On the Vandermonde Matrix

Tom Copeland ([email protected]) May 2014

Part I

The n-simplices and the Vandermonde matrix

There are several denitions of the Vandermonde matrix Vn. Here

 1 1 ··· 1   x1 x2 ··· xn   . . .  Vn = Vn(x1, x2, ..., xn) =  . . .. .  .  . . . .  n−1 n−1 n−1 x1 x2 ··· xn with the determinant Y |Vn| = |Vn(x1, . . . , xn)| = (xj − xi). 1≤i

The coordinates of the n vertices Cn−1(x1),Cn−1(x2), ..., Cn−1(xn) of the 2 n−1 (n − 1)-simplex along the unique moment curve Cn−1(x) = (x, x , ..., x ) (ra- tional normal curve) that passes through the vertices comprise the columns of

Vn (except for the rst row of ones). With one of the vertices located at the origin, say vertex m with xm = 0, the determinant gives the (possibly signed) hyper-volume of the hyper-parallelepiped framed by the position vectors from the origin to the other vertices, which bound the facet opposing the vertex at the origin (see the V 3 example below). Note that the xj are unit-less parameters, i.e., have no physical units or measures attached to them and in that sense are dimensionless. In other words, k should not be directly interpreted as a quantity with dimension a length, xj k an area, a volume, etc. It is a dimensionless parameter that must be interpreted in context. Ignoring this leads to confusion, such as believing that |V4| with six factors in its product formula must somehow have dimension six.

The associated |Vn| is also an eigenfunction of a generalized shift (dilation) operator with the face polynomial of the (n − 1)-simplex as the eigenvalue (cf. the MathOverow Q. A sum involving derivatives of Vandermonde?), i.e.,

1 Pn k=1 exp[t· : xkDk :]|Vn| = |Vn((1 + t) · x1, x2, ..., xn)| + ... + |Vn(x1, x2, ..., (1 + t) · xn)| (1+t)n−1 , = [ t ] · |Vn|

where m m dm by denition for notational convenience and sug- : xkDk : = xk m dxk gestive parallelism with the form of the shift operator exp(tDk)f(xk) = f(t + xk). For example, look at the 3-simplex, the tetrahedron with face polynomial [(1 + t)4 − 1]/t = 4 + 6t + 4t2 + t3 (4 vertices/0-D faces,6 edges/1-D faces,4 triangles/2-D faces,1 tetrahedron/3-D face) and with the twisted cubic curve 2 3 as the parametrized moment curve with the th vertex at 2 3 (x, x , x ) k (xk, xk, xk) along the curve (cf. MO-Q165283 The twisted kiss of the curvaceous cubic and the staid tetrahedron). The coordinates of n vertices of one (n − 1)-dimensional face, or facet, of the n-simplex along the moment curve comprise the moment matrix   x1 x2 ··· xn 2 2 2  x1 x2 ··· xn   . . .  Mn = Mn(x1, x2, ..., xn) =  . . .. .   . . . .  n n n x1 x2 ··· xn

The Vandermonde and moment matrices are related by Mn = Vn · diag(x1, x2, ...xn). Using this relation and the eigenvalue relation with t = −1, a recursion relation follows:

n+1 |Vn(x1, x2, ..., xn)| = |Mn−1(x2, x3, ..., xn)|−|Mn−1(x1, x3, ..., xn)|+...+(−1) |Mn−1(x1, x2, ..., xn−1)|

with, e.g., |Vˆn−1(x2)| = |Vn−1(x1, x3, ..., xn)| and eˆn−1(x1) = en−1(x2, x3, ..., xn) = (x2 · x3 ··· xn), the highest order elementary symmetric polynomial for the (n − 1) indetermi- nates.

The |Mn−1| are the "hyper-volumes" of the hyper-parallelepipeds "framed" by the coordinate vectors of the (n − 1)-dimensional faces (facets) of the n - simplex. With one vertex placed at the origin (i.e., one xm = 0), only one |Mn−1| is non-vanishing, giving the (possibly signed) hyper-volume of the hyper- parallelepiped framed by the position vectors to the vertices, which are those of the facet opposing the origin.

