【线性代数基础进阶】向量

烧灯续昼2002
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nn维向量

nn个数a1,a2,⋯ ,ana_{1},a_{2},\cdots,a_{n}构成的有序数组称为nn维向量

 

(a1a2⋮an)\begin{pmatrix}a_{1} \\ a_{2} \\ \vdots \\ a_{n}\end{pmatrix}或(a1a2⋯an)T\begin{pmatrix}a_{1} & a_{2} & \cdots & a_{n}\end{pmatrix}^{T}列向量

(a1a2⋯an)\begin{pmatrix}a_{1} & a_{2} & \cdots & a_{n}\end{pmatrix}行向量

其中aia_{i}称为向量的第ii个分量(i=1,2,⋯ ,n)(i=1,2,\cdots,n)

 

如果向量的所有分量都是00，就称其为零向量，记作O=(0,0,⋯ ,0)TO=(0,0,\cdots,0)^{T}

 

设nn维向量α=(a1,a2,⋯ ,an)T,β=(b1,b2,⋯ ,bn)T\alpha=(a_{1},a_{2},\cdots,a_{n})^{T},\beta=(b_{1},b_{2},\cdots,b_{n})^{T}

  • α=β⇔a1=b1,a2=b2,⋯ ,an=bn\alpha=\beta\Leftrightarrow a_{1}=b_{1},a_{2}=b_{2},\cdots,a_{n}=b_{n}

  • α+β=(a1+b1,a2+b2,⋯ ,an+bn)T\alpha+\beta=(a_{1}+b_{1},a_{2}+b_{2},\cdots,a_{n}+b_{n})^{T}

  • kα=(ka1,ka2,⋯ ,kan)Tk \alpha=(ka_{1},ka_{2},\cdots,ka_{n})^{T}

         特别的，0α=(0,0,⋯ ,0)T0\alpha=(0,0,\cdots,0)^{T}

 

运算规律

α+β=β+α\alpha+\beta=\beta+\alpha

(α+β)+γ=α+(β+γ)(\alpha+\beta)+\gamma=\alpha+(\beta+\gamma)

α+0=α\alpha+0=\alpha

α+(−α)=0\alpha+(-\alpha)=0

1⋅α=α1\cdot \alpha=\alpha

k(lα)=(kl)αk(l \alpha)=(kl)\alpha

k(α+β)=kα+kβk(\alpha+\beta)=k \alpha+k \beta

(k+l)α=kα+lα(k+l)\alpha=k \alpha+l \alpha

 

解方程组

例：{x1+x2+x3=53x1+2x2+x3=15x2+2x3=2\begin{cases}x_{1}+x_{2}+x_{3}=5 \\3x_{1}+2x_{2}+x_{3}=15\\x_{2}+2x_{3}=2\end{cases}

 

对增广矩阵作初等行变换

Aˉ=(1115321 130122)→(11150−1−2−20122)→(111501220000)→(10−1301220000) \begin{aligned} \bar{A}&=\begin{pmatrix} 1 & 1 & 1 & 5 \\ 3 & 2 & 1  & 13 \\ 0 & 1 & 2 & 2 \end{pmatrix}\rightarrow \begin{pmatrix} 1 & 1 & 1 & 5 \\ 0 & -1 & -2 & -2 \\ 0 & 1 & 2 & 2 \end{pmatrix}\\ &\rightarrow \begin{pmatrix} 1 & 1 & 1 & 5 \\ 0 & 1 & 2 & 2 \\ 0 & 0 & 0 & 0 \end{pmatrix}\rightarrow \begin{pmatrix} 1 & 0 & -1 & 3 \\ 0 & 1 & 2 & 2 \\ 0 & 0 & 0 & 0 \end{pmatrix} \end{aligned}

同解方程组为

{x1−x3=3x2+2x3=2 \begin{cases} x_{1}-x_{3}=3 \\ x_{2}+2x_{3}=2 \end{cases}

∀x3=t\forall x_{3}=t，方程组有∞\infty解

{x1=3+tx2=2−2tx3=t或(x1x2x3)=(320)+t(1−21) \begin{cases} x_{1}=3+t \\ x_{2}=2-2t \\ x_{3}=t \end{cases}或\begin{pmatrix} x_{1} \\ x_{2} \\ x_{3} \end{pmatrix}=\begin{pmatrix} 3 \\ 2 \\ 0 \end{pmatrix}+t \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}

 

令x3=0⇒x1=3,x2=2x_{3}=0\Rightarrow x_{1}=3,x_{2}=2

特解

α=(3,2,0) \alpha=(3,2,0)

令x3=1⇒x1=1,x2=−2x_{3}=1\Rightarrow x_{1}=1,x_{2}=-2

基础解系

η=(1,−2,1)T \eta=(1,-2,1)^{T}

因此x=α+kη=(320)+k(1−21)x=\alpha+k \eta=\begin{pmatrix}3 \\ 2 \\ 0\end{pmatrix}+k \begin{pmatrix}1 \\ -2 \\ 1\end{pmatrix}

 

解方程组只能行变换

 

例：{x1−2x2+x3+2x4=0x1+x2−2x3−4x4=0x1−5x2+4x3+8x4=0\begin{cases}x_{1}-2x_{2}+x_{3}+2x_{4}=0\\x_{1}+x_{2}-2x_{3}-4x_{4}=0\\x_{1}-5x_{2}+4x_{3}+8x_{4}=0\end{cases}

 

对系数矩阵作初等行变换

A=(1−21211−2−41−548)→(10−1−201−1−20000) A=\begin{pmatrix} 1 & -2 & 1 & 2 \\ 1 & 1 & -2 & -4 \\ 1 & -5 & 4 & 8 \end{pmatrix}\rightarrow \begin{pmatrix} 1 & 0 & -1 & -2 \\ 0 & 1 & -1 & -2 \\ 0 & 0 & 0 & 0 \end{pmatrix}

