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[machine learning | fundamentals of mathematics] linear algebra of Mathematics for machine learning series (22): definition and properties of linear space

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The articles

【 machine learning | Mathematical basis 】Mathematics for Machine Learning Linear algebra of series (1): Second and third order determinants 、 Total permutation and its reverse order number

【 machine learning | Mathematical basis 】Mathematics for Machine Learning Linear algebra of series (2):n Step determinant 、 exchange

【 machine learning | Mathematical basis 】Mathematics for Machine Learning Linear algebra of series (3): The nature of determinants

【 machine learning | Mathematical basis 】Mathematics for Machine Learning Linear algebra of series (4): Determinant by line ( Column ) an

【 machine learning | Mathematical basis 】Mathematics for Machine Learning Linear algebra of series (5): Kramer's law

【 machine learning | Mathematical basis 】Mathematics for Machine Learning Linear algebra of series (6): Matrix operation

【 machine learning | Mathematical basis 】Mathematics for Machine Learning Linear algebra of series (7): Inverse matrix

【 machine learning | Mathematical basis 】Mathematics for Machine Learning Linear algebra of series (8): Elementary transformation of matrix

【 machine learning | Mathematical basis 】Mathematics for Machine Learning Linear algebra of series (9): The rank of a matrix 、 Solutions of linear equations

【 machine learning | Mathematical basis 】Mathematics for Machine Learning Linear algebra of series (10): Vector group and its linear combination

【 machine learning | Mathematical basis 】Mathematics for Machine Learning Linear algebra of series (11): Linear correlation of vector groups

【 machine learning | Mathematical basis 】Mathematics for Machine Learning Linear algebra of series (12): The rank of a vector group

【 machine learning | Mathematical basis 】Mathematics for Machine Learning Linear algebra of series (13): Structure of solutions of linear equations

【 machine learning | Mathematical basis 】Mathematics for Machine Learning Linear algebra of series (14): Vector space

【 machine learning | Mathematical basis 】Mathematics for Machine Learning Linear algebra of series (15): The inner product of a vector 、 Length and orthogonality

【 machine learning | Mathematical basis 】Mathematics for Machine Learning Linear algebra of series (16): Eigenvalues and eigenvectors of square matrices

【 machine learning | Mathematical basis 】Mathematics for Machine Learning Linear algebra of series (17): Similarity matrix

【 machine learning | Mathematical basis 】Mathematics for Machine Learning Linear algebra of series (18): Diagonalization of symmetric matrices

【 machine learning | Mathematical basis 】Mathematics for Machine Learning Linear algebra of series (19): Quadratic form and its canonical form

【 machine learning | Mathematical basis 】Mathematics for Machine Learning Linear algebra of series (20): The quadratic form is transformed into standard form by formula method

【 machine learning | Mathematical basis 】Mathematics for Machine Learning Linear algebra of series (21): Positive definite quadratic form

6.1 Definition and properties of linear space

Definition 1: Linear space

set up V V V It's a Nonempty set , R \mathbb{R} R by Real number field

If for any two elements α , β ∈ V \alpha,\beta \in V α,βV, There is always one element γ ∈ V \gamma \in V γV With the corresponding , be called α \alpha α And β \beta β And , Write it down as γ = α + β \gamma=\alpha+ \beta γ=α+β

For any number λ ∈ R , α ∈ V \lambda\in \mathbb{R},\alpha \in V λR,αV, There is always one element δ ∈ V \delta\in V δV With the corresponding , be called λ \lambda λ And α \alpha α Product of , Write it down as δ = λ α \delta=\lambda \alpha δ=λα

