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2021高等代数【南昌大学】

  1. 证明多项式 f ( x ) = 1 + x + x 2 2 ! + ⋯ + x n n ! f(x) = 1 + x + \frac{x^2}{2!} + \cdots + \frac{x^n}{n!} f(x)=1+x+2!x2++n!xn 无重根。

f ( x ) − f ′ ( x ) = x n n ! f(x) - f'(x) = \frac{x^n}{n!} f(x)f(x)=n!xn

( f ( x ) , f ′ ( x ) ) = ( f ( x ) , f ( x ) − f ′ ( x ) ) = ( f ( x ) , x n n ! ) = x k \left( f(x), f'(x) \right) = \left( f(x), f(x) - f'(x) \right) = \left( f(x), \frac{x^n}{n!} \right) = x^k (f(x),f(x))=(f(x),f(x)f(x))=(f(x),n!xn)=xk

其中 k k k 是小于等于 n n n 的非负整数,由于 0 显然不是 f ( x ) f(x) f(x) 的根,所以只能是 k = 0 k=0 k=0,即 ( f ( x ) , f ′ ( x ) ) = 1 \left( f(x), f'(x) \right) =1 (f(x),f(x))=1

f ( x ) f(x) f(x) 无重根

  1. 求行列式

    D n = ∣ x + a 1 a 2 a 3 ⋯ a n a 1 x + a 2 a 3 ⋯ a n a 1 a 2 x + a 3 ⋯ a n ⋮ ⋮ ⋮ ⋱ ⋮ a 1 a 2 a 3 ⋯ x + a n ∣ D_n = \begin{vmatrix} x + a_1 & a_2 & a_3 & \cdots & a_n \\ a_1 & x + a_2 & a_3 & \cdots & a_n \\ a_1 & a_2 & x + a_3 & \cdots & a_n \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ a_1 & a_2 & a_3 & \cdots & x + a_n \\ \end{vmatrix} Dn= x+a1a1a1a1a2x+a2a2a2a3a3x+a3a3anananx+an

D n = ∣ x + a 1 a 2 a 3 ⋯ a n a 1 x + a 2 a 3 ⋯ a n a 1 a 2 x + a 3 ⋯ a n ⋮ ⋮ ⋮ ⋱ ⋮ a 1 a 2 a 3 ⋯ x + a n ∣ = ∣ x E n + [ 1 1 1 ⋮ 1 ] [ a 1 a 2 a 3 ⋯ a n ] ∣ = x n − 1 ∣ x + [ a 1 a 2 a 3 ⋯ a n ] [ 1 1 1 ⋮ 1 ] ∣ = x n − 1 ( x + ∑ k = 1 n a k ) . \begin{aligned} D_n &= \left| \begin{array}{ccccc} x + a_1 & a_2 & a_3 & \cdots & a_n \\ a_1 & x + a_2 & a_3 & \cdots & a_n \\ a_1 & a_2 & x + a_3 & \cdots & a_n \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ a_1 & a_2 & a_3 & \cdots & x + a_n \end{array} \right| \\ &= \left| xE_n + \left[ \begin{array}{c} 1 \\ 1 \\ 1 \\ \vdots \\ 1 \end{array} \right] \left[ \begin{array}{ccccc} a_1 & a_2 & a_3 & \cdots & a_n \end{array} \right] \right| \\ &= x^{n-1} \left| x + \left[ \begin{array}{ccccc} a_1 & a_2 & a_3 & \cdots & a_n \end{array} \right] \left[ \begin{array}{c} 1 \\ 1 \\ 1 \\ \vdots \\ 1 \end{array} \right] \right| \\ &= x^{n-1} \left( x + \sum\limits_{k = 1}^n a_k \right). \end{aligned} Dn= x+a1a1a1a1a2x+a2a2a2a3a3x+a3a3anananx+an = xEn+ 1111 [a1a2a3an] =xn1 x+[a1a2a3an] 1111 =xn1(x+k=1nak).