通解方程组

{x1−x3−2x4=0x2−x3−2x4=0 \begin{cases} x_{1}-x_{3}-2x_{4}=0 \\ x_{2}-x_{3}-2x_{4}=0 \end{cases}

其中x3,x4x_{3},x_{4}为自由变量，令x3=t,x4=ux_{3}=t,x_{4}=u

{x1=t+2ux2=t+2ux3=tx4=u或x=t(1110)+u(2201) \begin{cases} x_{1}=t+2u \\ x_{2}=t+2u \\ x_{3}=t \\ x_{4}=u \end{cases}或x=t \begin{pmatrix} 1 \\ 1 \\ 1 \\ 0 \end{pmatrix}+u \begin{pmatrix} 2 \\ 2 \\ 0 \\ 1 \end{pmatrix}

 

根据n−r(A)=4−2=2n-r(A)=4-2=2，即有两个自由变量

令x3=1,x4=0⇒x1=1,x2=1x_{3}=1,x_{4}=0\Rightarrow x_{1}=1,x_{2}=1

η1=(1,1,1,0)T \eta_{1}=(1,1,1,0)^{T}

令x3=0,x4=1⇒x1=2,x2=2x_{3}=0,x_{4}=1\Rightarrow x_{1}=2,x_{2}=2

η2=(2,2,0,1)T \eta_{2}=(2,2,0,1)^{T}

基础解系

x=k1(1110)+k2(2201) x=k_{1}\begin{pmatrix} 1 \\ 1 \\ 1 \\ 0 \end{pmatrix}+k_{2}\begin{pmatrix} 2 \\ 2 \\ 0 \\ 1 \end{pmatrix}

 

线性表出（组合）

线性组合定义：mm个nn维向量α1,α2,cdots,αm\alpha_{1},\alpha_{2},cdots,\alpha_{m}及mm个实数k1,k2,⋯ ,kmk_{1},k_{2},\cdots,k_{m}，称

k1α1+k2α2+⋯+kmαm k_{1}\alpha_{1}+k_{2}\alpha_{2}+\cdots+k_{m}\alpha_{m}

是向量α1,α2,⋯ ,αm\alpha_{1},\alpha_{2},\cdots,\alpha_{m}的一个线性组合，k1,k2,⋯ ,kmk_{1},k_{2},\cdots,k_{m}称为这个线性组合的系数

 

线性表出定义：如果向量β\beta能表示为α1,α2,⋯ ,αm\alpha_{1},\alpha_{2},\cdots,\alpha_{m}的线性组合，即存在一组数k1,k2,⋯ ,kmk_{1},k_{2},\cdots,k_{m}，使

β=k1α1+k2α2+⋯+kmαm \beta=k_{1}\alpha_{1}+k_{2}\alpha_{2}+\cdots+k_{m}\alpha_{m}

则称向量β\beta可以由α1,α2,⋯ ,αm\alpha_{1},\alpha_{2},\cdots,\alpha_{m}线性表出（示）

 

例：设α1=(1,2,3)T,α2=(1,3,4)T,α3=(2,−1,1)T,β=(2,5,t)T\alpha_{1}=(1,2,3)^{T},\alpha_{2}=(1,3,4)^{T},\alpha_{3}=(2,-1,1)^{T},\beta=(2,5,t)^{T}，问tt取何值时

  • 向量β\beta不能由α1,α2,α3\alpha_{1},\alpha_{2},\alpha_{3}线性表示

  • 向量β\beta能由α1,α2,α3\alpha_{1},\alpha_{2},\alpha_{3}线性表示，并且写出表达式

 

设x1α1+x2α2+x3α3=βx_{1}\alpha_{1}+x_{2}\alpha_{2}+x_{3}\alpha_{3}=\beta

即

x1(1,2,3)T+x2(1,3,4)T+x3(2,−1,1)T=(2,5,t)T(x1+x2+2x3,2x1+3x2−x3,3x1+4x2+x3)T=(2,5,t)T{x1+x2+2x3=22x1+3x2−x3=53x1+4x2+x3=t \begin{gathered} x_{1}(1,2,3)^{T}+x_{2}(1,3,4)^{T}+x_{3}(2,-1,1)^{T}=(2,5,t)^{T}\\ (x_{1}+x_{2}+2x_{3},2x_{1}+3x_{2}-x_{3},3x_{1}+4x_{2}+x_{3})^{T}=(2,5,t)^{T}\\ \begin{cases} x_{1}+x_{2}+2x_{3}=2 \\ 2x_{1}+3x_{2}-x_{3}=5 \\ 3x_{1}+4x_{2}+x_{3}=t \end{cases} \end{gathered}

对增广矩阵Aˉ=(α1,α2,α3,β)\bar{A}=(\alpha_{1},\alpha_{2},\alpha_{3},\beta)作初等行变换

Aˉ=(112223−15341t)→(112201−51000t−7) \begin{aligned} \bar{A}=\begin{pmatrix} 1 & 1 & 2 & 2 \\ 2 & 3 & -1 & 5 \\ 3 & 4 & 1 & t \end{pmatrix}\rightarrow \begin{pmatrix} 1 & 1 & 2 & 2 \\ 0 & 1 & -5 & 1 \\ 0 & 0 & 0 & t-7 \end{pmatrix} \end{aligned}

当t≠7t\ne7时，方程组无解，即β\beta不能由α1,α2,α3\alpha_{1},\alpha_{2},\alpha_{3}线性表示

当t=7t=7时，

Aˉ→(112201−510000)→(107101−510000) \bar{A}\rightarrow \begin{pmatrix} 1 & 1 & 2 & 2 \\ 0 & 1 & -5 & 1 \\ 0 & 0 & 0 & 0 \end{pmatrix}\rightarrow \begin{pmatrix} 1 & 0 & 7 & 1 \\ 0 & 1 & -5 & 1 \\ 0 & 0 & 0 & 0 \end{pmatrix}

