Example

Tag 064P, permanent. Cite it as [Bourbaki, Tag 064P].

(1) Vandermonde determinant. Given a sequence $(\zeta_i)_{1 \leq i \leq n}$ of $n$ elements of $\mathbf{A}$ , the Vandermonde determinant of this sequence is the determinant

V(\zeta_1, \zeta_2, \ldots, \zeta_n) = \begin{vmatrix} 1 & 1 & \ldots & 1 \\ \zeta_1 & \zeta_2 & \ldots & \zeta_n \\ \zeta_1^2 & \zeta_2^2 & \ldots & \zeta_n^2 \\ \vdots & \vdots & \ddots & \vdots \\ \zeta_1^{n-1} & \zeta_2^{n-1} & \ldots & \zeta_n^{n-1} \end{vmatrix}

We shall show that

(29)

V(\zeta_1, \zeta_2, \ldots, \zeta_n) = \prod_{i < j} (\zeta_j - \zeta_i).

Since the proposition is immediate for $n = 1$ , we argue by induction on $n$ .

For each index $k \geq 2$ , we subtract from the row of index $k$ the row of index $k - 1$ multiplied by $\zeta_1$ ; the value of the determinant is unaltered and hence

V(\zeta_1, \zeta_2, \ldots, \zeta_n) = \begin{vmatrix} 1 & 1 & \cdots & 1 \\ 0 & \zeta_2 - \zeta_1 & \cdots & \zeta_n - \zeta_1 \\ 0 & \zeta_2(\zeta_2 - \zeta_1) & \cdots & \zeta_n(\zeta_n - \zeta_1) \\ \cdots & \cdots & \cdots & \cdots \\ 0 & \zeta_2^{n-2}(\zeta_2 - \zeta_1) & \cdots & \zeta_n^{n-2}(\zeta_n - \zeta_1) \end{vmatrix}

whence, expanding by the first column and then taking out the factor $\zeta_k - \zeta_1$ from the column of index $k - 1$ from the minor thus obtained ( $2 \leq k \leq n$ )

V(\zeta_1, \ldots, \zeta_n) = (\zeta_2 - \zeta_1)(\zeta_3 - \zeta_1) \cdots (\zeta_n - \zeta_1) V(\zeta_2, \ldots, \zeta_n)

which establishes (29) by induction.

(2) Consider a square matrix of order $n$ which is presented in the form of an "upper triangular matrix of matrices" (II, § 10, no. 7, Example IV)

X = \begin{pmatrix} Y & T \\ 0 & Z \end{pmatrix}

We show that

\det X = (\det Y)(\det Z).

Let $n$ be the order of the matrix $X$ , $h$ that of $Y$ , $(e_i)_{1 \leq i \leq n}$ the canonical basis of $\mathbf{A}^n$ and $x_i$ ( $1 \leq i \leq n$ ) the columns of $X$ ; the hypothesis implies that the columns $x_1, \ldots, x_h$ belong to the submodule of $\mathbf{A}^n$ with basis $e_1, \ldots, e_h$ and then by definition (no. 3, formula (6))

x_1 \wedge x_2 \wedge \cdots \wedge x_n = (\det Y)e_1 \wedge e_2 \wedge \cdots \wedge e_h.

On the other hand, for every index $i > h$ , we can write $x_i = y_i + z_i$ , where $y_i$ is a linear combination of $e_1, e_2, \ldots, e_h$ and $z_i$ is a linear combination of $e_{h+1}, \ldots, e_n$ . By (30), $x_1 \wedge x_2 \wedge \cdots \wedge x_n \wedge y_i = 0$ for all $i > h$ , therefore

x_1 \wedge x_2 \wedge \cdots \wedge x_n = (\det Y)e_1 \wedge e_2 \wedge \cdots \wedge e_h \wedge z_{h+1} \wedge \cdots \wedge z_n.

But by definition

z_{h+1} \wedge z_{h+2} \wedge \cdots \wedge z_n = (\det Z)e_{h+1} \wedge e_{h+2} \wedge \cdots \wedge e_n

whence formula (30).

By induction on $p$ , it follows that if $X$ is of the form of an upper triangular matrix of matrices:

X = \begin{pmatrix} X_{11} & X_{12} & \cdots & X_{1p} \\ 0 & X_{22} & \cdots & X_{2p} \\ \cdots & \cdots & \cdots & \cdots \\ 0 & 0 & \cdots & X_{pp} \end{pmatrix}

(31)

\det X = (\det X_{11})(\det X_{22}) \ldots (\det X_{pp}).

This can be applied in particular to a triangular matrix (where all the $X_{ii}$ are of order 1) and more particularly to a diagonal matrix:

(32)

\det(\operatorname{diag}(\alpha_1, \alpha_2, \ldots, \alpha_n)) = \alpha_1 \alpha_2 \ldots \alpha_n.

(3) Let $M, M'$ be two free $A$ -modules of respective dimensions $n, n'$ , $u$ an endomorphism of $M$ and $u'$ an endomorphism of $M'$ . Then

(33)

\det(u \otimes u') = (\det u)^{n'} (\det u')^n.

For we can write $u \otimes u' = (u \otimes 1_{M'}) \circ (1_M \otimes u')$ and are then led to the case where one of the two endomorphisms $u, u'$ is the identity. For example if $u' = 1_{M'}$ and $X$ is the matrix of $u$ with respect to a basis $(e_i)$ of $M$ , then the matrix of $u \otimes 1_{M'}$ with respect to the tensor product of $(e_i)$ and a basis of $M'$ can be written as a matrix (with $n'$ rows and $n'$ columns) of matrices with $n$ rows and $n$ columns

\begin{pmatrix} X & 0 & \ldots & 0 \\ 0 & X & \ldots & 0 \\ \ldots & \ldots & \ldots & \ldots \\ 0 & 0 & \ldots & X \end{pmatrix}

whence, by virtue of Example 2,

\det(u \otimes 1_{M'}) = (\det X)^{n'} = (\det u)^{n'}

which immediately gives formula (33).

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

Cited by

Nothing.

Cites

II, § 10, no. 7