常见矩阵性质
常见矩阵性质
1. 转置矩阵
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( A T ) T = A \left(\boldsymbol{A}^{\mathrm{T}}\right)^{\mathrm{T}}=\boldsymbol{A} (AT)T=A
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( A B ) T = B T A T (\boldsymbol{A B})^{\mathrm{T}}=\boldsymbol{B}^{\mathrm{T}} \boldsymbol{A}^{\mathrm{T}} (AB)T=BTAT
设 A = ( a i j ) m × s , B = ( b i j ) s × n \boldsymbol{A}=\left(a_{i j}\right)_{m \times s}, \boldsymbol{B}=\left(b_{i j}\right)_{s \times n} A=(aij)m×s,B=(bij)s×n，记 A B = C = ( c i j ) m × n , B T A T = D = ( d i j ) m × m \boldsymbol{A B}=\boldsymbol{C}=\left(c_{i j}\right)_{m \times n}, \boldsymbol{B}^{T} \boldsymbol{A}^{T}=\boldsymbol{D}=\left(d_{i j}\right)_{m \times m} AB=C=(cij)m×n,BTAT=D=(dij)m×m，按照矩阵乘法的定义，有
c j i = ∑ k = 1 s a j k b k i c_{j i}=\sum_{k=1}^{s} a_{j k} b_{k i} cji=k=1∑sajkbki
而 B T \boldsymbol{B}^{T} BT的第 i i i行为 ( b 1 i , ⋯ , b s i ) , A T \left(b_{1 i}, \cdots, b_{s i}\right), \boldsymbol{A}^{T} (b1i,⋯,bsi),AT， A T A^{T} AT的第 j j j列为 ( a j 1 , ⋯ , a j s ) T \left(a_{j 1}, \cdots, a_{j s}\right)^{T} (aj1,⋯,ajs)T，因此
d i j = ∑ k = 1 s b k i a j k = ∑ k = 1 s a j k b k i d_{i j}=\sum_{k=1}^{s} b_{k i} a_{j k}=\sum_{k=1}^{s} a_{j k} b_{k i} dij=k=1∑sbkiajk=k=1∑sajkbki
所以
d i j = c j i ( i = 1 , 2 , ⋯ , n ; j = 1 , 2 , ⋯ , m ) d_{i j}=c_{j i} \quad(i=1,2, \cdots, n ; j=1,2, \cdots, m) dij=cji(i=1,2,⋯,n;j=1,2,⋯,m)
则 D = C T \boldsymbol{D}=\boldsymbol{C}^{\mathrm{T}} D=CT，即 ( A B ) T = B T A T (\boldsymbol{A B})^{\mathrm{T}}=\boldsymbol{B}^{\mathrm{T}} \boldsymbol{A}^{\mathrm{T}} (AB)T=BTAT
- ( k A ) T = k A T (k \boldsymbol{A})^{\mathrm{T}}=k \boldsymbol{A}^{\mathrm{T}} (kA)T=kAT
- R ( A ) = R ( A T ) R(\boldsymbol{A})=R\left(\boldsymbol{A}^{\mathrm{T}}\right) R(A)=R(AT)
- ∣ A T ∣ = ∣ A ∣ \left|\boldsymbol{A}^{\mathrm{T}}\right|=|\boldsymbol{A}| ∣∣∣AT∣∣∣=∣A∣
- ( A + B ) T = A T + B T (\boldsymbol{A}+\boldsymbol{B})^{\mathrm{T}}=\boldsymbol{A}^{\mathrm{T}}+\boldsymbol{B}^{\mathrm{T}} (A+B)T=AT+BT
- E T = E \boldsymbol{E}^{\mathrm{T}}=\boldsymbol{E} ET=E
2.可逆矩阵
- ( A − 1 ) − 1 = A \left(\boldsymbol{A}^{-1}\right)^{-1}=\boldsymbol{A} (A−1)−1=A