令x3=k⇒x2=1+5k,x1=1−7kx_{3}=k\Rightarrow x_{2}=1+5k,x_{1}=1-7k

因此β=(1−7k)α1+(1+5k)α2+kα3\beta=(1-7k)\alpha_{1}+(1+5k)\alpha_{2}+k \alpha_{3}

 

定理：向量β\beta可由α1,α2,⋯ ,αm\alpha_{1},\alpha_{2},\cdots,\alpha_{m}线性表出

⇔\Leftrightarrow存在实数k1,k2,⋯ ,kmk_{1},k_{2},\cdots,k_{m}使

k1α1+k2α2+⋯+kmαm=β k_{1}\alpha_{1}+k_{2}\alpha_{2}+\cdots+k_{m}\alpha_{m}=\beta

⇔\Leftrightarrow存在实数k1,k2,⋯ ,kmk_{1},k_{2},\cdots,k_{m}使

(α1α2⋯αm)(k1k2⋮km)=β \begin{pmatrix} \alpha_{1} & \alpha_{2} & \cdots & \alpha_{m} \end{pmatrix}\begin{pmatrix} k_{1} \\ k_{2} \\ \vdots \\ k_{m} \end{pmatrix}=\beta

⇔\Leftrightarrow方程组有解

(α1α2⋯αm)(x1x2⋮xm)=β \begin{pmatrix} \alpha_{1} & \alpha_{2} & \cdots & \alpha_{m} \end{pmatrix}\begin{pmatrix} x_{1} \\ x_{2} \\ \vdots \\ x_{m} \end{pmatrix}=\beta

⇔r(α1α2⋯αm)=r(α1α2⋯αmβ)\Leftrightarrow r\begin{pmatrix}\alpha_{1} & \alpha_{2} & \cdots & \alpha_{m}\end{pmatrix}=r\begin{pmatrix}\alpha_{1} & \alpha_{2} & \cdots & \alpha_{m} & \beta\end{pmatrix}

 

例：设α1=(1+λ,1,1)T,α2=(1,1+λ,1)T,α3=(1,1,1+λ)T,β=(0,λ,λ2)T\alpha_{1}=(1+\lambda,1,1)^{T},\alpha_{2}=(1,1+\lambda,1)^{T},\alpha_{3}=(1,1,1+\lambda)^{T},\beta=(0,\lambda,\lambda^{2})^{T}，当λ\lambda为何值时

  • β\beta不能由α1,α2,α3\alpha_{1},\alpha_{2},\alpha_{3}线性表示

  • β\beta能由α1,α2,α3\alpha_{1},\alpha_{2},\alpha_{3}线性表示，并写出该表达式

 

设x1α1++x2α2+x3α3=βx_{1}\alpha_{1+}+x_{2}\alpha_{2}+x_{3}\alpha_{3}=\beta按分量写出方程组

{(1+λ)x1+x2+x3=0x1+(1+λ)x2+x3=λx1+x2+x(1+λ)x3=λ2 \begin{cases} (1+\lambda)x_1+x_2+x_3=0 \\ x_1+(1+\lambda)x_2+x_3=\lambda\\ x_1+x_2 +x(1+\lambda)x_3=\lambda^2 \end{cases}

 

此处可以用正常处理方法，也可以用类似爪型行列式的思路

由于只是要计算r(A)r(A)和r(A∣β)r(A|\beta)之间的大小关系，所以AA的部分只要是三角矩阵即可，不一定必须是行最简矩阵

 

类似爪型行列式的思路

由

Aˉ=(1+λ11011+λ1λ111+λλ2)→(1+λ110−λλ0λ−λ0λλ2) \bar{A}=\begin{pmatrix}1+\lambda & 1 & 1 & 0 \\ 1 & 1+\lambda & 1 & \lambda \\ 1 & 1 & 1+\lambda & \lambda^{2}\end{pmatrix}\rightarrow \begin{pmatrix} 1+\lambda & 1 & 1 & 0 \\ -\lambda & \lambda & 0 & \lambda \\ -\lambda & 0 & \lambda & \lambda^{2} \end{pmatrix}

若λ=0\lambda=0，同解方程组

x1+x2+x3=0 x_{1}+x_{2}+x_{3}=0

令x2=t,x3=ux_{2}=t,x_{3}=u，得x1=−t−ux_{1}=-t-u

故

β=−(t+u)α1+tα2+uα3,t,u是任意常数 \beta=-(t+u)\alpha_{1}+t \alpha_{2}+u \alpha_{3},t,u是任意常数

若λ≠0\lambda\ne0

Aˉ→(1+λ110−1101−101λ)此处只要化成三角行列式即可,由于只能行变换因此根据爪型行列式考虑化成下三角形式(λ+300−λ−1−1101−101λ) \begin{aligned} \bar{A}&\rightarrow\begin{pmatrix} 1+\lambda & 1 & 1 & 0 \\ -1 & 1 & 0 & 1 \\ -1 & 0 & 1 & \lambda \end{pmatrix}\\ &此处只要化成三角行列式即可,由于只能行变换\\ &因此根据爪型行列式考虑化成下三角形式\\ &\begin{pmatrix} \lambda+3 & 0 & 0 & -\lambda-1\\ -1 & 1 & 0 & 1\\ -1 & 0 & 1 & \lambda \end{pmatrix} \end{aligned}

若λ=−3\lambda=-3，方程组无解，β\beta不能由α1,α2,α3\alpha_{1},\alpha_{2},\alpha_{3}线性表出

若λ≠−3且λ≠0\lambda\ne-3且\lambda\ne0，解出

x1=−λ+1λ+3,x2=2λ+3,x3=λ2+2λ−1λ+3 x_{1}=- \frac{\lambda+1}{\lambda+3},x_{2}=\frac{2}{\lambda+3},x_{3}=\frac{\lambda^{2}+2\lambda-1}{\lambda+3}

