= Solution
For distinct ordered knots, put $h_i=t_{i+1}-t_i>0$. The linear <B-spline> $N_i$ is the triangular hat supported on $[t_i,t_{i+2}]$, equal to $(t-t_i)/h_i$ on its rising segment and $(t_{i+2}-t)/h_{i+1}$ on its falling segment. Its integral-normalized partner is $M_i=2N_i/(h_i+h_{i+1})$. Thus
$$
\int N_i^2=\frac{h_i+h_{i+1}}3,\qquad \int N_iN_{i-1}=\frac{h_i}6,\qquad \int N_iN_{i+1}=\frac{h_{i+1}}6.
$$
For example, the left overlap is $\int_0^{h_i}(u/h_i)(1-u/h_i)\,du=h_i/6$. Nonadjacent hats have disjoint interiors of their <supports>, so their <inner product> is zero. The <linear-spline mixed Gram matrix> therefore has row
$$
\boxed{g_{ii}=\frac23,\quad g_{i,i-1}=\frac{h_i}{3(h_i+h_{i+1})},\quad g_{i,i+1}=\frac{h_{i+1}}{3(h_i+h_{i+1})},\quad g_{ij}=0\ (|i-j|\ge2)}.
$$
Retain only column indices in $1,\ldots,n$: the first or last row has no neighbor outside this range. Admissible repeated endpoint knots are obtained by a limit with zero spacings, provided $h_i+h_{i+1}>0$; a degenerate hat with zero support width is not a <basis> element.
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