范德蒙行列式应用

我们知道，n阶范德蒙德行列式

1x11x2Vn???1xnx12?x1n?12n?1x2?x2???xi?xj?，

??1≤j?i≤n2n?1xn?xn当这些xi两两互异时，Vn?0．这个事实有助于我们理解不少结果．

例1 证明一个n次多项式之多有n个互异根．

证 设f?x??a0?a1x?a2x2???anxn有n?1个互异的零点x1,x2,?,xn?1，则有

f?xi??a0?a1xi?a2xi2???anxin?0，1 ≤ i ≤ n?1．

即

?a0?x1a1?x12a2???x1nan?0,?2n?a0?x2a2?x2a2???x2an?0, ? ???a?xa?x2a???xna?0.n?1n?0n?1nn?12这个关于a0,a1,?,an的齐次线性方程组的系数行列式

11?x1x2?x122x2???x1nnx2???xi?xj??0， ?1≤j?i≤n?11xn?12nxn?xn?1?1因此a0?a1?a2???an?0．这个矛盾表明f?x?至多有n个互异根．

例2 设a1,a2,?,an是n个两两互异的数．证明对任意n个数b1,b2,?,bn，存在惟一的次数小于n的多项式L?x?：

L?x???bi?i?1j?inx?ajai?aj，

使得L?ai??bi，1 ≤ i ≤ n．

证 从定义容易看出L?x?的次数小于n，且L?ai??bi，故只需证明唯一性即可． 设f?x??c0?c1x?c2x???cn?1x2n?1满足

f?ai??bi，1 ≤ i ≤ n，

即

?c0?a1c1?a12c2???a1n?1cn?1?b1,?2n?1?c0?a2c1?a2c2???a2cn?1?b2, ?? ??c?ac?a2c???an?1c?b.nn?1n?0n1n2这个关于c0,c1,c2,?,cn?1的线性方程组的系数行列式

1a11a2??1ana12?a1n?12n?1a2?a2???ai?aj??0，

??1≤j?i≤n2n?1an?an故c0,c1,c2,?,cn?1是唯一的，必须f?x??L?x?． 这个例子就是有名的拉格朗日插值公式．

例3 设f1?x?,f2?x?,?,fn?1?x?是n?1个复系数多项式，满足

1?x???xn?1|f1?xn??xf2?xn????xn?2fn?1?xn?，

证明f1?1??f2?1????fn?1?1??0．

nnn?2nn?1证 设f1x?xf2x???xfn?1x?p?x?1?x???x，取??cos????????2?2??isin，nn分别以x??,?2,?,?n?1代入，可得

?f1?1???f2?1?????n?2fn?1?1??0,?2?n?2?fn?1?1??0,?f1?1???2f2?1????? ?? ???n?1??n?2?n?1f1??f1????fn?1?1??0.????2?1这个关于f1?1?,f2?1?,?,fn?1?1?的齐次线性方程组的系数行列式

11???2????n?2?2?n?2??n?1??n?2??0，

1?n?1???因此f1?1??f2?1????fn?1?1??0．

例4 设n是奇数，f1?x?,f2?x?,?,fn?1?x?是n?1个复系数多项式，满足

xn?1?xn?2?xn?3???1|f1?xn?2??xf2?xn????xn?2fn?1?xn?，

证明f1??1??f2??1????fn?1??1??0．

证 注意到当n是奇数时，

xn?1??x?1??xn?1?xn?2?xn?3???1?，

可按照例3的思路完成证明．

例5 设A是个n阶矩阵，证明A的属于不同特征值的特征向量线性无关．

证 设?1,?2,?,?r是A的两两不同的r个特征值，非零向量?1,?2,?,?r适合

1 ≤ i ≤ r， A?i???ii，

假设

x1?1?x2?2???xr?r?0，

那么有

Aj?x1?1?x2?2???xr?r??0，1 ≤ j ≤ r?1．

即

r?r?rjA??xi?i???xiA?i???ij??xi?i??0，

i?1?i?1?i?1j注意到

???jir?r?0，

必须x1?1?x2?2???xr?r?0，于是x1?x2???xr?0，这证明了?1,?2,?,?r线性无关．

例6 计算行列式

11???1?1?x1?Dn??2?x1??其中?k?x??x?a1kxk?1?x2??2?x2???1?xn??2?xn?，

??n?1?x1??n?1?x2???n?1?xn?k?1???ank．

解 注意到下面的等式：

??11?1???1?x1??1?x2???1?xn?? ???2?x1??2?x2????2?xn??????????n?1?x1??n?1?x2?????n?1xn?????100?0??11?a1110?0??????x1x2?a22a121?0??x2x22???????1??????an?1n?1an?2n?1an?3n?1?1????xn?1xn?112即得

Dn???xi?xj?．

1≤j?i≤n例7 计算行列式

11?1??x1???x2??1?????xn?D??1???1??n???，??x1??n?1????x2??n?1?????xn??n?1??其中??x??x?x?1???x?k?1??k??k!．

解 直接利用例6可得

D1n?1!2!??n?1?!1≤??xi?xj?．j?i≤n例8 设a1,a2,?,an是正整数，证明n阶行列式

1a1a2n?11?a1V?1a2a2?an?122n????

1ana2n?1n?an能被1n?12n?2??n?2?2?n?1?整除．

证 直接运用例6、例7可得

?1?

?x?nx2??n??? ,??xn?1n?? 1a1Vn?1a2??1ana1?a1?1??a1?a1?1??a1?2???a1?n?2?a2?a2?1??a2?a2?1??a2?2???a2?n?2???an?an?1??an?an?1??an?2???an?n?2??a1??a1???????2??n?1??a2??a2???????2??n?1???

?a1?1???1??a?1?2? ?1!2!??n?1?!?1????an??an??an?1????????1??2??n?1?n?1n?2能被1!2!??n?1?!?12??n?2?2?n?1?整除．

11??1例9 计算n阶范德蒙德行列式

11Vn?1???2?n?1?2?4??n?1??2?n?1?，

????n?1?21?其中??cos?2?n?1?2?2??i?sin． nnk解 注意到??1当且仅当n|k，可得

n0?0000?0n?n?1??n?2?2Vn?00?n0???1?2nn，

????0n?00?n?1??n?2?由此Vn??i故

2n，Vn的模Vn?n．现在来确定Vn的幅角：令??cosn2n2?n?isin?n，???，

2Vn? ? ?0≤j?k≤n?1?????k??j??0≤j?k≤n?1???2k??2j?

?k?j?k?j????k?j??k?j0≤j?k≤n?1??n0≤j?k≤n?10≤j?k≤n?1?k?j???2i?sin .

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