2 On the other hand, setting t = −1 in the eigenvalue equation for the deter- minants is equivalent to applying Cramers's rule to nd the rst column of the adjugate matrix of Vn, so dividing through by |Vn| gives

the inner product of the rst row of ones of Vn with the rst column of its inverse (see below for a 3-D example and the Appendix).

Examples: -

For V2(x1, x2), the associated 1-simplex is the line segment with vertices C1(x1) = x1 and C1(x2) = x2; therefore,

|V2(x1, x2)| = |M1(x2)| − |M1(x1)| = x2 − x1,

equaling a parametrized (unit-less) length, the distance between the 0- simplexes, the vertices C1(x1) and C1(x2).

For V3(x1, x2, x3), the associated 2-simplex is a triangle with vertices at 2 , 2 , and 2 ; therefore, C2(x1) = (x1, x1) C2(x2) = (x2, x2) C2(x3) = (x3, x3)

|V3(x1, x2, x3)| = |M2(x2, x3)| − |M2(x1, x3)| + |M2(x1, x2)|,

which equals the area of the parallelogram formed by the vectors to the endpoints of the 1-simplex, the side of the triangle from C2(x2) to C2(x3) minus that from C2(x1) to C2(x3) plus that from C2(x1) to C2(x2). 2 Consider x1 = −u, x2 = 0, and x3 = u > 0. Then |V3| = −|M2(x1, x3)| = 2u · u , the area ( u is a dimensionless parameter, i.e., it has no physical units) of the parallelogram framed by the position vectors (−u, u2) and (u, u2) from the vertex

3 at the origin to the other two vertices, the endpoints of the opposing facet to the vertex at the origin. The product formula gives (u − 0)(u − (−u))(0 − (−u)) = 2u · u2, the base ( 2u) times the height ( u2) of the triangle.

For V4(x1, x2, x3, x4), the associated 3-simplex is a tetrahedron with vertices at 2 3 , 2 3 2 3 , and C3(x1) = (x1, x1, x1) C3(x2) = (x2, x2, x2),C3(x3) = (x3, x3, x3) 2 3 ; therefore, C3(x4) = (x4, x4, x4)

Part II

A recursive product formula from matrix inversion

Looking at the inverse of V , i.e., nding the dual basis to V , leads to a recursive product formula. (The eigenvalue formula above actually encodes the column covectors of the inverse matrix. Compare the t = −1 example above with the inverse equation below and the Appendix. Also see MO-Q A sum involving derivatives of Vandermonde?.) The inversion process can be interpreted geometrically as in the MO-Q Cavalieri's principle and the inversion of the Vandermonde matrix or in terms of orthonormality conditions of vectors and covectors. For example, for 3-D space spanned by the non-orthogonal column vectors of V3, I = N · S3 · V3  ˆ      |V2(x1)| 0 0 eˆ2(x1) −eˆ1(x1) 1 1 1 1 1 ˆ =  0 −|V2(x2)| 0   eˆ2(x2) −eˆ1(x2) 1   x1 x2 x3  |V3| 2 2 2 0 0 |Vˆ2(x3)| eˆ2(x3) −eˆ1(x3) 1 x1 x2 x3

where I is the identity matrix, N is the rst normalizing diagonal matrix (incorporating the 1/|V3|), and S3 is the matrix of elementary symmetric poly- nomials, containing the non-normalized covectors to the column vectors of V3. A shorthand has been used as a mnemonic and for conciseness:

|Vˆ2(x1)| = det V2(x2, x3) = x3 −x2, |Vˆ2(x2)| = det V2(x1, x3) = x3 −x1, and |Vˆ2(x3)| = det V2(x1, x2) = x2−x1, similarly for the elementary symmetric poly- nomials, e.g., eˆ1(x2) = e1(x1, x3) = x1 + x3 and eˆ2(x2) = e2(x1, x3) = x1x3.

4 ∂(e1(x1,x2,x3),e2(...),e3(...)) S3 is essentially the transpose of the Jacobian (with ∂(x1,x2,x3) columns interchanged, mod signs) with |S3| = −|V3|, and since |I| = 1, it 2 3 follows that |N| = −1/|V3| , but also |N| = −|Vˆ2(x1)| · |Vˆ2(x2)| · |Vˆ2(x3)|/|V3| ; therefore,

|Vˆ2(x1)| · |Vˆ2(x2)| · |Vˆ2(x3)| = |V3|. For higher dimensions,

n−2 |Vˆn−1(x1)| · · · |Vˆn−1(xn)| = |Vn| .

