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Vector groups of linear algebra and their linear combinations

4.1 Vector groups and their linear combinations

Definition 1

1. N-dimensional vectors

Definition: n ordered numbers A1, A2,…. ,ana_1,a_2,…. ,a_na1,a2,…. ,an, where the n number becomes the n components of the vector, and the ith number aia_{I}ai is called the ith component

2. The real vector

Definition: All components of a vector are real numbers

3. The complex vector

Definition: At least one component of a vector is complex

4. N-dimensional column vector A

5. N-dimensional row vector B

$b = (b_1,b_2,… ,b_n)$

In general, column vectors are represented by boldface lowercase letters α, β\alpha, \betaα, β, etc., and row vectors are represented by αT, βT\alpha^{T}, \beta^{T}αT, βT, etc., which are generally regarded as column vectors without special instructions

6. Three-dimensional vector space

Definition: a collection of all three – dimensional vectors

$\mathbb{R}^3 = \{r = (x, y, z)^T | x, y, z \in \mathbb{R} \}$

It’s called a three-dimensional vector space

In the discussion of vector operations, vectors are regarded as directed line segments

In the discussion of vector set, the vector r is regarded as a point P with r as the diameter, and the trajectory of point P is regarded as the graph of vector set

Radial: Generally refers to position vector. A directed line segment starting at the origin of coordinates and ending at the position of a moving particle at a certain time

Such as point set Π = {P (x, y, z) ∣ ax + by + cz = d \ Pi = \ {P (x, y, z) | Π ax + by + cz = d = {P (x, y, z) ∣ ax + by + cz = d is a plane (a, b, c, insufficiency of 0)

Let’s say that a is equal to b is equal to c is equal to 1 and d is equal to 0 so x plus y plus z is equal to 0 and I’ll distort it a little bit z is equal to minus x minus y so it’s easy to see that this is a plane

So the vector set {r = (x, y, z) T ∣ ax + by + cz = d} \ {r = (x, y, z) ^ T | ax + by + cz = d \} {r = (x, y, z) T ∣ ax + by + cz = d}

Also called a plane in the vector space R3\mathbb{R}^3, and has π \Pi π as its graph

7. N-dimensional vector space

All of the n-dimensional vectors

$\mathbb{R}^n = \{ x = (x_1, x_2,…. , x_n)^T | x_1,x_2,… ,x_n \in \mathbb{R} \}$

{x=(x1,x2… , xn) T ∣ a1x1 + a2x2 +… +anxn=b\{ x = (x_1,x_2,… ,x_n)^T | a_1x_1 + a_2x_2 +…. + a_nx_n = b{x=(x1,x2,… , xn) T ∣ a1x1 + a2x2 +… +anxn=b is called the n-1-dimensional hyperplane in the n-dimensional vector space Rn \mathbb{R}^n The n-1-dimensional hyperplane in Rn

8. Vector group

Definition: a set of column vectors (or row vectors) of the same dimension

The entire column vector of an M by n matrix is a set of n m-dimensional column vectors

The column vector group and row vector group of matrix are vector groups containing only finite vectors. Conversely, a set of finite vectors can always form a matrix

For example, A vector group consisting of m n-dimensional column vectors A: A1, A2… ,am(ai, I ∈[1,m] represents an n-dimensional column vector)A:a_1,a_2… ,a_m(a_i, I \in [1,m] represents an n-dimensional column vector)A:a1,a2… , am (ai, I ∈ [m] 1, said A n d column) can make A n * m matrix A = (a1, a2,… ,am)A = (a_1,a_2,… ,a_m)A=(a1,a2,… ,am)

Vector B composed of m n-dimensional row vectors :β1T,β2T… , beta mTB: \ beta_1 ^ T, \ beta_2 ^ T,… , \ beta_m ^ TB: beta 1 t, beta 2 t… ,βmT forms an M * N matrix

β I \beta_iβ I is the column vector

In short, an ordered set of vectors containing a finite number of vectors can correspond to a matrix one to one

