线性代数复习公式整理(自用/持续更新)

第一章 行列式

设A、B为n阶矩阵

∣ A T ∣ = ∣ A ∣ \left | A^T \right | =\left | A \right | AT =A

∣ A m ∣ = ∣ A ∣ m \left | A^m \right | =\left | A \right | ^m Am=Am

∣ k A ∣ = k n ∣ A ∣ \left | kA \right | =k^n\left | A \right | kA=knA

∣ A B ∣ = ∣ A ∣ ∣ B ∣ \left | AB \right | =\left | A \right | \left | B \right | AB=AB

若 A 可逆,则 ∣ A − 1 ∣ = 1 ∣ A ∣ 若A可逆,则\left | A^{-1} \right | =\frac{1}{\left | A\right | } A可逆,则 A1 =A1

∣ A ∗ ∣ = ∣ A ∣ n − 1 \left | A^* \right | =\left | A \right | ^{n-1} A=An1

A A ∗ = A ∗ A = ∣ A ∣ E AA^*=A^*A=\left | A \right | E AA=AA=AE

A ∗ = ∣ A ∣ A − 1 ( 若 A 可逆 ) A^*=\left | A \right | A^{-1}(若A可逆) A=AA1(A可逆)

A = ∣ A ∣ ( A ∗ ) − 1 A=\left | A \right | (A^*)^{-1} A=A(A)1

∣ A 1 A 2 A 3 ∣ = A 1 A 2 A 3 , ∣ A 1 A 2 A 3 ∣ = − A 1 A 2 A 3 \begin{vmatrix}A_1 & & \\ & A_2 & \\ & &A_3 \end{vmatrix}=A_1A_2A_3, \begin{vmatrix} & &A_1 \\ & A_2 & \\A_3 & & \end{vmatrix}=-A_1A_2A_3 A1A2A3 =A1A2A3, A3A2A1 =A1A2A3

设A为n阶矩阵,B为m阶矩阵,根据拉普拉斯展开定理有

∣ A 0 0 B ∣ = ∣ A C 0 B ∣ = ∣ A 0 C B ∣ = ∣ A ∣ ∣ B ∣ \begin{vmatrix}A & 0\\0 &B \end{vmatrix}=\begin{vmatrix}A & C\\0 &B \end{vmatrix}=\begin{vmatrix}A & 0\\C &B \end{vmatrix}=\left | A \right | \left | B \right | A00B = A0CB = AC0B =AB

∣ 0 A B 0 ∣ = ∣ C A B 0 ∣ = ∣ 0 A B C ∣ = ( − 1 ) m n ∣ A ∣ ∣ B ∣ \begin{vmatrix}0 & A\\B &0 \end{vmatrix}=\begin{vmatrix}C & A\\B &0 \end{vmatrix}=\begin{vmatrix}0 & A\\B &C \end{vmatrix}=(-1)^{mn}\left | A \right | \left | B \right | 0BA0 = CBA0 = 0BAC =(1)mnAB

化“叉”型行列式

∣ a 0 . . . 0 b . . . A . . . c 0 . . . 0 d ∣ = ( a d − b c ) ∣ A ∣ , 其中 A 是方阵 , 且除了主对角线和副对角线以外其余所有的元素均为 0 \begin{vmatrix} a& 0& ...& 0&b \\ ...& & A& &... \\ c& 0 & ... &0 &d \end{vmatrix}=(ad-bc)\left | A \right | ,其中A是方阵,且除了主对角线和副对角线以外其余所有的元素均为0 a...c00...A...00b...d =(adbc)A,其中A是方阵,且除了主对角线和副对角线以外其余所有的元素均为0

化“ab”型行列式

∣ a b b . . . b b a b . . . b b b a . . . b . . . . . . . . . . . . b b b b . . . a ∣ = [ a + ( n − 1 ) b ] ( a − b ) n − 1 \begin{vmatrix} a& b& b& ...&b \\ b& a& b& ...&b \\ b& b& a& ...&b \\ ...& ...& ...& ...&b \\ b& b& b& ...&a \end{vmatrix}=[a+(n-1)b](a-b)^{n-1} abb...bbab...bbba...b...............bbbba =[a+(n1)b](ab)n1

特征值求行列式

若题干可求得矩阵 A 的所有特征值 λ 1 , λ 2 . . . , λ n , 那么立即有 ∣ A ∣ = λ 1 λ 2 . . . λ n 若题干可求得矩阵A的所有特征值\lambda _1,\lambda _2...,\lambda _n,那么立即有\left | A \right | =\lambda _1\lambda _2...\lambda _n 若题干可求得矩阵A的所有特征值λ1,λ2...,λn,那么立即有A=λ1λ2...λn

第二章 矩阵

矩阵转置的性质

( A T ) T = A (A^T)^T=A (AT)T=A

( k A ) T = k A T (kA)^T=kA^T (kA)T=kAT

( A ± B ) T = A T ± B T (A\pm B)^T=A^T\pm B^T (A±B)T=AT±BT

( A B ) T = B T A T (AB)^T=B^TA^T (AB)T=BTAT

( A − 1 ) T = ( A T ) − 1 (A^{-1})^T=(A^T)^{-1} (A1)T=(AT)1

( A T ) m = ( A m ) T (A^T)^m=(A^m)^T (AT)m=(Am)T

矩阵伴随的性质

A ∗ = ∣ A ∣ A − 1 ( 若 A 可逆 ) A^*=\left | A \right | A^{-1}(若A可逆) A=AA1(A可逆)

A − 1 = 1 ∣ A ∣ A ∗ A^{-1}=\frac{1}{\left | A \right | } A^* A1=A1A

( A T ) ∗ = ( A ∗ ) T (A^T)^*=(A^*)^T (AT)=(A)T

( k A ) ∗ = k n − 1 A ∗ (kA)^*=k^{n-1}A^* (kA)=kn1A

( A B ) ∗ = B ∗ A ∗ (AB)^*=B^*A^* (AB)=BA

( a b c d ) ∗ = ( d − b − c a ) \begin{pmatrix} a & b\\ c&d \end{pmatrix}^*=\begin{pmatrix} d & -b\\ -c&a \end{pmatrix} (acbd)=(dcba)

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