• 【线性代数基础进阶】向量-part2

    文章目录

    线性相关

    定义：对 m m m个 n n n维向量 α 1 , α 2 , ⋯   , α m \alpha_{1},\alpha_{2},\cdots,\alpha_{m} α1​,α2​,⋯,αm​，若存在不全为 0 0 0的实数 k 1 , k 2 , ⋯   , k m k_{1},k_{2},\cdots,k_{m} k1​,k2​,⋯,km​使
    k 1 α 1 + k 2 α 2 + ⋯ + k m α m = 0 k_{1}\alpha_{1}+k_{2}\alpha_{2}+\cdots+k_{m}\alpha_{m}=0 k1​α1​+k2​α2​+⋯+km​αm​=0
    成立，则其向量组 α 1 , α 2 , ⋯   , α m \alpha_{1},\alpha_{2},\cdots,\alpha_{m} α1​,α2​,⋯,αm​线性相关，否则称其线性无关

    例：判断向量组 α 1 = ( 1 , 2 , − 1 , 4 ) T , α 2 = ( 0 , − 1 , − 5 , 3 ) T , α 3 = ( 2 , 5 , 3 , 5 ) T \alpha_{1}=(1,2,-1,4)^{T},\alpha_{2}=(0,-1,-5,3)^{T},\alpha_{3}=(2,5,3,5)^{T} α1​=(1,2,−1,4)T,α2​=(0,−1,−5,3)T,α3​=(2,5,3,5)T的线性相关性

    设 x 1 α 1 + x 2 α 2 + x 3 α 3 = 0 x_{1}\alpha_{1}+x_{2}\alpha_{2}+x_{3}\alpha_{3}=0 x1​α1​+x2​α2​+x3​α3​=0，即
    x 1 ( 1 2 − 1 4 ) + x 2 ( 0 − 1 − 5 3 ) + x 3 ( 2 5 3 5 ) = ( 0 0 0 0 ) x_{1}

    (12−14)" role="presentation" style="position: relative;">(12−14)
    +x_{2}
    (0−1−53)" role="presentation" style="position: relative;">(0−1−53)
    +x_{3}
    (2535)" role="presentation" style="position: relative;">(2535)
    =
    (0000)" role="presentation" style="position: relative;">(0000)
    x1​⎝ ⎛​12−14​⎠ ⎞​+x2​⎝ ⎛​0−1−53​⎠ ⎞​+x3​⎝ ⎛​2535​⎠ ⎞​=⎝ ⎛​0000​⎠ ⎞​
    按分量写出
    { x 1 + 2 x 3 = 0 2 x 1 − x 2 + 5 x 3 = 0 − x 1 − 5 x 2 + 3 x 3 = 0 4 x 1 + 3 x 2 + 5 x 2 = 0
    {x1+2x3=02x1−x2+5x3=0−x1−5x2+3x3=04x1+3x2+5x2=0" role="presentation" style="position: relative;">{x1+2x3=02x1−x2+5x3=0−x1−5x2+3x3=04x1+3x2+5x2=0
    ⎩ ⎨ ⎧​x1​+2x3​=02x1​−x2​+5x3​=0−x1​−5x2​+3x3​=04x1​+3x2​+5x2​=0​
    写出系数矩阵
    ( 1 0 2 2 − 1 5 − 1 − 5 3 4 3 5 ) → ( 1 0 2 0 1 − 1 0 0 0 0 0 0 )
    (1022−15−1−53435)" role="presentation" style="position: relative;">(1022−15−1−53435)
    \rightarrow
    (10201−1000000)" role="presentation" style="position: relative;">(10201−1000000)
    ⎝ ⎛​12−14​0−1−53​2535​⎠ ⎞​→⎝ ⎛​1000​0100​2−100​⎠ ⎞​
    同解方程组
    { x 1 + 2 x 3 = 0 x 2 − x 3 = 0
    {x1+2x3=0x2−x3=0" role="presentation" style="position: relative;">{x1+2x3=0x2−x3=0
    {x1​+2x3​=0x2​−x3​=0​
    有非零解

