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Nickname: Haihong

Tag: programmer monkey | C++ contestant | student

Introduction: because of C language to get acquainted with programming, then transferred to the computer major, had the honor to get some national awards, provincial awards… Has been confirmed. Currently learning C++/Linux/Python

Learning experience: solid foundation + more notes + more code + more thinking + learn English well!

 

Machine learning little White stage

The article is only used as my own study notes for the establishment of knowledge system and review

Know what is, know why!

6.1 Definition and properties of linear space

Definition 1: Linear space

Let VVV be a non-empty set and R\mathbb{R}R be a real number field

If for any two elements α,β∈V\alpha,\beta \in Vα,β∈V, there is always a unique element γ∈V\gamma \in Vγ∈V, which is called the sum of α\alphaα and β\betaβ. As gamma = alpha + beta, gamma = \ alpha +, beta gamma = alpha + beta

For any number λ∈R,α∈V\lambda\in \mathbb{R},\alpha \in Vλ∈R,α∈V, there is always a unique element δ∈V\delta\in Vδ∈V, which is called the product of λ\lambdaλ and α\alphaα. As the delta = lambda alpha \ delta = \ lambda \ alpha delta = lambda alpha

And these two kinds of transportation meet eight operation rules:

1. 
   $\alpha+\beta=\beta+\alpha$
2. 
   $(\alpha+\beta)+\gamma=\alpha+(\beta+\gamma)$
3. There exists zero element 0\ boldSymbol00 in VVV, and for any α∈V\alpha\in Vα∈V, there is α+0=α\alpha+ boldSymbol0 =\alphaα+0=α
4. For any alpha, alpha and alpha in V ∈ V ∈ V, have alpha \ alpha negative elements of alpha beta \ beta \ beta in V ∈ V ∈ V, the alpha + beta = 0 \ \ beta alpha + = \ boldsymbol0 alpha + beta = 0
5. 
   $1\alpha=\alpha$
6. 
   $\lambda(\mu\alpha)=(\lambda \mu)\alpha$
7. 
   $(\lambda+\mu)\alpha=\lambda \alpha+\mu\alpha$
8. 
   $\lambda(\alpha+\beta)=\lambda \alpha+\lambda\beta$

Note: alpha, beta, gamma ∈ V. Lambda, u ∈ R \ alpha, \ beta, gamma \ in V. \ lambda, u \ \ mathbb in {R} alpha, beta, gamma ∈ V. Lambda, u ∈ R

Then VVV is called the vector space (or linear space) over the real number field R\mathbb{R}

In a nutshell

• Any addition and multiplication operation satisfying the above eight rules is called linear operation
• The set of linear operations defined is called a vector space

Properties of linear space

The nature of the 1

The zero element is unique

Proof (reduction)

Suppose there are two zero elements, 01,02∈V0_1,0_2 \in V01,02∈V

By definition of the zero element, there are

get

$0 _1 = 0 _2$

The zero element is unique

The nature of the 2

The negative element of any element is unique, and the negative element of α\alphaα is denoted −α-\alpha−α

Proof (reduction)

Suppose α∈V\alpha \in Vα∈V has two negative elements, denoted as β,γ\beta, gammaβ,γ

By the definition of a negative element, there are

again

$\beta=\beta+0=\beta+(\alpha+\gamma)=(\beta+\alpha)+\gamma=0+\gamma=\gamma$

namely

$\beta=\gamma$

In summary, the negative element of any element is unique

The nature of the three

(1) alpha (1) = 0 0 0 \ alpha = \ boldsymbol0 alpha (1) 0 = 0 (2) (1) – alpha = – alpha (2) (1) \ alpha = – \ alpha (2) (1) – alpha = – alpha lambda 0 = 0 (3) (3) \ lambda \boldsymbol0=\boldsymbol0 (3) λ0=0

(1)

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

get

$0\alpha=\boldsymbol0$

(2)

$\alpha+(-1)\alpha=(1-1)\alpha=\boldsymbol0$

According to the definition of the negative element, we get

$(-1)\alpha=-\alpha$

(3)

$\lambda \boldsymbol0=\lambda(\alpha+(-1)\alpha)=\lambda\alpha+(-\lambda)\alpha=(\lambda+(-\lambda))\alpha=0\alpha=\boldsymbol0$

namely

$\lambda \boldsymbol0=\boldsymbol0$

The nature of the 4

If λα=0\lambda \alpha= boldsymbol0λα=0, then λ=0\lambda=0λ=0 or α=0\alpha= boldsymbol0α=0

prove

(1) Certificate Adequacy:

When λ=0\lambda=0λ=0, λα=0\lambda \alpha= boldsymbol0λα=0

When the lambda indicates 0 \ lambda \ neq0 lambda  = 0,

Lambda alpha= 0 lambda alpha= boldsymbol0 lambda alpha= 0 multiply both sides by 1 lambda frac{1}{lambda}λ1

