Editing Vandermonde matrix (section)

== Generalizations ==

If the columns of the Vandermonde matrix, instead of <math display="inline">1, x, x^2, ...</math>, are general polynomials <math display="inline">p_0, p_1, ..., p_n</math>, such that each one has degree <math display="inline">0, 1, ..., n</math>, that is, if <math>V = [p_i(x_j)]_{i, j \in 0:n}</math>, <math display="block">\det V(x_{0:n}) = \prod_k c_k \Delta(x)</math>where <math display="inline">c_0, ..., c_n</math> are the head coefficients of <math display="inline">p_0, p_1, ..., p_n</math>, and <math>\Delta(x) = \prod_{j>k} (x_j - x_k)</math> is the Vandermonde determinant.

{{Math proof|title=Proof|proof=
<math display="inline">\det V(x_{0:n})</math> is zero whenever <math display="inline">x_j = x_k</math>, and has degree <math display="inline">\frac 12 n(n+1)</math>, so it is a multiple of <math display="inline">\prod_{j>k} (x_j - x_k)</math>. To find the constant in front, simply calculate the coefficient of the term <math display="inline">x_0^0 \dots x_n^n</math>, which is <math display="inline">c_0 \dots c_n</math>.
}}

By multiplying with the Hermitian conjugate, we find that <math display="block">
 \det \left[\sum_l p_j(z_l) p_k(z_l^*)\right] = \prod_k |c_k|^2 |\Delta(z)|^2
 </math>

{{Math theorem
| math_statement = Fix <math display="inline">x</math>, and at the <math display="inline">y \to 0</math> limit, <math display="block">\det [e^{x_i y_j}] = \frac{1}{1! \dots n! (n+1)!}\Delta(x) \Delta(y) + o(\Delta(y))</math> uniformly for <math display="inline">y</math>
| note = Tao 2012, page 251
}}

{{hidden begin|style=width:100%|ta1=center|border=1px #aaa solid|title=Proof}}

{{Math proof|title=Proof|proof=

If any <math display="inline">x_j = x_k</math> or <math display="inline">y_j = y_k</math>, then the determinant is zero, so it has the form <math display="block">\det [e^{x_i y_j}] = C(x,y) \Delta(x) \Delta(y)</math> where <math display="inline">C(x,y)</math> is some power series in <math display="inline">x, y</math>.

The left side is a sum of the form <math display="block">
          \sum_{\sigma} (-1)^{|\sigma|} e^{\sum_i x_i y_{\sigma(i)}}
          </math> Expand them by Taylor expansion. For fixed <math display="inline">x</math>, the series is uniformly convergent in <math display="inline">y</math> in a neighborhood of zero.

To find the constant term of <math display="inline">C(x, y)</math>, simply calculate the coefficient of the term <math display="inline">x_0^0 y_0^0 \dots x_n^n y_n^n</math>, which is <math display="inline">\frac{1}{1! \cdots (n+1)!}</math>.

By the symmetry of the determinant, the next lowest-powered term of <math display="inline">C(x, y)</math> is of form <math display="inline">a(x_0y_0 + \dots + x_n y_n)</math>, which is <math display="inline">o(1)</math> as <math display="inline">y \to 0</math>.

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