Linear-Algebra-Done-Right · Unit 29 · Video 7 · Interactive Practice

The Vandermonde Determinant via Polynomial Evaluation

IKey Formulas

Formula	Name	Hypotheses
$\det \begin{pmatrix} 1 & \beta_1 & \cdots & \beta_1^{n-1} \\ \vdots & & & \vdots \\ 1 & \beta_n & \cdots & \beta_n^{n-1} \end{pmatrix} = \displaystyle\prod_{1 \le j < k \le n} (\beta_k - \beta_j)$	Vandermonde determinant	$n > 1$ , $\beta_1, \dots, \beta_n \in \mathbf{F}$
$A = \mathcal{M}\bigl(S, (1, z, \dots, z^{n-1}), (e_1, \dots, e_n)\bigr)$	Evaluation map	$Sp = (p(\beta_1), \dots, p(\beta_n))$ , $S \colon \mathcal{P}_{n-1}(\mathbf{F}) \to \mathbf{F}^n$
$Tz^k = (z - \beta_1) \cdots (z - \beta_k)$ , $\ \det B = 1$	Change of basis	$B = \mathcal{M}(T)$ upper triangular, diagonal entries $1$

Key Insight: $\det A = \det(AB)$ , and $AB$ is lower triangular because $(z - \beta_1)\cdots(z - \beta_k)$ vanishes at $\beta_1, \dots, \beta_k$ ; its diagonal entries are the products of differences.

IIOne Polynomial Through Three Points

A nonzero Vandermonde determinant means $S$ is invertible: exactly one $p \in \mathcal{P}_2(\mathbf{R})$ takes the values $w_k$ at the nodes $\beta_k$ .

IIIA, B and the Lower-Triangular AB

For $n = 3$ , multiplying the Vandermonde matrix $A$ by the unit upper-triangular $B$ produces a lower-triangular $C = AB$ .

β₁ −1 β₂ 1 β₃ 3

IVQuiz Questions

Problem 1 · Three Nodes

Find $\det \begin{pmatrix} 1 & 1 & 1 \\ 1 & 2 & 4 \\ 1 & 4 & 16 \end{pmatrix}$ .

✅ Correct! With

\beta = (1, 2, 4)

(2 - 1)(4 - 1)(4 - 2) = 1 \cdot 3 \cdot 2 = 6

❌ Check the order. Each factor is

\beta_k - \beta_j

with

j < k

(later minus earlier); with three factors, reversing them all flips the sign.

❌ Not quite.

1 \cdot 2 \cdot 4

multiplies the nodes; the formula multiplies their differences.

❌ Not quite. Include every pair

j < k

, not only neighbours: the pair

(1, 3)

contributes

4 - 1 = 3

Show solution

The rows are $(1, \beta_k, \beta_k^2)$ with $\beta_1 = 1$ , $\beta_2 = 2$ , $\beta_3 = 4$ . By the Vandermonde formula,

\det = (\beta_2 - \beta_1)(\beta_3 - \beta_1)(\beta_3 - \beta_2) = (1)(3)(2) = 6.

Problem 2 · Four Nodes, Two Negative Factors

Given: $\beta_1, \beta_2, \beta_3, \beta_4 = 0, 1, -1, 2$ . Find the $4$ -by- $4$ Vandermonde determinant $\prod_{j < k}(\beta_k - \beta_j)$ .

✅ Correct! The six factors are

1, -1, 2, -2, 1, 3

; two are negative.

❌ Check the sign. Count the negative factors:

\beta_3 - \beta_1 = -1

and

\beta_3 - \beta_2 = -2

, an even number.

❌ Not quite. The determinant is

0

only when two nodes coincide; these four are distinct.

❌ Not quite. There are

\binom{4}{2} = 6

factors; check that none was left out.

Show solution

List $\beta_k - \beta_j$ for all six pairs $j < k$ :

\underbrace{(1 - 0)}_{1}\underbrace{(-1 - 0)}_{-1}\underbrace{(2 - 0)}_{2}\underbrace{(-1 - 1)}_{-2}\underbrace{(2 - 1)}_{1}\underbrace{(2 - (-1))}_{3} = 12.

Problem 3 · When Does It Vanish?

Given: $M_t = \begin{pmatrix} 1 & t & t^2 \\ 1 & 2 & 4 \\ 1 & 3 & 9 \end{pmatrix}$ with $t \in \mathbf{R}$ . Find every $t$ for which $M_t$ is not invertible.

✅ Correct!

\det M_t = (2 - t)(3 - t)(3 - 2)

, zero exactly when

t

equals another node.

❌ Not quite.

t = \tfrac{5}{2}

is a new, distinct node, so the determinant is nonzero there.

❌ Check the sign. The factors are

2 - t

and

3 - t

, which vanish at

t = 2

and

t = 3

❌ Not quite. At

t = 2

rows

1

and

2

are equal.

Show solution

$M_t$ is the Vandermonde matrix with $\beta_1 = t$ , $\beta_2 = 2$ , $\beta_3 = 3$ , so

\det M_t = (2 - t)(3 - t)(3 - 2) = (2 - t)(3 - t).

A square matrix is invertible exactly when its determinant is nonzero, so $M_t$ fails to be invertible exactly for $t = 2$ and $t = 3$ , where two rows coincide.

Problem 4 · Interpolation at Four Nodes

Given: distinct nodes $-1, 0, 1, 2$ and any numbers $w_1, w_2, w_3, w_4 \in \mathbf{R}$ . How many $p \in \mathcal{P}_3(\mathbf{R})$ satisfy $p(-1) = w_1$ , $p(0) = w_2$ , $p(1) = w_3$ , $p(2) = w_4$ ?

✅ Correct! The evaluation map

S \colon \mathcal{P}_3(\mathbf{R}) \to \mathbf{R}^4

has matrix with determinant

\prod_{j < k}(\beta_k - \beta_j) \neq 0

, so

S

is invertible.

❌ Not quite. An invertible

S

is surjective: every

w \in \mathbf{R}^4

Sp

for some

p

❌ Not quite. An invertible

S

is injective: two such polynomials would have the same image, so they are equal.

❌ Not quite. An invertible linear map has exactly one preimage for each

w

Show solution

Let $Sp = (p(-1), p(0), p(1), p(2))$ . Its matrix in the bases $1, z, z^2, z^3$ and $e_1, \dots, e_4$ is the Vandermonde matrix of the nodes, with determinant

\prod_{j < k}(\beta_k - \beta_j) = (1)(2)(3)(1)(2)(1) = 12 \neq 0.

So $S$ is invertible, and each $w \in \mathbf{R}^4$ has exactly one preimage $p = S^{-1}w$ .

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