And these two kinds of transportation meet Eight operation rules

  1. α + β = β + α \alpha+\beta=\beta+\alpha α+β=β+α
  2. ( α + β ) + γ = α + ( β + γ ) (\alpha+\beta)+\gamma=\alpha+(\beta+\gamma) (α+β)+γ=α+(β+γ)
  3. stay V V V There is a zero element in 0 \boldsymbol0 0, To any α ∈ V \alpha\in V αV, There are α + 0 = α \alpha+\boldsymbol0=\alpha α+0=α
  4. To any α ∈ V \alpha \in V αV, There are α \alpha α The negative element of β ∈ V \beta \in V βV, send α + β = 0 \alpha+\beta=\boldsymbol0 α+β=0
  5. 1 α = α 1\alpha=\alpha 1α=α
  6. λ ( μ α ) = ( λ μ ) α \lambda(\mu\alpha)=(\lambda \mu)\alpha λ(μα)=(λμ)α
  7. ( λ + μ ) α = λ α + μ α (\lambda+\mu)\alpha=\lambda \alpha+\mu\alpha (λ+μ)α=λα+μα
  8. λ ( α + β ) = λ α + λ β \lambda(\alpha+\beta)=\lambda \alpha+\lambda\beta λ(α+β)=λα+λβ

notes : α , β , γ ∈ V ; λ , u ∈ R \alpha,\beta,\gamma \in V;\lambda,u\in \mathbb{R} α,β,γV;λ,uR

that V V V It's called the real number field R \mathbb{R} R Upper Vector space ( Or linear space )

in short

  • All addition and multiplication that satisfy the above eight laws operation , It's called Linear operation
  • Any set that defines a linear operation , It's called vector space

Properties of linear spaces

nature 1

The zero element is unique

prove ( Reduction to absurdity )

Suppose there are two zero elements 0 1 , 0 2 ∈ V 0_1,0_2 \in V 01,02V

According to the definition of zero element , Yes

{ 0 1 + 0 2 = 0 1 ( 0 2 see become zero element plain ) 0 1 + 0 2 = 0 2 ( 0 1 see become zero element plain ) \begin{cases} 0_1 + 0_2=0_1(0_2 As a zero element )\\ 0_1 + 0_2=0_2(0_1 As a zero element )\\ \end{cases} { 01+02=01(02 see become zero element plain )01+02=02(01 see become zero element plain )

obtain

0 1 = 0 2 0_1=0_2 01=02

It shows that the zero element is unique

nature 2

The negative element of any element is unique , α \alpha α The negative element of is written as − α -\alpha α

prove ( Reduction to absurdity )

hypothesis α ∈ V \alpha \in V αV There are two negative elements , Write it down as β , γ \beta,\gamma β,γ

According to the definition of negative element , Yes

{ α + β = 0 α + γ = 0 \begin{cases} \alpha + \beta = 0\\ \alpha + \gamma = 0 \end{cases} { α+β=0α+γ=0

also

β = β + 0 = β + ( α + γ ) = ( β + α ) + γ = 0 + γ = γ \beta=\beta+0=\beta+(\alpha+\gamma)=(\beta+\alpha)+\gamma=0+\gamma=\gamma β=β+0=β+(α+γ)=(β+α)+γ=0+γ=γ

namely

β = γ \beta=\gamma β=γ

Sum up , The negative element of any element is unique

nature 3

( 1 ) 0 α = 0 (1)0\alpha=\boldsymbol0 10α=0
( 2 ) ( − 1 ) α = − α (2)(-1)\alpha=-\alpha 2(1)α=α
( 3 ) λ 0 = 0 (3)\lambda \boldsymbol0=\boldsymbol0 3λ0=0

Prove (1)

α + 0 α = 1 α + 0 α = ( 1 + 0 ) α = α \alpha+0\alpha=1\alpha+0\alpha=(1+0)\alpha=\alpha α+0α=1α+0α=(1+0)α=α

obtain

0 α = 0 0\alpha=\boldsymbol0 0α=0

Prove (2)

α + ( − 1 ) α = ( 1 − 1 ) α = 0 \alpha+(-1)\alpha=(1-1)\alpha=\boldsymbol0 α+(1)α=(11)α=0

