is the matrix $A$ positive-definite?

Question

For $i=1,2,\ldots,n$, let $$\phi_i(x)=\begin{cases}\displaystyle\frac{x(L-x_i)}{L} \text{ , for } 0\leq x\leq x_i\\ \frac{\displaystyle x_i(L-x)}{L} \text{ , for } x_i\leq x\leq L \end{cases}$$ and $$\beta_{ij}=\int_0^L\phi_i(x)\phi_j(x)dx$$ where $L$ is a constant and $x_i$ are given values, with $0\leq x_i \leq L$. Is the matrix $A$, with elements $a_{ij}$ given by $a_{ij}=\beta_{ij}$, positive-definite?

Answer 1

yes this is a gram matrix for the inner product $$ \langle f,g\rangle =\int_0^L f(x)g(x) dx $$ so in particular your matrix is positive semi-definite, but it suffice to prove that our matrix is invertible, or equivalently that is determinant is nonzero, but using the fact that our matrix is a gram matrix this is equivalent to the fact that our functional $\phi_i(x)$ are linearly independent, and this is evident for $x_i\neq x_j $ for all $i,j\leq n$.