故

β=−λ+1λ+3α1+2λ+3α2+λ2+2λ−1λ+3α3 \beta=- \frac{\lambda+1}{\lambda+3}\alpha_{1}+\frac{2}{\lambda+3}\alpha_{2}+\frac{\lambda^{2}+2\lambda-1}{\lambda+3}\alpha_{3}

 

正常行最简矩阵思路

 

 > 为了化成行最简矩阵计算方便，尽量通过交换行，使第一行在化简前都是简单的数字，复杂的尽可能往第一行的后几列放

 

由

Aˉ=(1+λ11011+λ1λ111+λλ2)将13行互换,第一行有更多简单的数字→(111+λλ211+λ1λ1+λ111)正常从上向下化简→(111+λλ20λ−λλ−λ200−λ(λ+3)λ(1−2λ−λ2)) \begin{aligned} \bar{A}&=\begin{pmatrix}1+\lambda & 1 & 1 & 0 \\ 1 & 1+\lambda & 1 & \lambda \\ 1 & 1 & 1+\lambda & \lambda^{2}\end{pmatrix}\\ &将13行互换,第一行有更多简单的数字\\ &\rightarrow\begin{pmatrix} 1 & 1 & 1+\lambda & \lambda^{2} \\ 1 & 1+\lambda & 1 & \lambda \\ 1+\lambda & 1 & 1 & 1 \end{pmatrix}\\ &正常从上向下化简\\ &\rightarrow \begin{pmatrix} 1 & 1 & 1+\lambda & \lambda^{2} \\ 0 & \lambda & -\lambda & \lambda-\lambda^{2} \\ 0 & 0 & -\lambda(\lambda+3) & \lambda(1-2\lambda-\lambda^{2}) \end{pmatrix} \end{aligned}

如果要使β\beta能由α1,α2,α3\alpha_{1},\alpha_{2},\alpha_{3}线性表出且唯一，即r(A)=r(A∣β)r(A)=r(A|\beta)，故

−λ(λ+3)≠0 -\lambda(\lambda+3)\ne0

得λ≠0\lambda\ne0且λ≠−3\lambda\ne-3，解出

x1=−λ+1λ+3,x2=2λ+3,x3=λ2+2λ−1λ+3 x_{1}=- \frac{\lambda+1}{\lambda+3},x_{2}=\frac{2}{\lambda+3},x_{3}=\frac{\lambda^{2}+2\lambda-1}{\lambda+3}

故

β=−λ+1λ+3α1+2λ+3α2+λ2+2λ−1λ+3α3 \beta=- \frac{\lambda+1}{\lambda+3}\alpha_{1}+\frac{2}{\lambda+3}\alpha_{2}+\frac{\lambda^{2}+2\lambda-1}{\lambda+3}\alpha_{3}

如果要使β\beta能由α1,α2,α3\alpha_{1},\alpha_{2},\alpha_{3}线性表出且不唯一，即r(A)=r(A∣β)r(A)=r(A|\beta)，故

{−λ(λ+3)=0λ(1−2λ−λ2)=0 \begin{cases} -\lambda(\lambda+3)=0 \\ \lambda(1-2\lambda-\lambda^{2})=0 \end{cases}

得λ=0\lambda=0

如果要使β\beta不能由α1,α2,α3\alpha_{1},\alpha_{2},\alpha_{3}线性表出，即r(A)≠r(A∣β)r(A)\ne r(A|\beta)，故

{−λ(λ+3)=0λ(1−2λ−λ2)≠0 \begin{cases} -\lambda(\lambda+3)=0 \\ \lambda(1-2\lambda-\lambda^{2})\ne0 \end{cases}

得λ=−3\lambda=-3

 

例：已知α1=(1,2,1)T,α2=(2,3,a)T,β1=(1,3,0)T,β2=(1,a+2,−2)T\alpha_{1}=(1,2,1)^{T},\alpha_{2}=(2,3,a)^{T},\beta_{1}=(1,3,0)^{T},\beta_{2}=(1,a+2,-2)^{T}，β1\beta_{1}不能由α1,α2\alpha_{1},\alpha_{2}表出，β2\beta_{2}能由α1,α2\alpha_{1},\alpha_{2}表出，则a=()a=()

 

由题意得x1α1+x2α2=β1无解,x1α1+x2α2=β2有解x_{1}\alpha_{1}+x_{2}\alpha_{2}=\beta_{1}无解,x_{1}\alpha_{1}+x_{2}\alpha_{2}=\beta_{2}有解

 

由于两个的系数矩阵一样，所以可以凑成一个大的增广矩阵

 

(1211233a+21a0−2)→(12110−11a00a−3a2−2a−3) \begin{pmatrix} 1 & 2 & 1 & 1 \\ 2 & 3 & 3 & a+2 \\ 1 & a & 0 & -2 \end{pmatrix}\rightarrow \begin{pmatrix} 1 & 2 & 1 & 1 \\ 0 & -1 & 1 & a \\ 0 & 0 & a-3 & a^{2}-2a-3 \end{pmatrix}

有

{a−3≠0a2−2a−3=0→a=−1 \begin{cases} a-3\ne0 \\ a^{2}-2a-3=0 \end{cases}\rightarrow a=-1

 

例：nn维向量α,β1,β2,β3,γ1,γ2\alpha,\beta_{1},\beta_{2},\beta_{3},\gamma_{1},\gamma_{2}，若α\alpha可由β1,β2,β3\beta_{1},\beta_{2},\beta_{3}线性表出，β1,β2,β3\beta_{1},\beta_{2},\beta_{3}可由γ1,γ2\gamma_{1},\gamma_{2}线性表出，则α\alpha可由γ1,γ2\gamma_{1},\gamma_{2}线性表出

 

线性表出具有传递性

 