You can easily convince yourself that the LHS of this formula contains (n=2) products of the factors of |Vn| given that |Vn−1| is given by the product formula. (Hint: use a table with individual dierences listed on the vertical and horizontal sides.)

The relationships among the "coordinate systems" given by xi and ei, the characteristic polynomial, the Vandermonde matrix, and its inverse are clearly presented on pages 20 & 38 of "Hamiltonian 2-forms in Kahler geometry, I General Theory" by Apostolov, Calderbank, and Gauduchon. See also Part IV below.

Part III

Product formula for |Vn| and the reection group An−1

Note that the number of factors in the product formula is the same as the number of edges in the complete graph Kn, which can be viewed as an edge projection onto a plane of the (n − 1)-simplex. In fact, if the vertices are labeled by x1, x2, ..., xn, the factors correspond to the dierences in the labels of the endpoints of the edges. Why?

From Beta Integrals by Warnaar, the reection group An−1 is the symmetry group of the -simplex of order and is generated by the n hyperplanes (n − 1) n! 2 (xi − xj) = 0 for 1 ≤ i < j ≤ n. |Vn| gives the product of the distances of the n/2 point x = (x1, ..., xn) to the hyperplanes up to a factor of 2 . (The integral n of powers of |Vn| under the Gaussian measure on R is the Mehta integral, a special case of the Selberg integral.)

5 Part IV

Geometry of the characteristic polynomial and the product formula

The MO-Q Geometric interpretation of characteristic polynomial presents the coecients of a characteristic polynomial as generalized traces and then argues that these coecients can be related to dierential volumes loosely related to the faces of hyper-boxes (in three dimensions, the vertices (0-D faces), edges (1-D faces), facets/rectangles (2-D faces), and the polyhedron/box (3-D face)), volumes that can be subtracted and added together to obtain the volume associated to the characteristic polynomial of the linear matrix transformation A. More precisely for the 3-D case,

Q3 det(I − t · A) = i=1(1 − t · xi)

2 3 = 1 − t · e1(x1, x2, x3) + t e2(x1, x2, x3) − t e3(x1, x2, x3)

2 3 = 1 − t · (x1 + x2 + x3) + t (x1x2 + x1x3 + x2x3) − t (x1x2x3)

= 1·1·1−{1 · 1 · t · x1 + 1 · 1 · t · x2 + 1 · 1 · t · x3}+{1 · (t · x1)(t · x2) + 1 · (t · x1)(t · x3) + 1 · (t · x2)(t · x3)}

−(t · x1)(t · x2)(t · x3).

The rst term, one, in the last equality is the volume of the unperturbed identity matrix I, the volume of the cube C with sides all of length one with one vertex at the origin and the adjoining edges along the three coordinate axes. With A a diagonal matrix with positive eigenvalues less than one, you should be able to easily visualize the volume Q3 of the perturbed cube i=1(1 − t · xi) with side i of length (1 − t · xi) with t also positive and much less than one. The second term in the last equality represents the sum of the volumes of three slabs attached to the three inner faces of C not touching the origin, each of area 1 · 1 and thickness t · xi. This overcompensates for the perturbation of C

6 since the slabs overlap, so the volumes of three boxes, each of length one with small faces of area (t · xi)(t · xj), attached to the three edges of C touched by pairs of the slabs are added in the third term, i.e., the long slender boxes that are the intersections of pairs of the slabs. This is also an overcompensation, so the volume of a small box, (t · x1)(t · x2)(t · x3), attached to the vertex of C opposite the origin, i.e., the vertex touched by all three slabs, has to be subtracted, i.e., the volume of the small box representing the intersection of the three slabs. Alternatively, let

Q3 det(xI − A) = i=1(x − xi)

3 2 = x − x · e1(x1, x2, x3) + x · e2(x1, x2, x3) − e3(x1, x2, x3)

3 2 = x − x · (x1 + x2 + x3) + x · (x1x2 + x1x3 + x2x3) − (x1x2x3),

but here let C be a cube of length x. Then analogous arguments apply, and if we let x = x4, this also becomes the inner product of the column vector 2 3 T of with the row vector , (1, x4, x4, x4) V 4 (−eˆ3(x4), eˆ2(x4), −eˆ1(x4), 1) which (in agreement with the matrix inversion) is equal to |V4|/|Vˆ3(x4)| since

