Unisolvent point set

Remarks

Equivalently, the only $f \in V$ with $f(x_1) = \cdots = f(x_N) = 0$ is $f = 0$. Equivalently, for a basis $\varphi_1, \dots, \varphi_N$ of $V$, the $N \times N$ matrix $(\varphi_j(x_i))$ is invertible. The usual case is $V$ equal to the polynomials in $d$ variables of total degree at most $n$, where $N = \binom{n + d}{d}$.

Example: in one variable, the points $0, 1, 2$ are unisolvent for the polynomials of degree at most $2$. A polynomial of degree at most $2$ with three roots is $0$, so the equivalent condition holds.

Non-example: for $V = \set{a + bx + cy : a, b, c \in \R}$ on $\R^2$, the points $(0, 0)$, $(1, 1)$, $(2, 2)$ are not unisolvent. The nonzero function $x - y \in V$ vanishes at all three points.