按已知条件，可设

α=k1β1+k2β2+k3β3β1=l11γ1+l12γ2β2=l21γ1+l22γ2β3=l31γ1+l32γ2 \begin{aligned} \alpha&=k_{1}\beta_{1}+k_{2}\beta_{2}+k_{3}\beta_{3}\\ \beta_{1}&=l_{11}\gamma_{1}+l_{12}\gamma_{2}\\ \beta_{2}&=l_{21}\gamma_{1}+l_{22}\gamma_{2}\\ \beta_{3}&=l_{31}\gamma_{1}+l_{32}\gamma_{2} \end{aligned}

于是

α=(k1l11+k2l21+k3l31)γ1+(k1l12+k2l22+k3l32)γ2 \alpha=(k_{1}l_{11}+k_{2}l_{21}+k_{3}l_{31})\gamma_{1}+(k_{1}l_{12}+k_{2}l_{22}+k_{3}l_{32})\gamma_{2}

或

α=(β1β2β3)(k1k2k3)=(γ1γ2)(l11l21l31l12l22l32)(k1k2k3)=(γ1γ2)(l11k1+l21k2+l31k3l12k1+l22k2+l32k3) \begin{aligned} \alpha&=\begin{pmatrix} \beta_{1} & \beta_{2} & \beta_{3} \end{pmatrix}\begin{pmatrix} k_{1} \\ k_{2} \\ k_{3} \end{pmatrix}\\ &=\begin{pmatrix} \gamma_{1} & \gamma_{2} \end{pmatrix}\begin{pmatrix} l_{11} & l_{21} & l_{31} \\ l_{12} & l_{22} & l_{32} \end{pmatrix}\begin{pmatrix} k_{1} \\ k_{2} \\ k_{3} \end{pmatrix}\\ &=\begin{pmatrix} \gamma_{1} & \gamma_{2} \end{pmatrix}\begin{pmatrix} l_{11}k_{1}+l_{21}k_{2}+l_{31}k_{3} \\ l_{12}k_{1}+l_{22}k_{2}+l_{32}k_{3} \end{pmatrix} \end{aligned}

线性相关

定义：对mm个nn维向量α1,α2,⋯ ,αm\alpha_{1},\alpha_{2},\cdots,\alpha_{m}，若存在不全为00的实数k1,k2,⋯ ,kmk_{1},k_{2},\cdots,k_{m}使

k1α1+k2α2+⋯+kmαm=0 k_{1}\alpha_{1}+k_{2}\alpha_{2}+\cdots+k_{m}\alpha_{m}=0

成立，则其向量组α1,α2,⋯ ,αm\alpha_{1},\alpha_{2},\cdots,\alpha_{m}线性相关，否则称其线性无关

 

例：判断向量组α1=(1,2,−1,4)T,α2=(0,−1,−5,3)T,α3=(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}的线性相关性

 

设x1α1+x2α2+x3α3=0x_{1}\alpha_{1}+x_{2}\alpha_{2}+x_{3}\alpha_{3}=0，即

x1(12−14)+x2(0−1−53)+x3(2535)=(0000) x_{1}\begin{pmatrix} 1 \\ 2 \\ -1 \\ 4 \end{pmatrix}+x_{2}\begin{pmatrix} 0 \\ -1 \\ -5 \\ 3 \end{pmatrix}+x_{3}\begin{pmatrix} 2 \\ 5 \\ 3 \\ 5 \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \end{pmatrix}

按分量写出

{x1+2x3=02x1−x2+5x3=0−x1−5x2+3x3=04x1+3x2+5x2=0 \begin{cases} x_{1}+2x_{3}=0 \\ 2x_{1}-x_{2}+5x_{3}=0 \\ -x_{1}-5x_{2}+3x_{3}=0 \\ 4x_{1}+3x_{2}+5x_{2}=0 \end{cases}

写出系数矩阵

(1022−15−1−53435)→(10201−1000000) \begin{pmatrix} 1 & 0 & 2 \\ 2 & -1 & 5 \\ -1 & -5 & 3 \\ 4 & 3 & 5 \end{pmatrix}\rightarrow \begin{pmatrix} 1 & 0 & 2 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}

同解方程组

{x1+2x3=0x2−x3=0 \begin{cases} x_{1}+2x_{3}=0 \\ x_{2}-x_{3}=0 \end{cases}

有非零解

 

定理：向量组α1,α2,⋯ ,αm\alpha_{1},\alpha_{2},\cdots,\alpha_{m}线性相关

⇔\Leftrightarrow存在不全为00的k1,k2,⋯ ,kmk_{1},k_{2},\cdots,k_{m}，使

k1α1+k2α2+⋯+kmαm=0 k_{1}\alpha_{1}+k_{2}\alpha_{2}+\cdots+k_{m}\alpha_{m}=0

⇔\Leftrightarrow存在不全为00的k1,k2,⋯ ,kmk_{1},k_{2},\cdots,k_{m}，使

(α1α2⋯αm)(k1k2⋮km)=0 \begin{pmatrix} \alpha_{1} & \alpha_{2} & \cdots & \alpha_{m} \end{pmatrix}\begin{pmatrix} k_{1} \\ k_{2} \\ \vdots \\ k_{m} \end{pmatrix}=0

⇔\Leftrightarrow齐次方程组有非零解

(α1α2⋯αm)(x1x2⋮xm)=0 \begin{pmatrix} \alpha_{1} & \alpha_{2} & \cdots & \alpha_{m} \end{pmatrix}\begin{pmatrix} x_{1} \\ x_{2} \\ \vdots \\ x_{m} \end{pmatrix}=0

⇔r(α1α2⋯αm)<m\Leftrightarrow r \begin{pmatrix}\alpha_{1} & \alpha_{2} & \cdots & \alpha_{m}\end{pmatrix}<m，mm为未知数的个数

 

推论：

  1. nn个nn维向量α1,α2,⋯ ,αm\alpha_{1},\alpha_{2},\cdots,\alpha_{m}线性相关⇔∣α1α2⋯αm∣=0\Leftrightarrow \begin{vmatrix}\alpha_{1}&\alpha_{2}&\cdots&\alpha_{m}\end{vmatrix}=0