    定理：向量组 α 1 , α 2 , ⋯   , α m \alpha_{1},\alpha_{2},\cdots,\alpha_{m} α1​,α2​,⋯,αm​线性相关
    ⇔ \Leftrightarrow ⇔存在不全为 0 0 0的 k 1 , k 2 , ⋯   , k m k_{1},k_{2},\cdots,k_{m} k1​,k2​,⋯,km​，使
    k 1 α 1 + k 2 α 2 + ⋯ + k m α m = 0 k_{1}\alpha_{1}+k_{2}\alpha_{2}+\cdots+k_{m}\alpha_{m}=0 k1​α1​+k2​α2​+⋯+km​αm​=0
    ⇔ \Leftrightarrow ⇔存在不全为 0 0 0的 k 1 , k 2 , ⋯   , k m k_{1},k_{2},\cdots,k_{m} k1​,k2​,⋯,km​，使
    ( α 1 α 2 ⋯ α m ) ( k 1 k 2 ⋮ k m ) = 0

    (α1α2⋯αm)" role="presentation" style="position: relative;">(α1α2⋯αm)
    (k1k2⋮km)" role="presentation" style="position: relative;">(k1k2⋮km)
    =0 (α1​​α2​​⋯​αm​​)⎝ ⎛​k1​k2​⋮km​​⎠ ⎞​=0
    ⇔ \Leftrightarrow ⇔齐次方程组有非零解
    ( α 1 α 2 ⋯ α m ) ( x 1 x 2 ⋮ x m ) = 0
    (α1α2⋯αm)" role="presentation" style="position: relative;">(α1α2⋯αm)
    (x1x2⋮xm)" role="presentation" style="position: relative;">(x1x2⋮xm)
    =0     (α1​​α2​​⋯​αm​​)⎝ ⎛​x1​x2​⋮xm​​⎠ ⎞​=0
    ⇔ r ( α 1 α 2 ⋯ α m ) < m \Leftrightarrow r
    (α1α2⋯αm)" role="presentation" style="position: relative;">(α1α2⋯αm)
    ⇔r(α1​​α2​​⋯​αm​​)<m    ， m m m为未知数的个数

    推论：

    1. n n n个 n n n维向量 α 1 , α 2 , ⋯   , α m \alpha_{1},\alpha_{2},\cdots,\alpha_{m} α1​,α2​,⋯,αm​线性相关 ⇔ ∣ α 1 α 2 ⋯ α m ∣ = 0 \Leftrightarrow
      |α1α2⋯αm|" role="presentation" style="position: relative;">|α1α2⋯αm|
      =0       ⇔∣ ∣​α1​​α2​​⋯​αm​​∣ ∣​=0
    2. n + 1 n+1 n+1个 n n n维向量必线性相关

    例： A = ( 1 2 − 2 2 1 2 3 0 4 ) , α = ( a 1 1 ) A=

    (12−2212304)" role="presentation" style="position: relative;">(12−2212304)
    ,\alpha=
    (a11)" role="presentation" style="position: relative;">(a11)
    A=⎝ ⎛​123​210​−224​⎠ ⎞​,α=⎝ ⎛​a11​⎠ ⎞​，已知 A α , α A \alpha,\alpha Aα,α线性相关，则 a = ( ) a=() a=()

    A α = ( a 2 a + 3 3 a + 4 )

    Aα=(a2a+33a+4)" role="presentation" style="position: relative;">Aα=(a2a+33a+4)
    Aα=⎝ ⎛​a2a+33a+4​⎠ ⎞​​
    A α A \alpha Aα与 α \alpha α线性相关
    a a = 2 a + 3 1 = 3 a + 4 1 \frac{a}{a}= \frac{2a+3}{1}=\frac{3a+4}{1} aa​=12a+3​=13a+4​
    因此 a = − 1 a=-1 a=−1

    例： A = ( α 1 α 2 ⋯ α n ) , B = ( β 1 β 2 ⋯ β n ) , A B = ( γ 1 γ 2 ⋯ γ n ) A=

    (α1α2⋯αn)" role="presentation" style="position: relative;">(α1α2⋯αn)
    ,B=
    (β1β2⋯βn)" role="presentation" style="position: relative;">(β1β2⋯βn)
    ,AB=
    (γ1γ2⋯γn)" role="presentation" style="position: relative;">(γ1γ2⋯γn)
    A=(α1​​α2​​⋯​αn​​),B=(β1​​β2​​⋯​βn​​),AB=(γ1​​γ2​​⋯​γn​​)均为 n n n阶矩阵，记向量组 ( I ) α 1 , α 2 , ⋯   , α n ; ( II ) β 1 , β 2 , ⋯   , β n ; ( III ) γ 1 , γ 2 , ⋯   , γ n (\text{I})\alpha_{1},\alpha_{2},\cdots,\alpha_{n};(\text{II})\beta_{1},\beta_{2},\cdots,\beta_{n};(\text{III})\gamma_{1},\gamma_{2},\cdots,\gamma_{n} (I)α1​,α2​,⋯,αn​;(II)β1​,β2​,⋯,βn​;(III)γ1​,γ2​,⋯,γn​，若向量组 ( III ) (\text{III}) (III)线性相关，证明 ( I ) , ( I I ) (I),(II) (I),(II)中至少有一个线性相关