$\frac{1}{\lambda}(\lambda \alpha)=\frac{1}{\lambda}\boldsymbol0=\boldsymbol0$

And because

$\frac{1}{\lambda}(\lambda \alpha)=(\frac{1}{\lambda}\lambda)\alpha=1\alpha=\alpha$

launch

$\alpha=\boldsymbol0$

(2) Necessity of certification:

By lambda = 0 \ lambda = 0 lambda = 0 or alpha = 0 \ alpha = \ boldsymbol0 alpha = 0

It is easy to get λα=0\lambda \alpha=\boldsymbol0λα=0

In summary, if λα=0\lambda \alpha=\boldsymbol0λα=0, then λ=0\lambda=0λ=0 or α=0\alpha=\boldsymbol0α=0

For example,

Case 1

Note P[x]nP[x]_nP[x]n is a vector space, where P[x]nP[x]_nP[x]n represents all polynomials of degrees not exceeding n

$P[x]_n=\{a_nx^n+a_{n-1}x^{n-1}+… +a_1x+a_0|a_n,… ,a_1,a_0 \in \mathbb{R} \}$

---

prove

Proof closure of addition operation:

set

$\alpha\in P[x]_n,\beta\in P[x]_n$

There are

$\alpha + \beta \in P[x]_n$

α+β\alpha + betaα+β cannot have more than NNN, so the result also belongs to P[x]nP[x]_nP[x]n

Closure of syndrome multiplication operation:

set

$k\in \mathbb{R},\alpha \in P[x]_n$

There are

$k\alpha \in P[x]_n$

The number multiplication operation does not make the highest degree of P[x]nP[x]_nP[x]n exceed n, and the result is also in P[x]nP[x]_nP[x]n

Polynomial addition, number multiplication operation to meet the linear operation law, namely eight operation law, here will not be detailed

In summary, P[x]nP[x]_nP[x]n is vector space

Case 2

Note Q[x]nQ[x]_nQ[x]n is the vector space, where Q[x]nQ[x]_nQ[x]n is denoted by

$Q[x]_n=\{a_nx^n+a_{n-1}x^{n-1}+… +a_1x+a_0|a_n,… ,a_1,a_0 \in \mathbb{R}, and a_n\neq0 \}$

---

prove

Proof closure of addition operation:

set

$\alpha \in Q[x]_n,\beta \in Q[x]_n$

There are

$\alpha + \beta \in Q[x]_n$

Closure of syndrome multiplication operation:

set

$k\in \mathbb{R},\alpha \in Q[x]_n$

K αk\alphakα does not necessarily belong to Q[x]nQ[x]_nQ[x]n

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

$k\alpha=0\alpha=0$

Note that Q[x]nQ[x]_nQ[x]n an≠0a_n\neq0an=0, indicating that Q[x]nQ[x]_nQ[x]n must be non-zero

So the multiplication operation is not closed

In summary, Q[x]nQ[x]_nQ[x]n is not a vector space

Case 5

The whole of the positive real numbers is denoted as R+\mathbb{R}^+R+, where the addition and multiplication operations are defined as

It is proved that R+\mathbb{R}^+R+ constitute a linear space for the above addition and multiplication operations

prove

Proof closure of addition operation:

For any A,b∈R+a,b\in\mathbb{R}^+a,b∈R+

$a\oplus b=ab \in \mathbb{R}^+$

Closure of syndrome multiplication operation:

For any lambda ∈ R, a ∈ R + \ lambda \ \ mathbb in {R}, a \ \ mathbb in ^ + {R} lambda ∈ R, a ∈ R +, there is

$\lambda \odot a=a^{\lambda}\in \mathbb{R^+}$

Article 8 Operation rules:

I’m not going to prove it here, but it turns out to be all eight

But there’s a caveat

The zero element here is 1

For any A ∈R+a \in \mathbb{R^+}a∈R+, a⊕ =aa\oplus1= AA ⊕1=a

We just need to follow the definition of negative element and zero element, and then solve it according to our custom operation rules

summary

(1) To prove whether a set constitutes a vector space is definitely not to verify the closure of addition and number multiplication operations

(2) If the addition and number multiplication operation defined is not the usual addition and number multiplication operation between real numbers, it is necessary to prove whether the eight-point linear operation rule is satisfied

(3) To prove uniqueness, you can use the method of contradiction, assuming that more than one element exists at the same time, and then prove that these elements are equal.

conclusion

Description:

• Refer to textbook “linear algebra” fifth edition tongji University mathematics department
• With the book concept explanation combined with some of their own understanding and thinking

The essay is just a study note, recording a process from 0 to 1

Hope to help you, if there is a mistake welcome small partners to correct ~

I am haihong ଘ(੭, ᵕ)੭

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