the Vandermonde matrix
/ VAN-der-mond /
If you insist on finding the interpolating polynomial the most literal way — by solving for its coefficients a_0, a_1, ..., a_n directly — you write down one equation per data point that says 'the polynomial equals y_i at x_i' and solve the resulting linear system. The matrix of that system, whose entries are powers of the x-values, is the Vandermonde matrix.
Setting p(x) = a_0 + a_1 x + ... + a_n x^n and demanding p(x_i) = y_i for each i gives the system V a = y, where row i of V is (1, x_i, x_i^2, ..., x_i^n) — every row is the geometric run of powers of one node. Its determinant has a famous closed form: the product of (x_j - x_i) over all i < j. The punchline is a clean proof of the uniqueness theorem: as long as the nodes are DISTINCT, no two factors vanish, the determinant is nonzero, V is invertible, and the coefficients a are uniquely determined. So the Vandermonde viewpoint is the 'linear algebra explanation' of why exactly one interpolating polynomial exists.
Conceptually clarifying, computationally treacherous. The Vandermonde matrix becomes severely ILL-CONDITIONED as the degree grows: its condition number explodes exponentially for equispaced nodes, so solving V a = y in floating point loses many digits and the recovered coefficients can be garbage even though the interpolant itself is fine. This is why nobody actually computes interpolants this way for more than a handful of points — they use the Lagrange or Newton forms, or the barycentric formula, which sidestep the matrix entirely. The Vandermonde matrix is for understanding, not for solving.
For nodes 0, 1, 2 the Vandermonde matrix is rows (1, 0, 0), (1, 1, 1), (1, 2, 4). Its determinant equals (1-0)(2-0)(2-1) = 2, nonzero because the nodes differ, so the coefficients are uniquely solvable.
Distinct nodes make the determinant nonzero — hence unique coefficients.
The Vandermonde system proves uniqueness elegantly but is a poor way to compute: for equispaced nodes its condition number grows roughly like 2^n, so high-degree fits via V a = y are numerically hopeless. Use Newton, Lagrange, or barycentric forms instead.