- 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
故 f ( x ) f(x) f(x) 无重根
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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+a1a1a1⋮a1a2x+a2a2⋮a2a3a3x+a3⋮a3⋯⋯⋯⋱⋯ananan⋮x+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+a1a1a1⋮a1a2x+a2a2⋮a2a3a3x+a3⋮a3⋯⋯⋯⋱⋯ananan⋮x+an = xEn+ 111⋮1 [a1a2a3⋯an] =xn−1 x+[a1a2a3⋯an] 111⋮1 =xn−1(x+k=1∑nak).
- 已知 A A A 为实矩阵，证明方程组 A X = 0 AX = 0 AX=0 与 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
若 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
故 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 与 A T A X = 0 A^T AX = 0 ATAX=0 同解。
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设 A A A 是 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}. T−1AT=(ErOO−En−r).
设 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
由 A A A 是 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} T−1AT=(ErOO−En−r)
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已知 A A A 为 n n n 阶实反称矩阵，证明： E n − A 2 E_n - A^2 En−A2 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 E−A2=(E+A)(E−A)=(E+A)(E+A)T
从而 E n − A 2 E_n - A^2 En−A2 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(E−A2)x=xTx+xTATAx=xTx+(Ax)TAx>0
故 E n − A 2 E_n - A^2 En−A2 为正定矩阵。 -
设 A A A 为 n n n 阶实矩阵， B , C B, C B,C 为 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
设 A A A 特征值为 λ = a + b i \lambda=a+bi λ=a+bi，设 α \alpha α 为 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