    符合 n n n个 n n n维向量，则有
    ( III ) 线性相关 ⇔ ∣ A B ∣ = 0 ∣ A ∣ ⋅ ∣ B ∣ = 0 ∣ A ∣ = 0 或 ∣ B ∣ = 0

    (III)线性相关⇔|AB|=0|A|⋅|B|=0|A|=0或|B|=0" role="presentation">(III)线性相关⇔|AB|=0|A|⋅|B|=0|A|=0或|B|=0
    (III)线性相关⇔∣AB∣∣A∣⋅∣B∣∣A∣=0​=0=0或∣B∣=0​
    得证

    定理：如果 n n n维向量 α 1 , α 2 , ⋯   , α n \alpha_{1},\alpha_{2},\cdots,\alpha_{n} α1​,α2​,⋯,αn​线性无关， α 1 , α 2 , ⋯   , α n , β \alpha_{1},\alpha_{2},\cdots,\alpha_{n},\beta α1​,α2​,⋯,αn​,β线性相关，则向量 β \beta β可由 α 1 , α 2 , ⋯   , α n \alpha_{1},\alpha_{2},\cdots,\alpha_{n} α1​,α2​,⋯,αn​线性表出，且表示方法唯一
    证明：因为 α 1 , α 2 , ⋯   , α n , β \alpha_{1},\alpha_{2},\cdots,\alpha_{n},\beta α1​,α2​,⋯,αn​,β线性相关，故存在不全为 0 0 0的 k 1 , k 2 , ⋯   , k n , k k_{1},k_{2},\cdots,k_{n},k k1​,k2​,⋯,kn​,k，使得
    k 1 α 1 + k 2 α 2 + ⋯ + k n α n + k β = 0 (1) k_{1}\alpha_{1}+k_{2}\alpha_{2}+\cdots+k_{n}\alpha_{n}+k \beta=0\tag1 k1​α1​+k2​α2​+⋯+kn​αn​+kβ=0(1)
    （反证）如果 k = 0 k=0 k=0，则 k 1 , k 2 , ⋯   , k n k_{1},k_{2},\cdots,k_{n} k1​,k2​,⋯,kn​不全为 0 0 0，而
    k 1 α 1 + k 2 α 2 + ⋯ + k n α n = 0 k_{1}\alpha_{1}+k_{2}\alpha_{2}+\cdots+k_{n}\alpha_{n}=0 k1​α1​+k2​α2​+⋯+kn​αn​=0
    与条件 α 1 , α 2 , ⋯   , α n \alpha_{1},\alpha_{2},\cdots,\alpha_{n} α1​,α2​,⋯,αn​线性无关矛盾，从而 k ≠ 0 k\ne0 k=0，由 ( 1 ) (1) (1)得
    β = − k 1 k α 1 − k 2 k α 2 − ⋯ − k n k α n \beta=- \frac{k_{1}}{k}\alpha_{1}- \frac{k_{2}}{k}\alpha_{2}-\cdots- \frac{k_{n}}{k}\alpha_{n} β=−kk1​​α1​−kk2​​α2​−⋯−kkn​​αn​
    即 β \beta β一定能由 α 1 , α 2 , ⋯   , α n \alpha_{1},\alpha_{2},\cdots,\alpha_{n} α1​,α2​,⋯,αn​线性表出
    （反证）如果 β \beta β有两种不同的表示方法，设
    β = x 1 α 1 + x 2 α 2 + ⋯ + x n α n = y 1 α 1 + y 2 α 2 + ⋯ + y n α n

    β=x1α1+x2α2+⋯+xnαn=y1α1+y2α2+⋯+ynαn" role="presentation" style="position: relative;">β=x1α1+x2α2+⋯+xnαn=y1α1+y2α2+⋯+ynαn
    β​=x1​α1​+x2​α2​+⋯+xn​αn​=y1​α1​+y2​α2​+⋯+yn​αn​​
    两式相减
    ( x 1 − y 1 ) α 1 + ( x 2 − y 2 ) α 2 + ⋯ + ( x n − y n ) α n = 0 (x_{1}-y_{1})\alpha_{1}+(x_{2}-y_{2})\alpha_{2}+\cdots+(x_{n}-y_{n})\alpha_{n}=0 (x1​−y1​)α1​+(x2​−y2​)α2​+⋯+(xn​−yn​)αn​=0
    因有两种不同的表示 x 1 − y 1 , x 2 − y 2 , ⋯   , x n − y n x_{1}-y_{1},x_{2}-y_{2},\cdots,x_{n}-y_{n} x1​−y1​,x2​−y2​,⋯,xn​−yn​不全为 0 0 0与 α 1 , α 2 , ⋯   , α n \alpha_{1},\alpha_{2},\cdots,\alpha_{n} α1​,α2​,⋯,αn​线性无关相矛盾，从而 β \beta β的表示法唯一