  2. n+1n+1个nn维向量必线性相关

 

例：A=(12−2212304),α=(a11)A=\begin{pmatrix}1 & 2 & -2 \\ 2 & 1 & 2 \\ 3 & 0 & 4\end{pmatrix},\alpha=\begin{pmatrix}a \\ 1 \\ 1\end{pmatrix}，已知Aα,αA \alpha,\alpha线性相关，则a=()a=()

 

Aα=(a2a+33a+4) \begin{aligned} A \alpha=\begin{pmatrix} a \\ 2a+3 \\ 3a+4 \end{pmatrix} \end{aligned}

AαA \alpha与α\alpha线性相关

aa=2a+31=3a+41 \frac{a}{a}= \frac{2a+3}{1}=\frac{3a+4}{1}

因此a=−1a=-1

 

例：A=(α1α2⋯αn),B=(β1β2⋯βn),AB=(γ1γ2⋯γn)A=\begin{pmatrix}\alpha_{1} & \alpha_{2} & \cdots & \alpha_{n}\end{pmatrix},B=\begin{pmatrix}\beta_{1} & \beta_{2} & \cdots & \beta_{n}\end{pmatrix},AB=\begin{pmatrix}\gamma_{1} & \gamma_{2} & \cdots & \gamma_{n}\end{pmatrix}均为nn阶矩阵，记向量组(I)α1,α2,⋯ ,αn;(II)β1,β2,⋯ ,βn;(III)γ1,γ2,⋯ ,γn(\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}，若向量组(III)(\text{III})线性相关，证明(I),(II)(I),(II)中至少有一个线性相关

 

符合nn个nn维向量，则有

(III)线性相关⇔∣AB∣=0∣A∣⋅∣B∣=0∣A∣=0或∣B∣=0 \begin{aligned} (\text{III})线性相关\Leftrightarrow|AB|&=0\\ |A|\cdot|B|&=0\\|A|=0&或|B|=0 \end{aligned}

得证

 

定理：如果nn维向量α1,α2,⋯ ,αn\alpha_{1},\alpha_{2},\cdots,\alpha_{n}线性无关，α1,α2,⋯ ,αn,β\alpha_{1},\alpha_{2},\cdots,\alpha_{n},\beta线性相关，则向量β\beta可由α1,α2,⋯ ,αn\alpha_{1},\alpha_{2},\cdots,\alpha_{n}线性表出，且表示方法唯一

证明：因为α1,α2,⋯ ,αn,β\alpha_{1},\alpha_{2},\cdots,\alpha_{n},\beta线性相关，故存在不全为00的k1,k2,⋯ ,kn,kk_{1},k_{2},\cdots,k_{n},k，使得

k1α1+k2α2+⋯+knαn+kβ=0(1) k_{1}\alpha_{1}+k_{2}\alpha_{2}+\cdots+k_{n}\alpha_{n}+k \beta=0\tag1

（反证）如果k=0k=0，则k1,k2,⋯ ,knk_{1},k_{2},\cdots,k_{n}不全为00，而

k1α1+k2α2+⋯+knαn=0 k_{1}\alpha_{1}+k_{2}\alpha_{2}+\cdots+k_{n}\alpha_{n}=0

与条件α1,α2,⋯ ,αn\alpha_{1},\alpha_{2},\cdots,\alpha_{n}线性无关矛盾，从而k≠0k\ne0，由(1)(1)得

β=−k1kα1−k2kα2−⋯−knkαn \beta=- \frac{k_{1}}{k}\alpha_{1}- \frac{k_{2}}{k}\alpha_{2}-\cdots- \frac{k_{n}}{k}\alpha_{n}

即β\beta一定能由α1,α2,⋯ ,αn\alpha_{1},\alpha_{2},\cdots,\alpha_{n}线性表出

（反证）如果β\beta有两种不同的表示方法，设

β=x1α1+x2α2+⋯+xnαn=y1α1+y2α2+⋯+ynαn \begin{aligned} \beta&=x_{1}\alpha_{1}+x_{2}\alpha_{2}+\cdots+x_{n}\alpha_{n}\\ &=y_{1}\alpha_{1}+y_{2}\alpha_{2}+\cdots+y_{n}\alpha_{n} \end{aligned}

两式相减

(x1−y1)α1+(x2−y2)α2+⋯+(xn−yn)αn=0 (x_{1}-y_{1})\alpha_{1}+(x_{2}-y_{2})\alpha_{2}+\cdots+(x_{n}-y_{n})\alpha_{n}=0

因有两种不同的表示x1−y1,x2−y2,⋯ ,xn−ynx_{1}-y_{1},x_{2}-y_{2},\cdots,x_{n}-y_{n}不全为00与α1,α2,⋯ ,αn\alpha_{1},\alpha_{2},\cdots,\alpha_{n}线性无关相矛盾，从而β\beta的表示法唯一

 

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

 

必要性

如α1,α2,⋯ ,αs\alpha_{1},\alpha_{2},\cdots,\alpha_{s}线性相关，则存在不全为00的k1,k2,⋯ ,ksk_{1},k_{2},\cdots,k_{s}

k1α1+k2α2+⋯+ksαs=0 k_{1}\alpha_{1}+k_{2}\alpha_{2}+\cdots+k_{s}\alpha_{s}=0

不妨设k1≠0k_{1}\ne0，则有

k1α1=−k2α2−⋯−ksαs k_{1}\alpha_{1}=-k_{2}\alpha_{2}-\cdots-k_{s}\alpha_{s}

于是

α1=−k2k1α2−⋯−ksk1αs \alpha_{1}=- \frac{k_{2}}{k_{1}}\alpha_{2}-\cdots- \frac{k_{s}}{k_{1}}\alpha_{s}

