# Is the trapezoid rule integral operator $A$ a non singular matrix?

Let $A$ be the matrix of a linear bounded integral operator $A : L^2 \rightarrow L^2$, discretized by the trapezoid rule defined on the function space $BV([0, L])$, $L \in \mathbb{R}$. So, if $u \in BV([0, L])$ then the operator is

$A(u(x)) = \int_0^x u(s) \, ds$ and its $L^2([0, L])$ adjoint operator $A^T(u) = \int_x^L u(s) \, ds$.

The matrix $A^TA$ is symmetric with all entries real and non-negative. Is this matrix positive definite?