According to the definition of negative element , obtain

( − 1 ) α = − α (-1)\alpha=-\alpha (1)α=α

Prove (3)

λ 0 = λ ( α + ( − 1 ) α ) = λ α + ( − λ ) α = ( λ + ( − λ ) ) α = 0 α = 0 \lambda \boldsymbol0=\lambda(\alpha+(-1)\alpha)=\lambda\alpha+(-\lambda)\alpha=(\lambda+(-\lambda))\alpha=0\alpha=\boldsymbol0 λ0=λ(α+(1)α)=λα+(λ)α=(λ+(λ))α=0α=0

namely

λ 0 = 0 \lambda \boldsymbol0=\boldsymbol0 λ0=0

nature 4

If λ α = 0 \lambda \alpha=\boldsymbol0 λα=0, be λ = 0 \lambda=0 λ=0 or α = 0 \alpha=\boldsymbol0 α=0

prove

(1) Proof of adequacy :

When λ = 0 \lambda=0 λ=0 when , λ α = 0 \lambda \alpha=\boldsymbol0 λα=0

When λ ≠ 0 \lambda\neq0 λ=0 when ,

equation λ α = 0 \lambda \alpha=\boldsymbol0 λα=0 Ride on both sides 1 λ \frac{1}{\lambda} λ1, have to

1 λ ( λ α ) = 1 λ 0 = 0 \frac{1}{\lambda}(\lambda \alpha)=\frac{1}{\lambda}\boldsymbol0=\boldsymbol0 λ1(λα)=λ10=0

Again because

1 λ ( λ α ) = ( 1 λ λ ) α = 1 α = α \frac{1}{\lambda}(\lambda \alpha)=(\frac{1}{\lambda}\lambda)\alpha=1\alpha=\alpha λ1(λα)=(λ1λ)α=1α=α

Introduction

α = 0 \alpha=\boldsymbol0 α=0

(2) The necessity of proof :

from λ = 0 \lambda=0 λ=0 or α = 0 \alpha=\boldsymbol0 α=0

It's easy to get λ α = 0 \lambda \alpha=\boldsymbol0 λα=0

Sum up , If λ α = 0 \lambda \alpha=\boldsymbol0 λα=0, be λ = 0 \lambda=0 λ=0 or α = 0 \alpha=\boldsymbol0 α=0

give an example

example 1

explain P [ x ] n P[x]_n P[x]n It's vector space , among P [ x ] n P[x]_n P[x]n Indicates that the number of times does not exceed n All of the polynomials of

P [ x ] n = { a n x n + a n − 1 x n − 1 + . . . + a 1 x + a 0 ∣ a n , . . . , a 1 , a 0 ∈ R } P[x]_n=\{a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0|a_n,...,a_1,a_0 \in \mathbb{R} \} P[x]n={ anxn+an1xn1+...+a1x+a0an,...,a1,a0R}


prove

Prove the closure of addition operation :

set up

α ∈ P [ x ] n , β ∈ P [ x ] n \alpha\in P[x]_n,\beta\in P[x]_n αP[x]n,βP[x]n

Yes

α + β ∈ P [ x ] n \alpha + \beta \in P[x]_n α+βP[x]n

α + β \alpha + \beta α+β The number of times of any item in the will not exceed n n n, So the result also belongs to P [ x ] n P[x]_n P[x]n

The closure of the multiplication of the number of certificates :

set up

k ∈ R , α ∈ P [ x ] n k\in \mathbb{R},\alpha \in P[x]_n kR,αP[x]n

Yes

k α ∈ P [ x ] n k\alpha \in P[x]_n kαP[x]n

Multiplication does not make P [ x ] n P[x]_n P[x]n The maximum number of times exceeded n, The result is also in P [ x ] n P[x]_n P[x]n in

The addition of polynomials 、 The number multiplication operation satisfies the law of linear operation , That is, eight operation laws , No more details here