    定理：向量组 α 1 , α 2 , ⋯   , α s ( s ≥ 2 ) \alpha_{1},\alpha_{2},\cdots,\alpha_{s}(s\geq2) α1​,α2​,⋯,αs​(s≥2)线性相关 ⇔ \Leftrightarrow ⇔存在 a i a_{i} ai​可由其余的向量线性表出

    必要性
    如 α 1 , α 2 , ⋯   , α s \alpha_{1},\alpha_{2},\cdots,\alpha_{s} α1​,α2​,⋯,αs​线性相关，则存在不全为 0 0 0的 k 1 , k 2 , ⋯   , k s k_{1},k_{2},\cdots,k_{s} k1​,k2​,⋯,ks​
    k 1 α 1 + k 2 α 2 + ⋯ + k s α s = 0 k_{1}\alpha_{1}+k_{2}\alpha_{2}+\cdots+k_{s}\alpha_{s}=0 k1​α1​+k2​α2​+⋯+ks​αs​=0
    不妨设 k 1 ≠ 0 k_{1}\ne0 k1​=0，则有
    k 1 α 1 = − k 2 α 2 − ⋯ − k s α s k_{1}\alpha_{1}=-k_{2}\alpha_{2}-\cdots-k_{s}\alpha_{s} k1​α1​=−k2​α2​−⋯−ks​αs​
    于是
    α 1 = − k 2 k 1 α 2 − ⋯ − k s k 1 α s \alpha_{1}=- \frac{k_{2}}{k_{1}}\alpha_{2}-\cdots- \frac{k_{s}}{k_{1}}\alpha_{s} α1​=−k1​k2​​α2​−⋯−k1​ks​​αs​
    充分性
    如果 α i \alpha_{i} αi​可由 α 1 , ⋯   , α i − 1 , α i + 1 , ⋯   , α s \alpha_{1},\cdots, \alpha_{i-1},\alpha_{i+1},\cdots ,\alpha_{s} α1​,⋯,αi−1​,αi+1​,⋯,αs​线性表出，设
    α i = k 1 α 1 + ⋯ + k i − 1 α i − 1 + k i + 1 α i + 1 + ⋯ + k s α s \alpha_{i}=k_{1}\alpha_{1}+\cdots+k_{i-1}\alpha_{i-1}+k_{i+1}\alpha_{i+1}+\cdots+k_{s}\alpha_{s} αi​=k1​α1​+⋯+ki−1​αi−1​+ki+1​αi+1​+⋯+ks​αs​
    即有
    k 1 α 1 + ⋯ + k i − 1 α i − 1 − α i + k i + 1 α i + 1 + ⋯ + k s α s = 0 k_{1}\alpha_{1}+\cdots+k_{i-1}\alpha_{i-1}-\alpha_{i}+k_{i+1}\alpha_{i+1}+\cdots+k_{s}\alpha_{s}=0 k1​α1​+⋯+ki−1​αi−1​−αi​+ki+1​αi+1​+⋯+ks​αs​=0
    组合系数
    k 1 , ⋯   , k i − 1 , − 1 , k i + 1 , ⋯   , k s k_{1},\cdots,k_{i-1},-1,k_{i+1},\cdots,k_{s} k1​,⋯,ki−1​,−1,ki+1​,⋯,ks​
    不全为 0 0 0

    定理：如果 α 1 , α 2 , ⋯   , α s \alpha_{1},\alpha_{2},\cdots,\alpha_{s} α1​,α2​,⋯,αs​可由 β 1 , β 2 , ⋯   , β t \beta_{1},\beta_{2},\cdots,\beta_{t} β1​,β2​,⋯,βt​线性表出，且 s > t s>t s>t，则 α 1 , α 2 , ⋯   , α s \alpha_{1},\alpha_{2},\cdots,\alpha_{s} α1​,α2​,⋯,αs​必然线性相关
    即多数向量能够用少数向量表示，则多数向量一定线性相关
    推论：如果 α 1 , α 2 , ⋯   , α s \alpha_{1},\alpha_{2},\cdots,\alpha_{s} α1​,α2​,⋯,αs​线性无关，且 α 1 , α 2 , ⋯   , α s \alpha_{1},\alpha_{2},\cdots,\alpha_{s} α1​,α2​,⋯,αs​可由 β 1 , β 2 , ⋯   , β t \beta_{1},\beta_{2},\cdots,\beta_{t} β1​,β2​,⋯,βt​线性表出，则 s ≤ t s\leq t s≤t

