本系列博客汇总在这里:考研数学知识点汇总系列博客
n个数 a 1 , a 2 , ⋅ ⋅ ⋅ , a n a_1,a_2,···,a_n a1,a2,⋅⋅⋅,an构成的有序数组称为n维向量。
[ a 1 a 2 ⋮ a n ] \begin{bmatrix} a_1 \\ a_2 \\ \vdots \\ a_n \end{bmatrix} ⎣⎢⎢⎢⎡a1a2⋮an⎦⎥⎥⎥⎤或 [ a 1 a 2 ⋅ ⋅ ⋅ a n ] T \begin{bmatrix} a_1 & a_2 & ··· & a_n \end{bmatrix}^T [a1a2⋅⋅⋅an]T称为列向量。
[ a 1 a 2 ⋅ ⋅ ⋅ a n ] \begin{bmatrix} a_1 & a_2 & ··· & a_n \end{bmatrix} [a1a2⋅⋅⋅an]称为行向量。
a i a_i ai称为向量的第i个分量(i=1,2,···,n)
如果向量的所有分量都是零,就称其为零向量,记作 [ 0 0 ⋅ ⋅ ⋅ 0 ] T \begin{bmatrix} 0 & 0 & ··· & 0 \end{bmatrix}^T [00⋅⋅⋅0]T
设α= [ a 1 a 2 ⋅ ⋅ ⋅ a n ] T \begin{bmatrix} a_1 & a_2 & ··· & a_n \end{bmatrix}^T [a1a2⋅⋅⋅an]T,β= [ b 1 b 2 ⋅ ⋅ ⋅ b n ] T \begin{bmatrix} b_1 & b_2 & ··· & b_n \end{bmatrix}^T [b1b2⋅⋅⋅bn]T
给定向量组A: α 1 , α 2 , ⋅ ⋅ ⋅ , α m α_1,α_2,···,α_m α1,α2,⋅⋅⋅,αm,对于任意一组实数 k 1 , k 2 , ⋅ ⋅ ⋅ , k m k_1,k_2,···,k_m k1,k2,⋅⋅⋅,km,表达式 k 1 α 1 + k 2 α 2 + ⋅ ⋅ ⋅ + k m α m k_1α_1+k_2α_2+···+k_mα_m k1α1+k2α2+⋅⋅⋅+kmαm为向量组A的一个线性组合, k 1 , k 2 , ⋅ ⋅ ⋅ , k m k_1,k_2,···,k_m k1,k2,⋅⋅⋅,km称为这个线性组合的系数。
给定向量组A: α 1 , α 2 , ⋅ ⋅ ⋅ , α m α_1,α_2,···,α_m α1,α2,⋅⋅⋅,αm和向量β,若存在一组数 k 1 , k 2 , ⋅ ⋅ ⋅ , k m k_1,k_2,···,k_m k1,k2,⋅⋅⋅,km,使 β = k 1 α 1 + k 2 α 2 + ⋅ ⋅ ⋅ + k m α m β=k_1α_1+k_2α_2+···+k_mα_m β=k1α1+k2α2+⋅⋅⋅+kmαm,则称向量β是向量组A的线性组合,又称向量β能由向量组A线性表示。若向量组B中任一向量均可由向量组A线性表示,则称向量组B可由向量组A线性表示。若向量组B能由向量组A线性表示,则 r B ⩽ r A r_B \leqslant r_A rB⩽rA。
若向量组A与向量组B能相互线性表示,称两向量组等价。向量组A与向量组B等价⟺ r ( A ) = r ( B ) = r ( A , B ) r(A)=r(B)=r(A,B) r(A)=r(B)=r(A,B),其中A,B是向量组A,B所构成的矩阵。
注:
给定向量组A: α 1 , α 2 , ⋅ ⋅ ⋅ , α m α_1,α_2,···,α_m α1,α2,⋅⋅⋅,αm,如果存在不全为零的数 k 1 , k 2 , ⋅ ⋅ ⋅ , k m k_1,k_2,···,k_m k1,k2,⋅⋅⋅,km,使得 k 1 α 1 + k 2 α 2 + ⋅ ⋅ ⋅ + k m α m = 0 k_1α_1+k_2α_2+···+k_mα_m=0 k1α1+k2α2+⋅⋅⋅+kmαm=0,则称向量组A线性相关;否则称线性无关,即当且仅当 k 1 = k 2 = ⋅ ⋅ ⋅ = k m = 0 k_1=k_2=···=k_m=0 k1=k2=⋅⋅⋅=km=0时上式才成立。
α 1 , α 2 , ⋅ ⋅ ⋅ , α r α_1,α_2,···,α_r α1,α2,⋅⋅⋅,αr是向量组T的部分向量组,如果它满足
则称向量组 α 1 , α 2 , ⋅ ⋅ ⋅ , α r α_1,α_2,···,α_r α1,α2,⋅⋅⋅,αr是向量组T的一个极大线性无关组,简称极大无关组。
向量组的极大无关组所含向量的个数称为这个向量组的秩。
设V为n维向量的集合,若集合V非空,且集合V对于n维向量的加法及数乘两种运算封闭,即
则称集合V为R上的向量空间。
记 R n R^n Rn为n维向量空间。
注:
设V是向量空间,若有r个向量 α 1 , α 2 , ⋅ ⋅ ⋅ , α r ∈ V α_1,α_2,···,α_r∈V α1,α2,⋅⋅⋅,αr∈V,且满足
V中任一向量都可由 α 1 , α 2 , ⋅ ⋅ ⋅ , α r α_1,α_2,···,α_r α1,α2,⋅⋅⋅,αr线性表示。