- ( A B ) − 1 = B − 1 A − 1 (\boldsymbol{A B})^{-1}=\boldsymbol{B}^{-1} \boldsymbol{A}^{-1} (AB)−1=B−1A−1
因为
( A B ) ( B − 1 A − 1 ) = A ( B B − 1 ) A − 1 = A E A − 1 = E (\boldsymbol{A B})\left(\boldsymbol{B}^{-1} \boldsymbol{A}^{-1}\right)=\boldsymbol{A}\left(\boldsymbol{B B}^{-1}\right) \boldsymbol{A}^{-1}=\boldsymbol{A E A}^{-1}=\boldsymbol{E} (AB)(B−1A−1)=A(BB−1)A−1=AEA−1=E
( B − 1 A − 1 ) ( A B ) = B − 1 ( A − 1 A ) B = B − 1 E B = E \left(\boldsymbol{B}^{-1} \boldsymbol{A}^{-1}\right)(\boldsymbol{A} \boldsymbol{B})=\boldsymbol{B}^{-1}\left(\boldsymbol{A}^{-1} \boldsymbol{A}\right) \boldsymbol{B}=\boldsymbol{B}^{-1} \boldsymbol{E} \boldsymbol{B}=\boldsymbol{E} (B−1A−1)(AB)=B−1(A−1A)B=B−1EB=E
所以
( A B ) − 1 = B − 1 A − 1 (\boldsymbol{A} \boldsymbol{B})^{-1}=\boldsymbol{B}^{-1} \boldsymbol{A}^{-1} (AB)−1=B−1A−1
- ( λ A ) − 1 = λ − 1 A − 1 (\lambda \boldsymbol{A})^{-1}=\lambda^{-1} \boldsymbol{A}^{-1} (λA)−1=λ−1A−1
- ( A T ) − 1 = ( A − 1 ) T \left(\boldsymbol{A}^{T}\right)^{-1}=\left(\boldsymbol{A}^{-1}\right)^{T} (AT)−1=(A−1)T
- ( A B ) − 1 = B − 1 A − 1 (\boldsymbol{A} \boldsymbol{B})^{-1}=\boldsymbol{B}^{-1} \boldsymbol{A}^{-1} (AB)−1=B−1A−1
- R ( A ) = R ( A − 1 ) R(\boldsymbol{A})=R\left(\boldsymbol{A}^{-1}\right) R(A)=R(A−1)
- ( A + B ) − 1 ≠ A − 1 + B − 1 (\boldsymbol{A}+\boldsymbol{B})^{-1} \neq \boldsymbol{A}^{-1}+\boldsymbol{B}^{-1} (A+B)−1=A−1+B−1
3.伴随矩阵
- ( A ∗ ) ∗ = ∣ A ∣ n − 2 A \left(\boldsymbol{A}^{*}\right)^{*}=|\boldsymbol{A}|^{n-2} \boldsymbol{A} (A∗)∗=∣A∣n−2A
- ( A B ) ∗ = B ∗ A ∗ (\boldsymbol{A} \boldsymbol{B})^{*}=\boldsymbol{B}^{*} \boldsymbol{A}^{*} (AB)∗=B∗A∗
- ( k A ) ∗ = k n − 1 A ∗ (k \mathbf{A})^{*}=k^{n-1} \mathbf{A}^{*} (kA)∗=kn−1A∗
- R ( A ∗ ) = { n , R ( A ) = n 1 , R ( A ) = n − 1 0 , R ( A ) < n − 1 R\left(\boldsymbol{A}^{*}\right)=\left\{\begin{array}{l}n, R(\boldsymbol{A})=n \\ 1, R(\boldsymbol{A})=n-1 \\ 0, R(\boldsymbol{A})
- ∣ A ∗ ∣ = ∣ A ∣ n − 1 \left|\boldsymbol{A}^{*}\right|=|\boldsymbol{A}|^{n-1} ∣A∗∣=∣A∣n−1
- ( A + B ) ∗ ≠ A ∗ + B ∗ (\boldsymbol{A}+\boldsymbol{B})^{*} \neq \boldsymbol{A}^{*}+\boldsymbol{B}^{*} (A+B)∗=A∗+B∗