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求Vandemonde determinant
Vandemonde determinant :
V=∣∣∣∣∣∣∣∣∣∣111⋮1a1a2a3⋮ama12a22a32⋮am2⋯⋯⋯⋱⋯a1n−1a2n−1a3n−1⋮amn−1∣∣∣∣∣∣∣∣∣∣
V的行列式的值为:</br>
假设 m=n ,那么:
V=1≤i<j<≤n∏(aj−ai)
下面是演算过程:
- 根据行列式的性质(将一行(列)的k倍加进另一行(列)里,行列式的值不变):
先把每一列减上其前一列的a1倍得:
V=∣∣∣∣∣∣∣∣∣∣111⋮10a2−a1a3−a1⋮am−a10a2(a2−a1)a3(a3−a1)⋮am(am−a1)⋯⋯⋯⋱⋯0a2n−2(a2−a1)a3n−2(a3−a1)⋮amn−2(am−a1)∣∣∣∣∣∣∣∣∣∣
- 此时把上述行列式按照第一行展开,得:
V=1×∣∣∣∣∣∣∣∣a2−a1a3−a1⋮am−a1a2(a2−a1)a3(a3−a1)⋮am(am−a1)⋯⋯⋱⋯a2n−2(a2−a1)a3n−2(a3−a1)⋮amn−2(am−a1)∣∣∣∣∣∣∣∣
- 再把上述行列式的公因数提出得(在行列式中,某一行(列)有公因子k,则可以提出k):
V=(a2−a1)(a3−a1)⋯(am−a1)∣∣∣∣∣∣∣∣11⋮1a2a3⋮am⋯⋯⋱⋯a2n−1a3n−1⋮amn−1∣∣∣∣∣∣∣∣
此时的V变成了 ∏1<i≤m(ai−a1) 乘一个 (n-1)×(m-1) 的Vandemonde determinant , 将 (n-1)×(m-1) 的Vandemonde determinant重复1-3操作.
最终得结果为:
V=1≤i<j<≤n∏(aj−ai)