则称 α 1 , α 2 , ⋅ ⋅ ⋅ , α r α_1,α_2,···,α_r α1,α2,⋅⋅⋅,αr为向量空间V的一个基,数r称为V的维数,记为dimV=r,并称V为r维向量空间。
注:
如果向量空间V中取定一个基 α 1 , α 2 , ⋅ ⋅ ⋅ , α r α_1,α_2,···,α_r α1,α2,⋅⋅⋅,αr,那么V中任一向量x可唯一地表示为 x = λ 1 α 1 + λ 2 α 2 + ⋅ ⋅ ⋅ + λ r α r x=λ_1α_1+λ_2α_2+···+λ_rα_r x=λ1α1+λ2α2+⋅⋅⋅+λrαr,数组 λ 1 , λ 2 , ⋅ ⋅ ⋅ , λ r λ_1,λ_2,···,λ_r λ1,λ2,⋅⋅⋅,λr称为向量x在基 α 1 , α 2 , ⋅ ⋅ ⋅ , α r α_1,α_2,···,α_r α1,α2,⋅⋅⋅,αr下的坐标。
n维向量空间 R n R^n Rn中任两个向量 α = ( a 1 , a 2 , ⋅ ⋅ ⋅ , a n ) T α=(a_1,a_2,···,a_n)^T α=(a1,a2,⋅⋅⋅,an)T, β = ( b 1 , b 2 , ⋅ ⋅ ⋅ , b n ) T β=(b_1,b_2,···,b_n)^T β=(b1,b2,⋅⋅⋅,bn)T的内积定义为 ( α , β ) = α T β = β T α = a 1 b 1 + a 2 b 2 + ⋅ ⋅ ⋅ + a n b n (α,β)=α^Tβ=β^Tα=a_1b_1+a_2b_2+···+a_nb_n (α,β)=αTβ=βTα=a1b1+a2b2+⋅⋅⋅+anbn,并称定义了内积的向量空间 R n R^n Rn为欧式空间。
设 α = ( a 1 , a 2 , ⋅ ⋅ ⋅ , a n ) T α=(a_1,a_2,···,a_n)^T α=(a1,a2,⋅⋅⋅,an)T, ∣ α ∣ = ( α , α ) = a 1 2 + a 2 2 + ⋅ ⋅ ⋅ + a n 2 |α|=\sqrt{(α,α)}=\sqrt{a_1^2+a_2^2+···+a_n^2} ∣α∣=(α,α)=a12+a22+⋅⋅⋅+an2,称为向量α的长度。
长度为1的向量称为单位向量。
若 ( α , β ) = 0 (α,β)=0 (α,β)=0,称α与β正交,记 α ⊥ β α⊥β α⊥β。
设 α 1 , α 2 , ⋅ ⋅ ⋅ , α r α_1,α_2,···,α_r α1,α2,⋅⋅⋅,αr为正交向量组,则 α 1 , α 2 , ⋅ ⋅ ⋅ , α r α_1,α_2,···,α_r α1,α2,⋅⋅⋅,αr是线性无关的。
设 α 1 , α 2 , ⋅ ⋅ ⋅ , α r α_1,α_2,···,α_r α1,α2,⋅⋅⋅,αr是线性无关的向量组,寻找一个标准正交向量组 ε 1 , ε 2 , ⋅ ⋅ ⋅ , ε r ε_1,ε_2,···,ε_r ε1,ε2,⋅⋅⋅,εr,使其与 α 1 , α 2 , ⋅ ⋅ ⋅ , α r α_1,α_2,···,α_r α1,α2,⋅⋅⋅,αr等价,其作法分两步:
1.正交化
β 1 = α 1 β_1=α_1 β1=α1
β 2 = α 2 − ( α 2 , β 1 ) ( β 1 , β 1 ) β 1 β_2=α_2-\frac{(α_2,β_1)}{(β_1,β_1)}β_1 β2=α2−(β1,β1)(α2,β1)β1
β 3 = α 3 − ( α 3 , β 1 ) ( β 1 , β 1 ) β 1 − ( α 3 , β 2 ) ( β 2 , β 2 ) β 2 β_3=α_3-\frac{(α_3,β_1)}{(β_1,β_1)}β_1-\frac{(α_3,β_2)}{(β_2,β_2)}β_2 β3=α3−(β1,β1)(α3,β1)β1−(β2,β2)(α3,β2)β2
⋮ \vdots ⋮
β r = α r − ( α r , β 1 ) ( β 1 , β 1 ) β 1 − ( α r , β 2 ) ( β 2 , β 2 ) β 2 − ⋅ ⋅ ⋅ − ( α r , β r − 1 ) ( β r − 1 , β r − 1 ) β r − 1 β_r=α_r-\frac{(α_r,β_1)}{(β_1,β_1)}β_1-\frac{(α_r,β_2)}{(β_2,β_2)}β_2-···-\frac{(α_r,β_{r-1})}{(β_{r-1},β_{r-1})}β_{r-1} βr=αr−(β1,β1)(αr,β1)β1−(β2,β2)(αr,β2)β2−⋅⋅⋅−(βr−1,βr−1)(αr,βr−1)βr−1
2.单位化(规范化)
取 ε 1 = β 1 ∣ β 1 ∣ , ε 2 = β 2 ∣ β 2 ∣ , ⋅ ⋅ ⋅ , ε r = β r ∣ β r ∣ ε_1=\frac{β_1}{|β_1|},ε_2=\frac{β_2}{|β_2|},···,ε_r=\frac{β_r}{|β_r|} ε1=∣β1∣β1,ε2=∣β2∣β2,⋅⋅⋅,εr=∣βr∣βr