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Matrix Theory study notes[5]

文章目录

  • linear space
  • references

linear space

  1. the dimension of a linear space is maximal number of vectors among all vetor groups of linearly indepedent in the linear space.
  2. let n is the dimension of A which is a linear space and A is {3a,4b,6a−5∣aandbarecoprime,aandbareintegers}\{3a,4b,6a-5|a\quad and\quad b\quad are \quad coprime,a \quad and \quad b \quad are \quad integers\}{3a,4b,6a5∣aandbarecoprime,aandbareintegers},the nnn is 2 because that the maximal numbers of vectors such as {3a,4b}\{3a,4b\}{3a,4b},{4b,6a−5}\{4b,6a-5\}{4b,6a5} are two,dim A=2.
  3. the linear space A with n dimension is called as n dimension linear space AnA^nAn in the number field,if n=+∞n=+\inftyn=+,then A is called as unlimited dimension linear space.
  4. a basis of linear space meets following conditions.
  • let V is a linear space in the number field K and v1,v2,v3,...,vr∈Vv_1,v_2,v_3,...,v_r \in Vv1,v2,v3,...,vrV,VVV satsifies two situations.

    • v1,v2,v3,...,vrv_1,v_2,v_3,... ,v_rv1,v2,v3,...,vr are linear independent.
    • each vector of V is the linear combination of v1,v2,v3,...,vrv_1,v_2,v_3,... ,v_rv1,v2,v3,...,vr.

    the v1,v2,v3,...,vrv_1,v_2,v_3,... ,v_rv1,v2,v3,...,vrcan be called as the basis of linear space VVV,viv_ivi is basic vector.

  • the basis of a linear space is not one and only.

  1. dimension of a linear space is the number of vectors that the basis of the linear space includes.
  2. the vectors which be included in fundamental resolution set of homogeneous linear equations system Ax=0Ax=0Ax=0 is a basis of solution space.
  • homogeneous linear equations system,all the constant terms of it are zero,can be formed as follows.

{a11x1+a12x2+⋯+a1nxn=0a21x1+a22x2+⋯+a2nxn=0⋮am1x1+am2x2+⋯+amnxn=0\begin{cases} a_{11}x_1 + a_{12}x_2 + \cdots + a_{1n}x_n = 0 \\ a_{21}x_1 + a_{22}x_2 + \cdots + a_{2n}x_n = 0 \\ \vdots \\ a_{m1}x_1 + a_{m2}x_2 + \cdots + a_{mn}x_n = 0 \end{cases} a11x1+a12x2++a1nxn=0a21x1+a22x2++a2nxn=0am1x1+am2x2++amnxn=0
the aija_{ij}aijis are coefficient,xjx_jxj are unkown number,the right items of all equations are zero.

  • the zero solution exists necessarily, x1=x2=⋯=xn=0x_1 = x_2 = \cdots = x_n = 0x1=x2==xn=0 , to be called as trivial solution.
  • the non-zero solution exists.
    • if the rank rrr of a equations system’s coefficient matrix is less than the number of unkown number nnn ,r<nr<nr<n,then there are infinited number of non-zero solutions exists in the equations system,also is called as non-trivial solution.
    • all solutions make up a vector space,can be called as solution space, which dimension is n−rn-rnr.
    • if the equations system has non-zero solution,then the basis of solution space can be called as fundamental system of solutions.the general solution can be expressed as a linear combination of fundamental system of solution set .
    • when r=nr = nr=n,the equations system have only zero solution.
  • the following example explain that how to get solution of homogeneous linear equations system.

{x1+2x2−x3=02x1−x2+x3=0\begin{cases} x_1 + 2x_2 - x_3 = 0 \\ 2x_1 - x_2 + x_3 = 0 \end{cases} {x1+2x2x3=02x1x2+x3=0
A=(12−12−11)A = \begin{pmatrix} 1 & 2 & -1 \\ 2 & -1 & 1 \end{pmatrix} A=(122111)
(12−10−53)\begin{pmatrix} 1 & 2 & -1 \\ 0 & -5 & 3 \end{pmatrix} (102513)
(101501−35)\begin{pmatrix} 1 & 0 & \frac{1}{5} \\ 0 & 1 & -\frac{3}{5} \end{pmatrix} (10015153)
the rank r=2r = 2r=2,the number of unkown numbern=3n = 3n=3r<nr < nr<n,the non-zero solution exists.
the free variable is x3x_3x3,let x3=5x_3 = 5x3=5
ξ=(−135)\mathbf{\xi} = \begin{pmatrix} -1 \\ 3 \\ 5 \end{pmatrix} ξ=135
finally,the general solution has computed.
k(−135)(k∈R)k \begin{pmatrix} -1 \\ 3 \\ 5 \end{pmatrix} \quad(k \in \mathbb{R}) k135(kR)

references

  1. deepseek
  2. 《矩阵论》
http://www.dtcms.com/a/299173.html

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