    简单的线性无关证明题思路
    当 k 1 α 1 + k 2 α 2 + ⋯ + k s α s = 0 k_{1}\alpha_{1}+k_{2}\alpha_{2}+\cdots+k_{s}\alpha_{s}=0 k1​α1​+k2​α2​+⋯+ks​αs​=0时，必有 k 1 = 0 , k 2 = 0 , ⋯   , k s = 0 k_{1}=0,k_{2}=0,\cdots,k_{s}=0 k1​=0,k2​=0,⋯,ks​=0，则称向量组 α 1 , α 2 , ⋯   , α s \alpha_{1},\alpha_{2},\cdots,\alpha_{s} α1​,α2​,⋯,αs​线性无关

    例：已知 A A A为 n n n阶可逆矩阵， α 1 , α 2 , α 3 \alpha_{1},\alpha_{2},\alpha_{3} α1​,α2​,α3​是 n n n维线性无关向量，证明 A α 1 , A α 2 , A α 3 A \alpha_{1},A \alpha_{2},A \alpha_{3} Aα1​,Aα2​,Aα3​线性无关

    设
    k 1 A α 1 + k 2 A α 2 + k 3 A α 3 = 0 A ( k 1 α 1 + k 2 α 2 + k 3 α 3 ) = 0 k 1 α 1 + k 2 α 2 + k 3 α 3 = A − 1 ⋅ 0 k 1 α 1 + k 2 α 2 + k 3 α 3 = 0

    k1Aα1+k2Aα2+k3Aα3=0A(k1α1+k2α2+k3α3)=0k1α1+k2α2+k3α3=A−1⋅0k1α1+k2α2+k3α3=0" role="presentation" style="position: relative;">k1Aα1+k2Aα2+k3Aα3=0A(k1α1+k2α2+k3α3)=0k1α1+k2α2+k3α3=A−1⋅0k1α1+k2α2+k3α3=0
    k1​Aα1​+k2​Aα2​+k3​Aα3​A(k1​α1​+k2​α2​+k3​α3​)k1​α1​+k2​α2​+k3​α3​k1​α1​+k2​α2​+k3​α3​​=0=0=A−1⋅0=0​
    由于 α 1 , α 2 , α 3 \alpha_{1},\alpha_{2},\alpha_{3} α1​,α2​,α3​是 n n n维线性无关向量，则
    k 1 = 0 , k 2 = 0 , k 3 = 0 k_{1}=0,k_{2}=0,k_{3}=0 k1​=0,k2​=0,k3​=0
    因此， A α 1 , A α 2 , A α 3 A \alpha_{1},A \alpha_{2},A \alpha_{3} Aα1​,Aα2​,Aα3​线性无关

    例：已知 α 1 , α 2 , α 3 \alpha_{1},\alpha_{2},\alpha_{3} α1​,α2​,α3​线性无关，证明 α 1 + α 2 , α 2 + α 3 , α 3 + α 1 \alpha_{1}+\alpha_{2},\alpha_{2}+\alpha_{3},\alpha_{3}+\alpha_{1} α1​+α2​,α2​+α3​,α3​+α1​线性无关

    设
    k 1 ( α 1 + α 2 ) + k 2 ( α 2 + α 3 ) + k 3 ( α 3 + α 1 ) = 0 ( k 1 + k 3 ) α 1 + ( k 1 + k 2 ) α 2 + ( k 2 + k 3 ) α 3 = 0