充分性

如果αi\alpha_{i}可由α1,⋯ ,αi−1,αi+1,⋯ ,αs\alpha_{1},\cdots, \alpha_{i-1},\alpha_{i+1},\cdots ,\alpha_{s}线性表出，设

αi=k1α1+⋯+ki−1αi−1+ki+1αi+1+⋯+ksαs \alpha_{i}=k_{1}\alpha_{1}+\cdots+k_{i-1}\alpha_{i-1}+k_{i+1}\alpha_{i+1}+\cdots+k_{s}\alpha_{s}

即有

k1α1+⋯+ki−1αi−1−αi+ki+1αi+1+⋯+ksαs=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

组合系数

k1,⋯ ,ki−1,−1,ki+1,⋯ ,ks k_{1},\cdots,k_{i-1},-1,k_{i+1},\cdots,k_{s}

不全为00

 

定理：如果α1,α2,⋯ ,αs\alpha_{1},\alpha_{2},\cdots,\alpha_{s}可由β1,β2,⋯ ,βt\beta_{1},\beta_{2},\cdots,\beta_{t}线性表出，且s>ts>t，则α1,α2,⋯ ,αs\alpha_{1},\alpha_{2},\cdots,\alpha_{s}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α1,α2,⋯ ,αs\alpha_{1},\alpha_{2},\cdots,\alpha_{s}线性无关，且α1,α2,⋯ ,αs\alpha_{1},\alpha_{2},\cdots,\alpha_{s}可由β1,β2,⋯ ,βt\beta_{1},\beta_{2},\cdots,\beta_{t}线性表出，则s≤ts\leq t

 

简单的线性无关证明题思路

当k1α1+k2α2+⋯+ksαs=0k_{1}\alpha_{1}+k_{2}\alpha_{2}+\cdots+k_{s}\alpha_{s}=0时，必有k1=0,k2=0,⋯ ,ks=0k_{1}=0,k_{2}=0,\cdots,k_{s}=0，则称向量组α1,α2,⋯ ,αs\alpha_{1},\alpha_{2},\cdots,\alpha_{s}线性无关

 

例：已知AA为nn阶可逆矩阵，α1,α2,α3\alpha_{1},\alpha_{2},\alpha_{3}是nn维线性无关向量，证明Aα1,Aα2,Aα3A \alpha_{1},A \alpha_{2},A \alpha_{3}线性无关

 

设

k1Aα1+k2Aα2+k3Aα3=0A(k1α1+k2α2+k3α3)=0k1α1+k2α2+k3α3=A−1⋅0k1α1+k2α2+k3α3=0 \begin{aligned} k_{1}A \alpha_{1}+k_{2}A \alpha_{2}+k_{3}A \alpha_{3}&=0\\ A(k_{1}\alpha_{1}+k_{2}\alpha_{2}+k_{3}\alpha_{3})&=0\\ k_{1}\alpha_{1}+k_{2}\alpha_{2}+k_{3}\alpha_{3}&=A^{-1}\cdot 0\\ k_{1}\alpha_{1}+k_{2}\alpha_{2}+k_{3}\alpha_{3}&=0 \end{aligned}

由于α1,α2,α3\alpha_{1},\alpha_{2},\alpha_{3}是nn维线性无关向量，则

k1=0,k2=0,k3=0 k_{1}=0,k_{2}=0,k_{3}=0

因此，Aα1,Aα2,Aα3A \alpha_{1},A \alpha_{2},A \alpha_{3}线性无关

 

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

 

设

k1(α1+α2)+k2(α2+α3)+k3(α3+α1)=0(k1+k3)α1+(k1+k2)α2+(k2+k3)α3=0 \begin{aligned} k_{1}(\alpha_{1}+\alpha_{2})+k_{2}(\alpha_{2}+\alpha_{3})+k_{3}(\alpha_{3}+\alpha_{1})&=0\\ (k_{1}+k_{3})\alpha_{1}+(k_{1}+k_{2})\alpha_{2}+(k_{2}+k_{3})\alpha_{3}&=0 \end{aligned}

因为α1,α2,α3\alpha_{1},\alpha_{2},\alpha_{3}线性无关

{k1+k3=0k1+k2=0k2+k3=0(1) \begin{cases} k_{1}+k_{3}=0 \\ k_{1}+k_{2}=0 \\ k_{2}+k_{3}=0 \end{cases}\tag{1}

由

∣101110011∣=2≠0 \begin{vmatrix} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{vmatrix}=2\ne0

齐次方程组(1)(1)只有00解，即必有k1=0,k2=0,k3=0k_{1}=0,k_{2}=0,k_{3}=0，因此α1+α2,α2+α3,α3+α1\alpha_{1}+\alpha_{2},\alpha_{2}+\alpha_{3},\alpha_{3}+\alpha_{1}线性无关

 

秩

极大线性无关组

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

  • αi1,αi2,⋯ ,αir\alpha_{i_{1}},\alpha_{i_{2}},\cdots,\alpha_{i_{r}}线性无关

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

 

则称 αi1,αi2,⋯ ,αir\alpha_{i_{1}},\alpha_{i_{2}},\cdots,\alpha_{i_{r}}是向量组α1,α2,⋯ ,αs\alpha_{1},\alpha_{2},\cdots,\alpha_{s}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αi1,αi2,⋯ ,αir\alpha_{i_{1}},\alpha_{i_{2}},\cdots,\alpha_{i_{r}}与αj1,αj2,⋯ ,αjt\alpha_{j_{1}},\alpha_{j_{2}},\cdots,\alpha_{j_{t}}都是向量组α1,α2,⋯ ,αs\alpha_{1},\alpha_{2},\cdots,\alpha_{s}的极大线性无关组，则r=tr=t

