= Solution
First take $k\ge2$, so the simple-knot <B-splines> are continuous, vanish at both <support> endpoints, and are strictly positive inside their <supports>. Thus the diagonal condition is $t_i<x_i<t_{i+k}$. We prove both directions of the <Schoenberg–Whitney theorem>, deriving the weighted zero-count fact needed for sufficiency rather than relying on an ambiguous <coefficient>-free hint.
For necessity, suppose $N_i(x_i)=0$. If $x_i\le t_i$, then for every row $r\le i$ and column $j\ge i$ we have $x_r\le x_i\le t_i\le t_j$, hence $N_j(x_r)=0$. The first $i$ rows have nonzero entries in at most $i-1$ columns and are linearly dependent. If instead $x_i\ge t_{i+k}$, then for $r\ge i$ and $j\le i$ we have $x_r\ge x_i\ge t_{i+k}\ge t_{j+k}$, so those entries vanish. The last $n-i+1$ rows have nonzero entries in at most $n-i$ columns and are again dependent. In either case the <B-spline collocation matrix> is singular. Thus invertibility implies the diagonal positivity condition.
We establish two elementary <spline> facts for the converse. For <local linear independence of B-splines>, extend the strictly increasing knots in both directions. On a cell $(t_j,t_{j+1})$ the $k$ active <basis> functions have indices $j-k+1,\ldots,j$. Their knot <polynomials> are $\psi_i(y)=\prod_{\ell=1}^{k-1}(y-t_{i+\ell})$. The <Cox-de Boor recurrence> proves <Marsden's identity>
$$
(y-x)^{k-1}=\sum_{i=j-k+1}^jN_i(x)\psi_i(y).
$$
Here is the induction behind that identity. For order one the indicator functions sum to one. On passing from order $k-1$ to $k$, collect the <coefficient> of each lower-order <basis> function. With $u=t_i$, $v=t_{i+k-1}$, its extra factor is
$$
\frac{(x-u)(y-v)+(v-x)(y-u)}{v-u}=y-x.
$$
Thus the identity at one order multiplies by $y-x$ at the next.
The $k$ <polynomials> $\psi_i$ are independent: evaluate successively at $y=t_j,t_{j-1},\ldots,t_{j-k+1}$. The first evaluation isolates $\psi_j$; after its <coefficient> has been eliminated the next isolates $\psi_{j-1}$, and so on, with nonzero diagonal factors because the knots are distinct. Comparing powers of $y$ in the identity then shows that the active $N_i$ span all <polynomials> of degree at most $k-1$ on the cell. There are $k$ of them, so they are independent. In particular, a <spline> vanishes on a whole open cell exactly when all its active <coefficients> vanish. Padding the finite <coefficient> vector with zeros makes this fact apply also near the outer knots.
Next let $s=\sum_{i=p}^qc_iN_i$ be a real <spline> on $I=(t_p,t_{q+k})$, with no identically zero knot cell there. The <simple-knot B-spline regularity> gives $s\in C^{k-2}$, and its <derivatives> through order $k-2$ vanish at both endpoints because it is supported on the closed interval. Suppose it has $Z$ distinct interior zeros. Its zero set initially has $Z+2$ <connected components>, counting the endpoints; the zeros are isolated because no cell vanishes identically.
A useful form of <Rolle's theorem> counts zero components, including any zero intervals that arise after differentiation. Between two successive zero components a real differentiable function is nonzero and has an interior extremum, supplying a zero of its <derivative>. These <derivative>-zero components are distinct: a zero interval connecting two such extrema would make the original function constant across an intervening zero and nonzero gap, which is impossible. If its <derivative> also vanishes at both endpoints, those supply two further distinct components. Thus differentiation increases the zero-component count by at least one. Apply this successively through <derivative> order $k-2$; every required endpoint <derivative> vanishes. No <derivative> in this process can vanish identically: that would make $s$ a global <polynomial> of degree below $k-1$, with all its endpoint <derivatives> through that degree zero, forcing $s=0$. The final <derivative> has at least $Z+k$ zero components.
That final <derivative> is continuous and piecewise affine on $q-p+k$ knot cells. Each closed affine cell can meet at most one zero component: two different zeros would force the whole cell to vanish, joining them into one component. Every zero component meets a cell, so there are at most $q-p+k$ of them. For $k=2$ this argument applies directly without differentiation. We have proved the <compact-support spline zero count>
$$
\boxed{Z(s;I)\le q-p.}
$$
This proof also accommodates zero intervals of the <derivatives>; counting them as infinitely many distinct zeros would invalidate a naive repeated-Rolle argument.
Now assume $N_i(x_i)>0$ for all $i$ and suppose $A_xc=0$ for a nonzero <coefficient> vector. Since $A_x$ is real, a nonzero real null vector can be chosen. Set $s=\sum_ic_iN_i$, so $s(x_i)=0$ for all $i$. Consider one <connected component> of the union of the open <supports> of the terms with $c_i\ne0$. If its first and last indices are $p,q$, the component is exactly $I=(t_p,t_{q+k})$. On this interval $s$ equals $s_0=\sum_{i=p}^qc_iN_i$. Each knot cell in $I$ has at least one active nonzero <coefficient>, so local independence ensures $s_0$ does not vanish on any whole cell.
For every $p\le i\le q$, diagonal positivity gives $t_i<x_i<t_{i+k}$, and this places $x_i$ inside $I$. Thus $s_0$ has the $q-p+1$ distinct zeros $x_p,\ldots,x_q$, contradicting the upper bound $q-p$. There is no nonzero null vector, proving sufficiency. Together the two directions give
$$
\boxed{A_x\text{ is invertible}\quad\Longleftrightarrow\quad N_i(x_i)>0\text{ for every }i.}
$$
For completeness, order $k=1$ consists of disjoint interval indicators, with the usual left-closed, right-open convention. An invertible ordered collocation requires exactly one site in each interval; ordering makes this equivalent to $N_i(x_i)>0$, and the <matrix> is then the identity. This handles possible sites at left knots without incorrectly replacing positivity by a strict interior inequality for order one.
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