  1. 已知 A A A 为实矩阵,证明方程组 A X = 0 AX = 0 AX=0 A T A X = 0 A^T AX = 0 ATAX=0 同解。

A x = 0 Ax=0 Ax=0 的解均是 A T A x = 0 A^T Ax = 0 ATAx=0 的解

A T A x = 0 A^T Ax = 0 ATAx=0,则 x T A T A x = ( A x ) T A x = 0 x^T A^T Ax=(Ax)^TAx = 0 xTATAx=(Ax)TAx=0

A x = 0 Ax=0 Ax=0,即 A T A x = 0 A^T Ax = 0 ATAx=0 的解均是 A x = 0 Ax=0 Ax=0 的解

因此方程组 A X = 0 AX = 0 AX=0 A T A X = 0 A^T AX = 0 ATAX=0 同解。

  1. A A A n n n 阶实对称矩阵,且 A 2 = E n A^2 = E_n A2=En,证明:存在正交矩阵 T T T,使得

    T − 1 A T = ( E r O O − E n − r ) . T^{-1} A T = \begin{pmatrix} E_r & O \\ O & -E_{n-r} \end{pmatrix}. T1AT=(ErOOEnr).

A A A 的特征值为 λ \lambda λ,从而 A 2 = E n A^2 = E_n A2=En 的特征值为 λ 2 = 1 \lambda^2=1 λ2=1,即 λ = ± 1 \lambda = \pm 1 λ=±1。由 λ \lambda λ 的任意性, A A A 的特征值都是 ± 1 \pm 1 ±1

A A A n n n 级实对称矩阵,故存在正交矩阵 T T T
T − 1 A T = ( E r O O − E n − r ) T^{-1} A T = \begin{pmatrix} E_r & O \\ O & -E_{n-r} \end{pmatrix} T1AT=(ErOOEnr)

  1. 已知 A A A n n n 阶实反称矩阵,证明: E n − A 2 E_n - A^2 EnA2 为正定矩阵。
    E − A 2 = ( E + A ) ( E − A ) = ( E + A ) ( E + A ) T E - A^2 = \left( E + A \right)\left( E - A \right) = \left( E + A \right)\left( E + A \right)^T EA2=(E+A)(EA)=(E+A)(E+A)T
    从而 E n − A 2 E_n - A^2 EnA2 为对称矩阵

    对于 n n n 维列向量 x ≠ 0 x\ne0 x=0 x T x > 0 x^T x > 0 xTx>0
    x T ( E − A 2 ) x = x T x + x T A T A x = x T x + ( A x ) T A x > 0 x^T \left( E - A^2 \right) x = x^T x + x^T A^T A x = x^T x + \left( Ax \right)^T Ax > 0 xT(EA2)x=xTx+xTATAx=xTx+(Ax)TAx>0
    E n − A 2 E_n - A^2 EnA2 为正定矩阵。

  2. A A A n n n 阶实矩阵, B , C B, C B,C n n n 阶正定矩阵,且

    A B + B A T = − C . AB + BA^T = -C. AB+BAT=C.

    证明: A A A 的特征值实部均小于零。

C 为正定矩阵,任意 x ≠ 0 x\ne0 x=0 x T C x > 0 x^TCx>0 xTCx>0,那么

x T ( A B + B A T ) x = x T A B x + x T B A T x = ( A T x ) T ( B x ) + ( B x ) T ( A T x ) = ( A T x ) T ( B x ) < 0 x^T \left( AB + B A^T \right) x = x^T ABx + x^T B A^T x = \left( A^T x \right)^T \left( Bx \right) + \left( Bx \right)^T \left( A^T x \right) = \left( A^T x \right)^T \left( Bx \right) < 0 xT(AB+BAT)x=xTABx+xTBATx=(ATx)T(Bx)+(Bx)T(ATx)=(ATx)T(Bx)<0
因此 A T x ≠ 0 A^Tx\ne0 ATx=0,即 A A A 可逆

A A A 特征值为 λ = a + b i \lambda=a+bi λ=a+bi,设 α \alpha α A T A^T AT 属于特征值 λ = a + b i \lambda=a+bi λ=a+bi 的特征向量
α ˉ T ( A B + B A T ) α = α ˉ T A B α + α ˉ T B A T α = ( A


原文地址:https://blog.csdn.net/m0_64548999/article/details/144297833

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