigenvalue of $\Gamma$ with eigenvector $\langle 0 , 1 \rangle$.
I think if $0 < t_{1} \leq \dots \leq t_{k} < 1$ then the claim might be true. In this case, you just have to show that $det(\Gamma) > 0$, and then you can use induction to claim that the upper left sub-matrices have positive determinant (induct on the dimension). Hope this helps.