Definition 2

(1) Given vector group A: A1, A2… ,amA:a_1,a_2,… ,a_mA:a1,a2,… Am, for any set of real numbers k1,k2… ,kmk_1,k_2,… ,k_mk1,k2,… ,km, expression k1a1+ k2A2 +… +kmamk_1a_1 + k_2a_2 + … + k_m a_mk1a1+k2a2+… +kmam is called A linear combination of vector group A, k1,k2…. ,kmk_1,k_2 …. , k_mk1,k2…. Km is called the coefficient of this linear combination

(2) Given vector group A: A1, A2… ,amA:a_1,a_2,… ,a_mA:a1,a2,… ,am and vector b, if there exists a set of numbers λ1,λ2… , lambda m \ lambda_1 \ lambda_2,… , \ lambda_m lambda 1, lambda. 2,… ,λm, such that b=λ 1A1 +λ2a2+…. + lambda_1a_1 + lambda_2a_2 +…. + \ lambda_ma_mb lambda = lambda 1 a1 + 2 a2 +… + lambda mam

(3) Then vector B is A linear representation of vector group A, that is, equations x1A1 + x2A2 +… +xmam=bx_1a_1 + x_2a_2 + … + x_ma_m = bx1a1+x2a2+… + xmam = b has a solution

Theorem 1

Vector BBB can be determined by vector A: A1, A2… ,amA:a_1,a_2,… ,a_mA:a1,a2,… The necessary and sufficient condition for the linear representation of am is that the matrix A=(a1,a2… ,am)A=(a_1,a_2,… ,a_m)A=(a1,a2,… The rank of am is equal to the matrix B=(A1, A2… am,b)B = (a_1,a_2,… a_m,b)B=(a1,a2,… The rank of am, b)

It’s just a system of equations x1A1 +x2a2+… +xmam=bx_1a_1 + x_2a_2 + … + x_ma_m = bx1a1+x2a2+… +xmam=b has A solution because the linear equations Ax=bAx=bAx=b have A solution, if and only if R(A)=R(A,b)R(A)=R(A,b)R(A)=R(A,b) (Theorem 5 in the previous chapter)

Definition 3

Let’s say two vector sets A: A1, A2… ,amA:a_1,a_2,… ,a_mA:a1,a2,… , am a and B: b1, b2,… ,blB:b_1,b_2,…. ,b_lB:b1,b2,…. B can be linearly represented by group A if every vector in group B can be linearly represented by group A.

Vector groups A and B are said to be equivalent if they are represented linearly with each other

Let the vector group A=(a1,a2… ,am)A = (a_1,a_2,… ,a_m)A=(a1,a2,… ,am), vector group B=(b1,b2… ,bl)B=(b_1,b_2,… ,b_l)B=(b1,b2,… ,bl)

If BBB can be linearly represented by AAA, then for each vector bj(j=1,2… , l) b_j (j = 1, 2,… , l) bj (j = 1, 2,… ,l)

There are several k1j, k2j,… ,kmjk_{1j},k_{2j},… ,k_{mj}k1j,k2j,… , KMJ, make

To get

Then B=AKB =AKB =AK, where the matrix Km∗ L =(KIj)K_{m * L} =(K_{ij})Km∗ L =(kij) is called the coefficient matrix of this linear representation

The above vector groups A and B are combined with column vector groups. Now we discuss row vector groups to form A and B

set

Because any vector in B can be linearly represented by AAA

So there are

$b_j^T = k_{j1}a_1^T + k_{j2}a_2^T + … + k_{jm}a_m^T =(k_{j1},k_{j2},… ,k_{jm})\begin{pmatrix} a_1^T\\ a_2^T\\ .\\ .\\ .\\ a_m^T \end{pmatrix} (j \in [1,l])$

And then get

B is equal to KAB is equal to K, AB is equal to KA

According to the above, if Cm∗n=Am∗lBl∗nC_{m * n} = A_{m * L} B_{L * n}Cm∗n=Am∗lBl∗n, the column vector group of matrix C can be linearly represented by the column vector group of matrix A, and B is the coefficient matrix of this representation

CCC is BBB, AAA is AAA, so BBB is K (B=AK), K (B=AK), K (B=AK).