    k1(α1+α2)+k2(α2+α3)+k3(α3+α1)=0(k1+k3)α1+(k1+k2)α2+(k2+k3)α3=0" role="presentation" style="position: relative;">k1(α1+α2)+k2(α2+α3)+k3(α3+α1)=0(k1+k3)α1+(k1+k2)α2+(k2+k3)α3=0
    k1​(α1​+α2​)+k2​(α2​+α3​)+k3​(α3​+α1​)(k1​+k3​)α1​+(k1​+k2​)α2​+(k2​+k3​)α3​​=0=0​
    因为 α 1 , α 2 , α 3 \alpha_{1},\alpha_{2},\alpha_{3} α1​,α2​,α3​线性无关
    { k 1 + k 3 = 0 k 1 + k 2 = 0 k 2 + k 3 = 0 (1)
    {k1+k3=0k1+k2=0k2+k3=0" role="presentation" style="position: relative;">{k1+k3=0k1+k2=0k2+k3=0
    \tag{1}     ⎩ ⎨ ⎧​k1​+k3​=0k1​+k2​=0k2​+k3​=0​(1)
    由
    ∣ 1 0 1 1 1 0 0 1 1 ∣ = 2 ≠ 0
    |101110011|" role="presentation" style="position: relative;">|101110011|
    =2\ne0     ∣ ∣​110​011​101​∣ ∣​=2=0
    齐次方程组 ( 1 ) (1) (1)只有 0 0 0解，即必有 k 1 = 0 , k 2 = 0 , k 3 = 0 k_{1}=0,k_{2}=0,k_{3}=0 k1​=0,k2​=0,k3​=0，因此 α 1 + α 2 , α 2 + α 3 , α 3 + α 1 \alpha_{1}+\alpha_{2},\alpha_{2}+\alpha_{3},\alpha_{3}+\alpha_{1} α1​+α2​,α2​+α3​,α3​+α1​线性无关

    秩

    极大线性无关组

    向量组 α i 1 , α i 2 , ⋯   , α i r ( i ≤ i r ) \alpha_{i_{1}},\alpha_{i_{2}},\cdots,\alpha_{i_{r}}(i\leq i_{r}) αi1​​,αi2​​,⋯,αir​​(i≤ir​)是向量组 α 1 , α 2 , ⋯   , α s \alpha_{1},\alpha_{2},\cdots,\alpha_{s} α1​,α2​,⋯,αs​的部分组，且满足

    • α i 1 , α i 2 , ⋯   , α i r \alpha_{i_{1}},\alpha_{i_{2}},\cdots,\alpha_{i_{r}} αi1​​,αi2​​,⋯,αir​​线性无关
    • 向量组中的任一一个向量 a i ( i ≤ i ≤ s ) a_{i}(i\leq i\leq s) ai​(i≤i≤s)均可由 α i 1 , α i 2 , ⋯   , α i r \alpha_{i_{1}},\alpha_{i_{2}},\cdots,\alpha_{i_{r}} αi1​​,αi2​​,⋯,αir​​线性表出

    则称 α i 1 , α i 2 , ⋯   , α i r \alpha_{i_{1}},\alpha_{i_{2}},\cdots,\alpha_{i_{r}} αi1​​,αi2​​,⋯,αir​​是向量组 α 1 , α 2 , ⋯   , α s \alpha_{1},\alpha_{2},\cdots,\alpha_{s} α1​,α2​,⋯,αs​的一个极大线性无关组

    同一向量组可以有多个极大线性无关组，其中的成员组成不一样，数量一定一样

    定理：如果 α i 1 , α i 2 , ⋯   , α i r \alpha_{i_{1}},\alpha_{i_{2}},\cdots,\alpha_{i_{r}} αi1​​,αi2​​,⋯,αir​​与 α j 1 , α j 2 , ⋯   , α j t \alpha_{j_{1}},\alpha_{j_{2}},\cdots,\alpha_{j_{t}} αj1​​,αj2​​,⋯,αjt​​都是向量组 α 1 , α 2 , ⋯   , α s \alpha_{1},\alpha_{2},\cdots,\alpha_{s} α1​,α2​,⋯,αs​的极大线性无关组，则 r = t r=t r=t
    证明：
    因为 α i 1 , α i 2 , ⋯   , α i r \alpha_{i_{1}},\alpha_{i_{2}},\cdots,\alpha_{i_{r}} αi1​​,αi2​​,⋯,αir​​是 α 1 , α 2 , ⋯   , α s \alpha_{1},\alpha_{2},\cdots,\alpha_{s} α1​,α2​,⋯,αs​的极大线性无关组，那么 α j 1 , α j 2 , ⋯   , α j t \alpha_{j_{1}},\alpha_{j_{2}},\cdots,\alpha_{j_{t}} αj1​​,αj2​​,⋯,αjt​​可由 α i 1 , α i 2 , ⋯   , α i r \alpha_{i_{1}},\alpha_{i_{2}},\cdots,\alpha_{i_{r}} αi1​​,αi2​​,⋯,αir​​线性表示
    又因为 α j 1 , α j 2 , ⋯   , α j t \alpha_{j_{1}},\alpha_{j_{2}},\cdots,\alpha_{j_{t}} αj1​​,αj2​​,⋯,αjt​​线性无关，则有 t ≤ r t\leq r t≤r
    同理 r ≤ t r\leq t r≤t，故有 r = t r=t r=t

