矩阵

矩阵_第1张图片

概念

  1. 正交矩阵:
    A T = A − 1 A^T=A^{-1} AT=A1
    ∣ A ∣ 2 = 1 |A|^2=1 A2=1
  2. 可逆矩阵:
    ∣ A ∣ ≠ 0 |A|\ne0 A=0
    r ( A ) = n r(A)=n r(A)=n
    A 的 列 向 量 组 线 性 无 关 A的列向量组线性无关 A线
    A = P 1 P 2 ⋯ P i A=P_1P_2\cdots P_i A=P1P2Pi
    A 与 单 位 矩 阵 等 价 A与单位矩阵等价 A
    0 不 是 特 征 值 0不是特征值 0

公式

  1. 可逆
    ( A n ) − 1 = ( A − 1 ) n (A^n)^{-1}=(A^{-1})^n (An)1=(A1)n
    ( A T ) − 1 = ( A − 1 ) T (A^T)^{-1}=(A^{-1})^T (AT)1=(A1)T
    ∣ A − 1 ∣ = ∣ A ∣ − 1 = 1 ∣ A ∣ A − 1 = 1 ∣ A ∣ A ∗ |A^{-1}|=|A|^{-1}=\frac{1}{|A|}A_{-1}=\frac{1}{|A|}A^* A1=A1=A1A1=A1A
  2. 伴随
    ∣ A ∗ ∣ = ∣ A ∣ n − 1 |A^*|=|A|^{n-1} A=An1
    ( A ∗ ) − 1 = ( A − 1 ) ∗ = 1 ∣ A ∣ A \left(A^*\right)^{-1}=\left(A^{-1}\right)^*=\frac{1}{|A|}A (A)1=(A1)=A1A
    ( A ∗ ) T = ( A T ) ∗ \left(A^*\right)^T=\left(A^T\right)^* (A)T=(AT)
    ( k A ) ∗ = k n − 1 A ∗ \left(kA\right)^*=k^{n-1}A^* (kA)=kn1A
    ( A ∗ ) ∗ = ∣ A ∣ n − 2 A \left(A^*\right)^*=|A|^{n-2}A (A)=An2A
    r ( A ∗ ) = { n , r ( A ) = n 1 , r ( A ) = n − 1 0 , r ( A ) < n − 1 r\left(A^*\right)= \left\{\begin{matrix} n, & r(A)=n \\ 1, & r(A)=n-1 \\ 0, & r(A)r(A)=n,1,0,r(A)=nr(A)=n1r(A)<n1

  3. r ( A ) = r ( A T ) r(A)=r(A^T) r(A)=r(AT)
    r ( A T A ) = r ( A ) r(A^TA)=r(A) r(ATA)=r(A)
    当 k ≠ 0 时 : r ( k A ) = r ( A ) 当k\ne 0时:r(kA)=r(A) k=0r(kA)=r(A)
    r ( A + B ) ≤ r ( A ) + r ( B ) r(A+B)\le r(A)+r(B) r(A+B)r(A)+r(B)
    r ( A B ) ≤ min ⁡ ( r ( A ) , r ( B ) ) r(AB)\le \min(r(A),r(B)) r(AB)min(r(A),r(B))
    若 A 是 m × n 矩 阵 , B 是 n × s 矩 阵 , A B = O , 则 : r ( A ) + r ( B ) ≤ n 若A是m\times n矩阵,B是n\times s矩阵,AB=O,则:r(A)+r(B)\le n Am×n,Bn×s,AB=O,r(A)+r(B)n
    r [ A O O B ] = r ( A ) + r ( B ) r\begin{bmatrix} A & O \\ O & B \end{bmatrix} =r(A)+r(B) r[AOOB]=r(A)+r(B)
    若 A ∼ B , 则 r ( A ) = r ( B ) , r ( A + k E ) = r ( B + k E ) 若A\sim B,则r(A)=r(B),r(A+kE)=r(B+kE) AB,r(A)=r(B),r(A+kE)=r(B+kE)
    A B 等 价 ⇔ r ( A ) = r ( A ∣ B ) = r ( B ) ⇔ 互 相 表 出 AB等价\Leftrightarrow r(A)=r(A|B)=r(B)\Leftrightarrow 互相表出 ABr(A)=r(AB)=r(B)
  4. 分块矩阵
    [ A B C D ] T = [ A T C T B T D T ] \begin{bmatrix} A & B \\ C & D \end{bmatrix} ^T= \begin{bmatrix} A^T & C^T \\ B^T & D^T \end{bmatrix} [ACBD]T=[ATBTCTDT]
    [ B O O C ] − 1 = [ B − 1 O O C − 1 ] \begin{bmatrix} B & O \\ O & C \end{bmatrix} ^{-1}= \begin{bmatrix} B^{-1} & O \\ O & C^{-1} \end{bmatrix} [BOOC]1=[B1OOC1]
    [ O B C O ] − 1 = [ O C − 1 B − 1 O ] \begin{bmatrix} O & B \\ C & O \end{bmatrix} ^{-1}= \begin{bmatrix} O & C^{-1} \\ B^{-1} & O \end{bmatrix} [OCBO]1=[OB1C1O]
    A = [ 0 a b 0 0 c 0 0 0 ] , 则 A 2 = [ 0 0 a c 0 0 0 0 0 0 ] A= \begin{bmatrix} 0 & a & b \\ 0 & 0 & c \\ 0 & 0 & 0 \end{bmatrix} ,则A^2= \begin{bmatrix} 0 & 0 & ac \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} A=000a00bc0,A2=000000ac00
    A = [ 0 d a b 0 0 e c 0 0 0 f 0 0 0 0 ] , 则 A 3 = [ 0 0 0 d e f 0 0 0 0 0 0 0 0 0 0 0 0 ] A= \begin{bmatrix} 0 & d & a & b \\ 0 & 0 & e & c \\ 0 & 0 & 0 & f \\ 0 & 0 & 0 & 0 \end{bmatrix} ,则A^3= \begin{bmatrix} 0 & 0 & 0 & def \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} A=0000d000ae00bcf0,A3=000000000000def000

运算法则

  1. A B = O AB=O AB=O推不出 A = O , B = O A=O,B=O A=O,B=O
  2. A B = A C , A ≠ C AB=AC,A\ne C AB=AC,A=C推不出 B = C B=C B=C
  3. ∣ A B ∣ = ∣ A ∣ ⋅ ∣ B ∣ |AB|=|A|\cdot|B| AB=AB

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