4 d Y |V |/|Vˆ (x )| = |V |/|V | = (x − x )(x − x )(x − x )= | (x − x ). 4 3 4 4 3 4 3 4 2 4 1 dx x=x4 i i=1

Generalizing these arguments gives a recursion formula for generating the product formula for |Vn| from |V2|. For easy comparison to the matrix inverse, the normalized inner product may be rewritten as

|V3(x1,x2,x3)| 3 2 x − x · e1(x1, x2, x3) + x4 · e2(x1, x2, x3) − 1 · e3(x1, x2, x3) |V4(x1,x2,x3,x4)| 4 4

|V3| 3 2 = x − x · eˆ1(x4) + x4 · eˆ2(x4) − 1 · eˆ3(x4) |V4| 4 4

7  1  x |V3|  4  . = − |V | eˆ3(x4) −eˆ2(x4)e ˆ1(x4) −1  2  = 1 4  x4  3 x4

Also in agreement with the matrix inversion discussion, the row vector −|V3| eˆ3(x4) −eˆ2(x4)e ˆ1(x4) −1 of the adjugate matrix is orthogonal to the other column vectors of V4 (just put x = xi ( i = 1, 2, 3) in the characteristic polynomial) and is the wedge (exterior) product of these remaining column vectors; therefore, its length/norm is the hyper-area spanned by the hyper-parallelogram framed by these vectors. The projection of the x4 column vector onto this row vector is the perpendicular distance of the endpoint of the column vector to the hyper-plane containing the hyper-parallelogram so that the inner product of the adjugate row vector and the x4 column vector gives the hyper-volume |V4|.

Appendix The eigenvalue relation and the inverse matrix Return to the eigenvalue solution to the generalized shift operator:

Pn k=1 exp[t· : xkDk :]|Vn| = |Vn((1 + t) · x1, x2, ..., xn)| + ... + |Vn(x1, x2, ..., (1 + t) · xn)|

(1+t)n−1 = [ t ] · |Vn|.

Setting t = −1, gives

1 · |Vn(0, x2, ..., xn)| + 1 · |Vn(x1, 0, ..., xn)| + ... + 1 · |Vn(x1, x2, ..., 0)|  1 1 ··· 1   1 1 ··· 1   0 x2 ··· xn   x1 x2 ··· 0   . . . .   . . . .   . . .. .   . . .. .  0   0   = x1 · |  k k  | + .... + xn · |  k k  | = |Vn|,  0 x2 ··· xn   x1 x2 ··· 0   . . .   . . .   . . .. .   . . .. .   . . . .   . . . .  n−1 n−1 n−1 n−1 0 x2 ··· xn x1 x2 ··· 0

which is the inner product of the rst row vector of ones of Vn with the rst column vector of the adjugate matrix as determined from Cramer's rule.

8 Now note that d k k (k) k d k ( dt ) |t=−1 f((1 + t)x) = x · f (0) = x · ( dx ) |x=0 f(x) n and d k (1+t) −1 for , so ( dt ) |t=−1 [ t ]) = k! k ≤ n

1 d k k! ( dt ) |t=−1 {|Vn((1 + t) · x1, x2, ..., xn)| + ... + |Vn(x1, x2, ..., (1 + t) · xn)|}

k k x1 d k xn d k = ( ) |x =0 |Vn(x1, x2, ..., xn)| + ... + ( ) |x =0 |Vn(x1, x2, ..., xn)| k! dx1 1 k! dxn n

 0 1 ··· 1   1 1 ··· 0   0 x2 ··· xn   x1 x2 ··· 0   . . . .   . . . .   . . .. .   . . .. .  k   k   = x1 · |  k k  | + .... + xn · |  k k  | = |Vn|,  1 x2 ··· xn   x1 x2 ··· 1   . . .   . . .   . . .. .   . . .. .   . . . .   . . . .  n−1 n−1 n−1 n−1 0 x2 ··· xn x1 x2 ··· 0

but this is the inner product of the kth row vector of Vn with the kth column vector of its adjugate matrix through Cramer's rule. Dividing through by |Vn|, we can identify the components of the covector dual to the row vector.