Meanwhile, if the row vector set of C can all be linearly represented by the row vector set of B, then A is the coefficient matrix of this representation

Using B=KAB =KAB =KA corresponding to C=ABC=ABC=AB you get A which is the matrix of coefficients

Theorem 2

Vector group B: b1, b2,… ,blB:b_1,b_2,… ,b_lB:b1,b2,… Bl can be determined by vector group A: A1, A2… ,amA:a_1,a_2,… ,a_mA:a1,a2,… The necessary and sufficient condition for the linear representation of am is that the matrix A=(a1,a2… ,am)A= (a_1,a_2,… ,a_m)A=(a1,a2,… The rank of am is equal to the matrix (A,B)=(A1… ,am,b1,… ,bl)(A,B) = (a_1,… ,a_m,b_1,… ,b_l)(A,B)=(a1,… ,am,b1,… ,bl) rank, i.e. R(A)=R(A,B)R(A) =R(A,B)R(A) =R(A,B)

instructions

Because BBB can be linearly represented by AAA

So there is a matrix of coefficients KKK, such that B is equal to AKB is equal to AKB is equal to AK

So you could say

There is at least one solution to AX=BAX =BAX =B

And because

System of linear equations Ax = bAx = bAx = b has A sufficient and necessary conditions of solutions is R (A) = R (A, b) R (A) = R (A, b) R (A) = R (A, b)

So R(A)=R(A,B)R(A) =R(A,B)

inference

Vector group A: a1, a2,… ,amA:a_1,a_2,… ,a_mA:a1,a2,… , AM and vector group B: B1, B2… ,blB:b_1,b_2,… ,b_lB:b1,b2,… , the sufficient and necessary conditions of bl equivalent is R = R (A) (B) = R (A, B) R = R (A) (B) = R (A, B) R = R (A) (B) = R (A, B)

Where, A and B are the matrices of vector groups A and B

Proof:

Because BBB can be linearly represented by AAA, then theorem 2 tells us that

$R(A) = R(A, B)$

And by the same token, AAA can be linearly represented by BBB, so it’s also going to be

$R(B) = R(B, A)$

And because

$R(A, B) = R(B, A)$

get

$R(A) = R(B) = R(A, B) = R(B, A)$

Proof done!

For example,

Case 1

A1 = [1122], a2 = [1212], a3 = (1-140], b = [1031] a_1 = \ begin {bmatrix} \ \ 1 \ \ 2 \ \ 2 \ end {bmatrix}, a_2 = \ begin {bmatrix} 1 2 1 \ \ \ \ \ \ 2 \end{bmatrix},a_3=\begin{bmatrix} 1\\ -1\\ 4\\ 0 \end{bmatrix},b=\begin{bmatrix} 1\\ 0\\ 3\\ 1 A1 = {bmatrix} \ end ⎣ ⎢ ⎢ ⎢ ⎡ 1122 ⎦ ⎥ ⎥ ⎥ ⎤, a2 = ⎣ ⎢ ⎢ ⎢ ⎡ 1212 ⎦ ⎥ ⎥ ⎥ ⎤, a3 = ⎣ ⎢ ⎢ ⎢ ⎡ 1-140 ⎦ ⎥ ⎥ ⎥ ⎤, b = ⎣ ⎢ ⎢ ⎢ ⎡ 1031 ⎦ ⎥ ⎥ ⎥ ⎤

It is proved that the vector BBB can be linearly represented by the vector groups A1, A2, A3a_1, A_2, a_3A1,a2,a3

Proof:

Set A = (a1, a2, a3), B = (A, B) = (a_1, a_2, a_3), B = (A, B) A = (a1, a2, a3), B = (A, B)

According to theorem 1, vector B can be linearly represented by vector groups A1, A2, A3a_1, A_2, a_3A1, A2, and A3

the

$R(A)=R(A,b)=R(B)$

So here we just need to simplify BBB to get the rank (R(B)R(B)R(B) R(A)R(A)R(A) R(A))