    例：已知向量组 α 1 = ( 1 , − 1 , 0 , 5 ) T , α 2 = ( 2 , 0 , 1 , 4 ) T , α 3 = ( 3 , 1 , 2 , 3 ) T , α 4 = ( 4 , 2 , 3 , a ) T \alpha_{1}=(1,-1,0,5)^{T},\alpha_{2}=(2,0,1,4)^{T},\alpha_{3}=(3,1,2,3)^{T},\alpha_{4}=(4,2,3,a)^{T} α1​=(1,−1,0,5)T,α2​=(2,0,1,4)T,α3​=(3,1,2,3)T,α4​=(4,2,3,a)T，其中 a a a是参数，求向量组的秩与一个极大线性无关组，并将其他向量用该极大线性无关组线性表示

    经初等行变换
    ( α 1 α 2 α 3 α 4 ) = ( 1 2 3 4 − 1 0 1 2 0 1 2 3 5 4 3 a ) → ( 1 0 − 1 − 2 0 1 2 3 0 0 0 a − 2 0 0 0 0 )

    (α1α2α3α4)" role="presentation" style="position: relative;">(α1α2α3α4)
    =
    (1234−10120123543a)" role="presentation" style="position: relative;">(1234−10120123543a)
    \rightarrow
    (10−1−20123000a−20000)" role="presentation" style="position: relative;">(10−1−20123000a−20000)
    (α1​​α2​​α3​​α4​​)=⎝ ⎛​1−105​2014​3123​423a​⎠ ⎞​→⎝ ⎛​1000​0100​−1200​−23a−20​⎠ ⎞​
    当 a = 2 a=2 a=2时，秩 r ( α 1 α 2 α 3 α 4 ) = 2 r
    (α1α2α3α4)" role="presentation" style="position: relative;">(α1α2α3α4)
    =2     r(α1​​α2​​α3​​α4​​)=2    ，极大线性无关组是 α 1 , α 2 \alpha_{1},\alpha_{2} α1​,α2​。 α 3 = − α 1 + 2 α 2 , α 4 = − 2 α 1 + 3 α 2 \alpha_{3}=-\alpha_{1}+2\alpha_{2},\alpha_{4}=-2\alpha_{1}+3\alpha_{2} α3​=−α1​+2α2​,α4​=−2α1​+3α2​
    （行最简形式中第一二列、一二行行列式不为 0 0 0；同理可以使 α 1 , α 3 \alpha_{1},\alpha_{3} α1​,α3​，即第一三列、一二行行列式不为 0 0 0）

    一般选择主元，即行最简每行主元所在的列，为了便于用极大线性无关组表示其他向量

    当 α ≠ 2 \alpha\ne2 α=2，秩 r ( α 1 α 2 α 3 α 4 ) = 3 r

    (α1α2α3α4)" role="presentation" style="position: relative;">(α1α2α3α4)
    =3 r(α1​​α2​​α3​​α4​​)=3，极大线性无关组是 α 1 , α 2 , α 4 \alpha_{1},\alpha_{2},\alpha_{4} α1​,α2​,α4​。 α 3 = − α 1 + 2 α 2 \alpha_{3}=-\alpha_{1}+2\alpha_{2} α3​=−α1​+2α2​

    矩阵的秩

    k k k阶子式： A A A为 m × n m\times n m×n的矩阵，任取 k k k行与 k k k列 ( k ≤ m , k ≤ n ) (k\leq m,k\leq n) (k≤m,k≤n)位于交叉点的 k 2 k^{2} k2元素，按 A A A中的位置次序而得到的 k k k阶行列式，称为矩阵 A A A的 k k k阶子式

    秩：矩阵 A A A中非 0 0 0子式的最高阶数称为矩阵 A A A的秩，记为 r ( A ) r(A) r(A)
    r ( A ) = r ⇔ A r(A)=r\Leftrightarrow A r(A)=r⇔A中有 r r r阶子式不为 0 0 0而所有 r + 1 r+1 r+1阶子式（若有）全为 0 0 0
    r ( A ) < r ⇔ A r(A)r(A)<r⇔A中 r r r阶子式全为 0 0 0
    r ( A ) ≥ r ⇔ A r(A)\geq r\Leftrightarrow A r(A)≥r⇔A中 r r r阶子式全不为 0 0 0
    A ≠ 0 ⇔ r ( A ) ≥ 1 A\ne0\Leftrightarrow r(A)\geq1 A=0⇔r(A)≥1
    A 为 n 阶矩阵 , r ( A ) = n ⇔ ∣ A ∣ ≠ 0 ⇔ A A为n阶矩阵,r(A)=n\Leftrightarrow|A|\ne0\Leftrightarrow A A为n阶矩阵,r(A)=n⇔∣A∣=0⇔A可逆