证明：

因为αi1,αi2,⋯ ,αir\alpha_{i_{1}},\alpha_{i_{2}},\cdots,\alpha_{i_{r}}是α1,α2,⋯ ,αs\alpha_{1},\alpha_{2},\cdots,\alpha_{s}的极大线性无关组，那么αj1,αj2,⋯ ,αjt\alpha_{j_{1}},\alpha_{j_{2}},\cdots,\alpha_{j_{t}}可由αi1,αi2,⋯ ,αir\alpha_{i_{1}},\alpha_{i_{2}},\cdots,\alpha_{i_{r}}线性表示

又因为αj1,αj2,⋯ ,αjt\alpha_{j_{1}},\alpha_{j_{2}},\cdots,\alpha_{j_{t}}线性无关，则有t≤rt\leq r

同理r≤tr\leq t，故有r=tr=t

 

例：已知向量组α1=(1,−1,0,5)T,α2=(2,0,1,4)T,α3=(3,1,2,3)T,α4=(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}，其中aa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

(α1α2α3α4)=(1234−10120123543a)→(10−1−20123000a−20000) \begin{pmatrix} \alpha_{1} & \alpha_{2} & \alpha_{3} & \alpha_{4} \end{pmatrix}=\begin{pmatrix} 1 & 2 & 3 & 4 \\ -1 & 0 & 1 & 2 \\ 0 & 1 & 2 & 3 \\ 5 & 4 & 3 & a \end{pmatrix}\rightarrow \begin{pmatrix} 1 & 0 & -1 & -2 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 0 & a-2 \\ 0 & 0 & 0 & 0 \end{pmatrix}

当a=2a=2时，秩r(α1α2α3α4)=2r \begin{pmatrix}\alpha_{1} & \alpha_{2} & \alpha_{3} & \alpha_{4}\end{pmatrix}=2，极大线性无关组是α1,α2\alpha_{1},\alpha_{2}。α3=−α1+2α2,α4=−2α1+3α2\alpha_{3}=-\alpha_{1}+2\alpha_{2},\alpha_{4}=-2\alpha_{1}+3\alpha_{2}

（行最简形式中第一二列、一二行行列式不为00；同理可以使α1,α3\alpha_{1},\alpha_{3}，即第一三列、一二行行列式不为00）

 

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，秩r(α1α2α3α4)=3r \begin{pmatrix}\alpha_{1} & \alpha_{2} & \alpha_{3} & \alpha_{4}\end{pmatrix}=3，极大线性无关组是α1,α2,α4\alpha_{1},\alpha_{2},\alpha_{4}。α3=−α1+2α2\alpha_{3}=-\alpha_{1}+2\alpha_{2}

 

矩阵的秩

kk阶子式：AA为m×nm\times n的矩阵，任取kk行与kk列(k≤m,k≤n)(k\leq m,k\leq n)位于交叉点的k2k^{2}元素，按AA中的位置次序而得到的kk阶行列式，称为矩阵AA的kk阶子式

 

秩：矩阵AA中非00子式的最高阶数称为矩阵AA的秩，记为r(A)r(A)

r(A)=r⇔Ar(A)=r\Leftrightarrow A中有rr阶子式不为00而所有r+1r+1阶子式（若有）全为00

r(A)<r⇔Ar(A)<r\Leftrightarrow A中rr阶子式全为00

r(A)≥r⇔Ar(A)\geq r\Leftrightarrow A中rr阶子式全不为00

A≠0⇔r(A)≥1A\ne0\Leftrightarrow r(A)\geq1

A为n阶矩阵,r(A)=n⇔∣A∣≠0⇔AA为n阶矩阵,r(A)=n\Leftrightarrow|A|\ne0\Leftrightarrow A可逆

 

r(12−1030 05100000600001)=3r\begin{pmatrix}1 & 2 & -1 & 0 & 3 \\ 0  & 0 & 5 & 1 & 0 \\ 0 & 0 & 0 & 0 & 6 \\ 0 & 0 & 0 & 0 & 1\end{pmatrix}=3，虽然第四行不全为00，但是所有四阶子式都为00

 

公式

  • r(AT)=r(A)r(A^{T})=r(A)

  • r(kA)=r(A),k≠0r(kA)=r(A),k\ne0

         r(0E−A)=r(A)r(0E-A)=r(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(AB)≤min{r(A),r(B)}r(AB)\leq min\{r(A),r(B)\}

         若AA可逆，则r(AB)=r(B),r(BA)=r(B)r(AB)=r(B),r(BA)=r(B)

  • r(ATA)=r(A)r(A^{T}A)=r(A)

  • A−m×n,B−n×sA-m\times n,B-n\times s且AB=0AB=0，则r(A)+r(B)≤nr(A)+r(B)\leq n

  • r(AOOB)=r(A)+r(B)r \begin{pmatrix}A & O \\ O & B\end{pmatrix}=r(A)+r(B)

  • 若A∼BA\sim B，则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=(111101−1b23a43517)r(A)=3,A=\begin{pmatrix}1 & 1 & 1 & 1 \\ 0 & 1 & -1 & b \\ 2 & 3 & a & 4 \\ 3 & 5 & 1 & 7\end{pmatrix}，求a,ba,b

 

对矩阵AA作初等变换

A=(111101−1b23a43517)→(111101−1b00a−12−b0004−2b) A=\begin{pmatrix}1 & 1 & 1 & 1 \\ 0 & 1 & -1 & b \\ 2 & 3 & a & 4 \\ 3 & 5 & 1 & 7\end{pmatrix}\rightarrow \begin{pmatrix} 1 & 1 & 1 & 1 \\ 0 & 1 & -1 & b \\ 0 & 0 & a-1 & 2-b \\ 0 & 0 & 0 & 4-2b \end{pmatrix}

有

{a−1=04−2b≠0或{a−1≠04−2b=0 \begin{cases} a-1=0 \\ 4-2b\ne0 \end{cases}或\begin{cases} a-1\ne0 \\ 4-2b=0 \end{cases}

则a≠1,b=2a\ne1,b=2或a=1,b≠2a=1,b\ne2

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