矩阵运算规律总结
矩阵相乘结合律
A ( B C ) = ( A B ) C A(BC)=(AB)C A(BC)=(AB)C
矩阵相乘交换律
A ( B + C ) = A B + A C ( A + B ) C = A C + B C A(B+C) = AB + AC \\ (A+B)C = AC + BC A(B+C)=AB+AC(A+B)C=AC+BC
矩阵的逆
设
A
A
A 为方阵，如果存在方阵
A
−
1
A^{-1}
A−1 使得
A
A
−
1
=
A
−
1
A
=
I
AA^{-1} = A^{-1}A = I
AA−1=A−1A=I则方阵
A
A
A 可逆，
A
−
1
A^{-1}
A−1 为
A
A
A 的逆矩阵。方阵的逆若存在，则是唯一的，即一个方阵不可能有两个或以上的逆。如果方阵
A
A
A 和
B
B
B 均可逆，则其乘积
A
B
AB
AB 也可逆
(
A
B
)
−
1
=
B
−
1
A
−
1
(
A
B
)
−
1
(
A
B
)
=
B
−
1
A
−
1
A
B
=
B
−
1
(
A
−
1
A
)
B
（
结
合
律
）
=
B
−
1
I
B
=
I
\begin{aligned} (AB)^{-1} =& \ B^{-1}A^{-1} \\ \\ (AB)^{-1}(AB) =& \ B^{-1}A^{-1}AB \\ =& \ B^{-1}(A^{-1}A)B （结合律）\\ =& \ B^{-1}IB \\ =& \ I \end{aligned}
(AB)−1=(AB)−1(AB)==== B−1A−1 B−1A−1AB B−1(A−1A)B（结合律） B−1IB I三个（或多个）方阵乘积的逆
(
A
B
C
)
−
1
=
C
−
1
B
−
1
A
−
1
(
A
1
A
2
…
A
n
)
−
1
=
A
n
−
1
…
A
2
−
1
A
1
−
1
\begin{aligned} (ABC)^{-1} =& \ C^{-1}B^{-1}A^{-1} \\ (A_1A_2 \dots A_n)^{-1} =& \ A_n^{-1} \dots A_2^{-1}A_1^{-1} \end{aligned}
(ABC)−1=(A1A2…An)−1= C−1B−1A−1 An−1…A2−1A1−1
若方阵
A
A
A 可逆，则方阵
A
A
A 满足消去律，即
A
B
=
A
C
⟹
B
=
C
AB = AC \implies B=C
AB=AC⟹B=C证明
A
B
=
A
C
⟹
A
−
1
A
B
=
A
−
1
A
C
⟹
I
B
=
I
C
⟹
B
=
C
AB = AC \implies A^{-1}AB = A^{-1}AC \implies IB=IC \implies B = C
AB=AC⟹A−1AB=A−1AC⟹IB=IC⟹B=C
矩阵转置
矩阵相加的转置
(
A
+
B
)
T
=
A
T
+
B
T
(A + B)^T = A^T + B^T
(A+B)T=AT+BT
矩阵乘积的转置
(
A
B
)
T
=
B
T
A
T
(
A
B
C
)
T
=
C
T
B
T
A
T
\begin{aligned} (AB)^T =& \ B^TA^T \\ (ABC)^T =& \ C^TB^TA^T \end{aligned}
(AB)T=(ABC)T= BTAT CTBTAT
矩阵的逆和转置操作可以交换，即
(
A
T
)
−
1
=
(
A
−
1
)
T
(A^{T})^{-1}=(A^{-1})^T
(AT)−1=(A−1)T证明：
A
T
(
A
−
1
)
T
=
(
A
−
1
A
)
T
=
I
A^T(A^{-1})^T = (A^{-1}A)^T = I
AT(A−1)T=(A−1A)T=I所以
(
A
−
1
)
T
(A^{-1})^T
(A−1)T 即是
A
T
A^T
AT 的逆，即
(
A
T
)
−
1
=
(
A
−
1
)
T
(A^{T})^{-1}=(A^{-1})^T
(AT)−1=(A−1)T