    r ( 1 2 − 1 0 3 0 0 5 1 0 0 0 0 0 6 0 0 0 0 1 ) = 3 r

    (12−103005100000600001)" role="presentation" style="position: relative;">(12−103005100000600001)
    =3 r⎝ ⎛​1000​2000​−1500​0100​3061​⎠ ⎞​=3，虽然第四行不全为 0 0 0，但是所有四阶子式都为 0 0 0

    公式

    • r ( A T ) = r ( A ) r(A^{T})=r(A) r(AT)=r(A)
    • r ( k A ) = r ( A ) , k ≠ 0 r(kA)=r(A),k\ne0 r(kA)=r(A),k=0
      r ( 0 E − A ) = r ( A ) r(0E-A)=r(A) r(0E−A)=r(A)
      r ( A − E ) = r ( E − A ) r(A-E)=r(E-A) r(A−E)=r(E−A)
    • r ( A + B ) ≤ r ( A ) + r ( B ) r(A+B)\leq r(A)+r(B) r(A+B)≤r(A)+r(B)
    • r ( A B ) ≤ m i n { r ( A ) , r ( B ) } r(AB)\leq min\{r(A),r(B)\} r(AB)≤min{r(A),r(B)}
      若 A A A可逆，则 r ( A B ) = r ( B ) , r ( B A ) = r ( B ) r(AB)=r(B),r(BA)=r(B) r(AB)=r(B),r(BA)=r(B)
    • r ( A T A ) = r ( A ) r(A^{T}A)=r(A) r(ATA)=r(A)
    • A − m × n , B − n × s A-m\times n,B-n\times s A−m×n,B−n×s且 A B = 0 AB=0 AB=0，则 r ( A ) + r ( B ) ≤ n r(A)+r(B)\leq n r(A)+r(B)≤n
    • r ( A O O B ) = r ( A ) + r ( B ) r
      (AOOB)" role="presentation" style="position: relative;">(AOOB)
      =r(A)+r(B)       r(AO​OB​)=r(A)+r(B)
    • 若 A ∼ B A\sim B A∼B，则 r ( A ) = r ( B ) , r ( A + k E ) = r ( B + k E ) r(A)=r(B),r(A+kE)=r(B+kE) r(A)=r(B),r(A+kE)=r(B+kE)

    例：已知 r ( A ) = 3 , A = ( 1 1 1 1 0 1 − 1 b 2 3 a 4 3 5 1 7 ) r(A)=3,A=

    (111101−1b23a43517)" role="presentation" style="position: relative;">(111101−1b23a43517)
    r(A)=3,A=⎝ ⎛​1023​1135​1−1a1​1b47​⎠ ⎞​，求 a , b a,b a,b

    对矩阵 A A A作初等变换
    A = ( 1 1 1 1 0 1 − 1 b 2 3 a 4 3 5 1 7 ) → ( 1 1 1 1 0 1 − 1 b 0 0 a − 1 2 − b 0 0 0 4 − 2 b ) A=

    (111101−1b23a43517)" role="presentation" style="position: relative;">(111101−1b23a43517)
    \rightarrow
    (111101−1b00a−12−b0004−2b)" role="presentation" style="position: relative;">(111101−1b00a−12−b0004−2b)
    A=⎝ ⎛​1023​1135​1−1a1​1b47​⎠ ⎞​→⎝ ⎛​1000​1100​1−1a−10​1b2−b4−2b​⎠ ⎞​
    有
    { a − 1 = 0 4 − 2 b ≠ 0 或 { a − 1 ≠ 0 4 − 2 b = 0
    {a−1=04−2b≠0" role="presentation" style="position: relative;">{a−1=04−2b≠0
    或
    {a−1≠04−2b=0" role="presentation" style="position: relative;">{a−1≠04−2b=0
    {a−1=04−2b=0​或{a−1=04−2b=0​
    则 a ≠ 1 , b = 2 a\ne1,b=2 a=1,b=2或 a = 1 , b ≠ 2 a=1,b\ne2 a=1,b=2

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  • 原文地址：https://blog.csdn.net/liu20020918zz/article/details/126462401