Sum up , P [ x ] n P[x]_n P[x]n It's vector space

example 2

explain Q [ x ] n Q[x]_n Q[x]n It's vector space , among Q [ x ] n Q[x]_n Q[x]n Expressed as

Q [ x ] n = { a n x n + a n − 1 x n − 1 + . . . + a 1 x + a 0 ∣ a n , . . . , a 1 , a 0 ∈ R , And a n ≠ 0 } Q[x]_n=\{a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0|a_n,...,a_1,a_0 \in \mathbb{R}, And a_n\neq0 \} Q[x]n={ anxn+an1xn1+...+a1x+a0an,...,a1,a0R, And an=0}


prove

Prove the closure of addition operation :

set up

α ∈ Q [ x ] n , β ∈ Q [ x ] n \alpha \in Q[x]_n,\beta \in Q[x]_n αQ[x]n,βQ[x]n

Yes

α + β ∈ Q [ x ] n \alpha + \beta \in Q[x]_n α+βQ[x]n

The closure of the multiplication of the number of certificates :

set up

k ∈ R , α ∈ Q [ x ] n k\in \mathbb{R},\alpha \in Q[x]_n kR,αQ[x]n

obtain k α k\alpha kα It doesn't necessarily belong to Q [ x ] n Q[x]_n Q[x]n

The special case is when k = 0 k=0 k=0 when

k α = 0 α = 0 k\alpha=0\alpha=0 kα=0α=0

Be careful Q [ x ] n Q[x]_n Q[x]n In the definition a n ≠ 0 a_n\neq0 an=0 , explain Q [ x ] n Q[x]_n Q[x]n It must be non-zero

So number multiplication is not closed

Sum up , Q [ x ] n Q[x]_n Q[x]n It's not vector space

example 5

The totality of positive real numbers , Write it down as R + \mathbb{R}^+ R+, In which the addition and multiplication operations are defined as

{ a ⊕ b = a b ( a , b ∈ R + ) λ ⊙ a = a λ ( λ ∈ R , a ∈ R + ) \begin{cases} a\oplus b=ab(a,b\in \mathbb{R}^+)\\ \lambda \odot a=a^{\lambda}(\lambda\in \mathbb{R},a\in \mathbb{R^+} ) \end{cases} { ab=ab(a,bR+)λa=aλ(λR,aR+)

Test certificate R + \mathbb{R}^+ R+ The above addition and multiplication operations form a linear space

prove

Prove the closure of addition operation :

To any a , b ∈ R + a,b\in\mathbb{R}^+ a,bR+, There are

a ⊕ b = a b ∈ R + a\oplus b=ab \in \mathbb{R}^+ ab=abR+

The closure of the multiplication of the number of certificates :

To any λ ∈ R , a ∈ R + \lambda\in \mathbb{R},a\in \mathbb{R^+} λR,aR+, Yes

λ ⊙ a = a λ ∈ R + \lambda \odot a=a^{\lambda}\in \mathbb{R^+} λa=aλR+

Prove eight operation laws :

It's not proved one by one here , The result is that all eight operation laws meet

But it should be noted that

At this time The zero element is 1

That is, for any a ∈ R + a \in \mathbb{R^+} aR+, Yes a ⊕ 1 = a a\oplus1=a a1=a

Just follow the negative element 、 Definition of zero element , Then we can solve it according to our custom operation rules

Summary

(1) Prove whether a set constitutes a vector space , Definitely not only To verify the addition 、 Closure of number multiplication operation

(2) When the defined addition and multiplication operations are not the usual addition and multiplication operations between real numbers , It also needs to prove whether eight linear operation laws are satisfied

(3) When proving uniqueness , have access to Reduction to absurdity , Suppose there are multiple elements at the same time , Then prove that these elements are equal .

Conclusion

explain :

  • Refer to textbook 《 linear algebra 》 The fifth edition Department of mathematics, Tongji University
  • Cooperate with the concept explanation in the book Combined with some of my own understanding and thinking

The article is only for study notes , Record from 0 To 1 A process of

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