Vector space

有關於向量空間的一些筆記，放在這裡以備不時之需

Vector space & subspaces

= The set of all n-dimensional vectors.
A subspace in the vector space must satisfies:
- If , in , is in
- If is in , is in
= The set all combinations of &
- Does equal to ?
  - 我們可以試著證明看看是否還滿足封閉性

A is a matrix
There are two important subspaces of A:
- Column space
  - which is in
  - any in
- Null space
  - which is in
We said that is solvable if and only if b can be expressed as combination of columns of A
- 換句話說，b要落在(的column space)中
- 就是vectors in
Nullspace of ()
- The set of all solutions to
- We know that is a subspace because:
  - Let's suppose , are two vector in , that is
  - has infinite solutions iff
    - in other words, has exactly one sol. if
  - Are all (which b is not zero)'s solution a subspace?
    - NO
    - If the solution set of has not path through the origin, that means we can combine two solutions of , then generate a new vector which is not in the solution set of

Is there a systematic way that can help us to find ?
- Yes.
- Suppose:
Reduce A by Gauss Elimination，then get the (upper triangular form) matirx
We notice that and are 1，which are pivot variables
Let's analyse by formula's perspective:
Let's say and are free variables
- ,
- So the pivot variable are:
  - ,
Finally, we can get the Complete solution
Let's say that Complete solution is linear combination of Special solution, which is:
= span of columns of
Number of special sol. = number of free variable
We know how to find N(A) now, but can we do it faster?
- Reduce row Echelon form
- Recall matrix, we'll keep reduce it by Jordan Elimination, finally get the matrix:
Follow the step below to get N from R:
1. we know the free variables are and , so fill the N like this(put the matrix into row 2 and row 4 of ): 2.Multiply remaining elements in by -1, then put them in N
= #pivot of A
- #free variable = n -

Suppose Ax = b, let's say x is General solution, which means:
- means complete homogeneous solution to
- is solution to which all free variables are 0
So,
這就代表只要存在，就一定會有解
如果存在，則有機會無限多解
如何檢驗?
- r = n r < n
  
  r = m 1. 3.
  
  r < m 2. 4.
1. 的維度與相同，且也與未知數個數相同(無free variable)
  - 唯一解，
2. 的維度小於，且沒有free variable
  - Full column rank
  - 無解或者唯一解
3. 的維度與相同，有free variable
  - Full row rank
  - 不僅存在，連也存在
  - 無限多解
4. 的維度小於，有free variable
  - 無解或者無限多解

	r = n	r < n
r = m	1.	3.
r < m	2.	4.

v1, v2, ..., vn are independent if
- only the trivial combination gives zero vector.
- otherwise, they are dependent.
How to find the independent vectors in a span?
- Reduce the matrix to ，the column vector which has pivot is independent vector.
We can also say that, a matrix is linear independet if
- full column rank
A basis of a subspace is a set of vecotr v1, ..., vn
- are independent
Dimension
- dim V = number of basis vecotrs in V
conclusion
- The columns of are independent
- ()
- rank = n(no free variable)
- - 也就是說A中的向量剛剛好就是basis中的向量

Column space
- Notation is
- In space
- The set of all combination of columns of
Null space
- Notation is
- In space
- The set of solutions to
Row space
- Notation is
- In space
- The set of all combination of rows of
Left Nullspace
- Notation is
- In space
- The set of solutions of
- The reason why we call it "Left":
  - , ,
  - 乘在A的左邊，使其為零，因此得名
Finding basis of these four space
The basis of
- Pivot cols of
The basis of
- Columns of the nullspace matrix
- = # free variables =
The basis of
- Pivot rows of
- Notice that but
The basis of
- There is a matrix call that can reduce to
Rows of generating zero rows in are basis of
上述的內容即為The rank theory
The invertible theorem
- is matrix(square)
  6. columns of form basis of
  7. columns of are independent
- 1, 2指的是沒有free variable，自然是唯一解
- 或從空間的角度來看，只要能夠在n為空間中提供n個pivot，那麼組成任意向量即可行
- 這一點與3,4,5,6,7說的是一樣的事情
- 應該說是，這7點其實都在講差不多的事情