由

available

Transposition,

So x3 is equal to cx_3 is equal to cx3 is equal to c

launch

$x = \begin{pmatrix}

x_1\ x_2\ x_3 \end{pmatrix}$

so

B = (a1, a2, a3) x = (- 3 c + 2) a1 + a2 (2 c – 1) + ca3b = (a_1, a_2, a_3) x = (- 3 c + 2) a_1 + a_2 + – 1 (2 c) Ca_3b = (a1, a2, a3) x = (- 3 c + 2) a1 + a2 + ca3 (2 c – 1), including c can take any value

Case 2

Set a1 = [1-11-1), a2 = [3113], b1 = [2011], b2 = [1102], b3 = a_1 = \ [3-120] begin {bmatrix} \ \ 1 \ \ 1 \ \ 1 \end{bmatrix},a_2=\begin{bmatrix} 3\\ 1\\ 1\\ 3 \end{bmatrix},b_1=\begin{bmatrix} 2\\ 0\\ 1\\ 1 \end{bmatrix},b_2=\begin{bmatrix} 1\\ 1\\ 0\\ 2 \end{bmatrix}, b_3=\begin{bmatrix} 3\\ -1\\ 2\\ 0 A1 = {bmatrix} \ end ⎣ ⎢ ⎢ ⎢ ⎡ 1-11-1 ⎦ ⎥ ⎥ ⎥ ⎤, a2 = ⎣ ⎢ ⎢ ⎢ ⎡ 3113 ⎦ ⎥ ⎥ ⎥ ⎤, b1 = ⎣ ⎢ ⎢ ⎢ ⎡ 2011 ⎦ ⎥ ⎥ ⎥ ⎤, b2 = ⎣ ⎢ ⎢ ⎢ ⎡ 1102 ⎦ ⎥ ⎥ ⎥ ⎤, b3 = ⎣ ⎢ ⎢ ⎢ ⎡ 3-120 ⎦ ⎥ ⎥ ⎥ ⎤

Prove that vector groups A1, A2A_1, A_2A1, A2 are equivalent to vector groups B1, B2, B3B_1, B_2, B_3B1, B2,b3

Proof:

Set A = (a1, a2), B = (b1, b2, b3) A = (a_1, a_2), B = (b_1 b_2, b_3) A = (a1, a2), B = (b1, b2, b3)

The corollary of theorem 2 shows that AAA is equivalent to BBB

instructions

$R(A)=R(B)=R(A,B)$

So let’s simplify (A,B)(A,B)(A,B)

get

$R(A)=R(A,B)=2$

And it’s obvious that BBB has a second order subexpression that is not equal to 0

instructions

$R(B) \geq 2$

And because

$R(B) \leq R(A,B)=2$

So there are

$2 \leq R(B) \leq 2$

launch

$R(B)=2$

To sum up

$R(A)=R(B)=R(A,B)$

So AAA is the same thing as BBB

Theorem 3

Let vector group B=b1,b2… ,blB = b_1, b_2,… , b_lB=b1,b2,… Bl can be determined by vector group A: A1, A2… ,amA:a_1,a_2,… ,a_mA:a1,a2,… ,am is linear, then R(b1,b2… , bl) or less R (a1, a2,… ,an)R(b_1,b_2,… ,b_l) \leq R(a_1,a_2,… ,a_n)R(b1,b2,… , bl) or less R (a1, a2,… ,an)

Proof:

Make A = (a1, a2,… ,am)A=(a_1,a_2,… ,a_m)A=(a1,a2,… , am), B = (b1, b2,… ,bl)B=(b_1,b_2,… ,b_l)B=(b1,b2,… ,bl)

Because BBB can be linearly represented by AAA

Then there is

$R(A) = R(A, B)(theorem 2)$

Because R(B)<=R(A,B)R(B) <=R(A,B)R(B) <=R(A,B)

so

$R(B) <= R(A)$

summary

By the above law, inference can be obtained

Vector group B: b1, b2,… ,blB:b_1,b_2,… ,b_lB:b1,b2,… Bl can be determined by vector group A: A1, A2… ,amA:a_1,a_2,… ,a_mA:a1,a2,… The matrix KKK is linear to indicate Leftrightarrow so that the equation AK=BAK=BAK=B has a solution

conclusion

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