(来自 succeme) $A$是给定的方阵，特征值已知，其他小写字母为复数，用$A$的特征值表出下列行列式的值： [ egin{pmatrix} b_0E & b_1A &b_2A^2 &cdots &b_{n-1}A^{n-1} \ ab_{n-1}A^{n-1} &b_0E & b_1A &cdots & b_{n-2}A^{n-2} \ ab_{n-2}A^{n-2} & ab_{n-1}A^{n-1} & b_0E & cdots & b_{n-3}A^{n-3} \ cdots &cdots &cdots &cdots & cdots \ ab_1A & ab_2A^2 & ab_3A^3 &cdots &b_0E end{pmatrix} ]
解答: (来自 torsor) 首先，若 $a=0$, 则所求矩阵是分块上三角阵，容易看出其行列式等于 $b_0^{n^2}$. 以下设 $a eq 0$, 并且 $omega_1,cdots,omega_n$ 是 $x^n-a$ 的 $n$ 个不同的根. 设复系数多项式 $$f(x)=b_0+b_1x+cdots+b_{n-1}x^{n-1}.$$ 考虑如下分块矩阵的乘法： $$ egin{bmatrix} b_0 & b_1A & b_2A^2 & cdots & b_{n-1}A^{n-1} \ ab_{n-1}A^{n-1} & b_0 & b_1A & cdots & b_{n-2}A^{n-2} \ vdots & vdots & vdots & vdots & vdots \ ab_1A & ab_2A^2 & ab_3A^3 & cdots & b_0 end{bmatrix} egin{bmatrix} I_n & I_n & I_n & cdots & I_n \ omega_1I_n & omega_2I_n & omega_3I_n & cdots & omega_nI_n \ vdots & vdots & vdots & vdots & vdots \ omega_1^{n-1}I_n & omega_2^{n-1}I_n & omega_3^{n-1}I_n & cdots & omega_n^{n-1}I_n end{bmatrix} $$ $$=egin{bmatrix} f(omega_1A) & f(omega_2A) & f(omega_3A) & cdots & f(omega_nA) \ omega_1f(omega_1A) & omega_2f(omega_2A) & omega_3f(omega_3A) & cdots & omega_nf(omega_nA) \ vdots & vdots & vdots & vdots & vdots \ omega_1^{n-1}f(omega_1A) & omega_2^{n-1}f(omega_2A) & omega_3^{n-1}f(omega_3A) & cdots & omega_n^{n-1}f(omega_nA) end{bmatrix} $$ $$=egin{bmatrix} I_n & I_n & I_n & cdots & I_n \ omega_1I_n & omega_2I_n & omega_3I_n & cdots & omega_nI_n \ vdots & vdots & vdots & vdots & vdots \ omega_1^{n-1}I_n & omega_2^{n-1}I_n & omega_3^{n-1}I_n & cdots & omega_n^{n-1}I_n end{bmatrix} egin{bmatrix} f(omega_1A) & & & & \ & f(omega_2A) & & & \ & & ddots & & \ & & & & f(omega_nA) end{bmatrix}. $$ 由 Laplace 定理容易得到 $$egin{vmatrix} I_n & I_n & I_n & cdots & I_n \ omega_1I_n & omega_2I_n & omega_3I_n & cdots & omega_nI_n \ vdots & vdots & vdots & vdots & vdots \ omega_1^{n-1}I_n & omega_2^{n-1}I_n & omega_3^{n-1}I_n & cdots & omega_n^{n-1}I_n end{vmatrix}=prod_{1leq i<jleq n}(omega_j-omega_i)^n eq 0, $$ 故 $$ egin{vmatrix} b_0 & b_1A & b_2A^2 & cdots & b_{n-1}A^{n-1} \ ab_{n-1}A^{n-1} & b_0 & b_1A & cdots & b_{n-2}A^{n-2} \ vdots & vdots & vdots & vdots & vdots \ ab_1A & ab_2A^2 & ab_3A^3 & cdots & b_0 end{vmatrix}=|f(omega_1A)|cdot|f(omega_2A)|cdots|f(omega_nA)|, $$ 用 $A$ 的特征值不难将上述行列式的值写出来.