The main task of part I is to solve linear equation
A vector space is determined by its basis, for example :
Or in short, they are columns of identity matrix . Descartes coordinate is an excellent application that using this basis to describe elements in dimensional space. Also, they are orthogonal to each other. Orthogonality will dramatically simplify the calculation in many cases. I will use some geometry examples to explain.
1. Plane system
A plane equals to a two-dimensional space. Descartes coordinate ( system) is a classic way to describe the system. Interpretion into linear algebra, it chooses as basis. As we know, there are many other bases that can also successfully span , for example: . What are the relationship between different bases?
Here show the first basis and show the second one. Obviously, rotate the first basis anti-clockwisely for 45 degrees and then you can get the second basis. From the perspective of linear algebra, it is natural to show that matrix [22−222222] plays a role of "rotation" :
For the general case, for example vector , assume that the direction of mirror is perpendicular to a unit vector . Recall the projection matrix:
projects onto a line, projects onto the orthogonal complement of this line. That is:
Since the direction of mirror is perpendicular to unit vector , the image will be composed of: one component parallel to mirror and the other component anti-parallel to unit vector . Therefore:
This transformation is called reflection matrix (Householder transformation):
2. Area and Volume
Algebra is a powerful tool to understand geometry. In lecture 1, I proved two equations by considering two vectors :
where is the rotation angle from to .
LHS(Left-hand side) of the first equation is the inner product between and LHS of the second equation is the determinant of . Notice that the absolute value of RHS of the second equation is the two times area of the triangle.
Consider a general case. There are three points forming a triangle. To calculate the area of this triangle, translate one of its vertex to the origin. Therefore:
And the area:
A much elegant way:
(Prove that two deteminants are equal.)
If you exchange and in columns, the determinant comes to the opposite. Without taking absolute value, does it have some meaning? The answer is yes, but quite artificial: Define the positive or negative value as the "direction of area" along z-axis:
Now I have extended the plane to three-dimensional space. are the unit vectors along positive x-, y-, z- axis. Rewrite the determinant in symbolically:
and are two points lying in x-y plane.
Now consider two arbitrary points in space: These two points with origin will form a plane, which is also a subspace.
Defnition 2.1: The cross product of and is a vector:
The vector is perpendicular tp and . The cross product .
The length of this vector satisfy:
is the space angle between and .
Since is a vector, we can take its inner product with a third vector . This triple product is in fact a determinant:
It is remarkable that the triple product equals to the volume of the box with sides .
3. Orthogonal Bases and Gram-Schmidt
All bases discussed in section 1 are orthogonal, for example rotation matrix:
Moveover:
They are unit vectors.
Definition 3.1: The vectors are orthonormal if:
Here is an indicator function: If , , else .
Let the vectors be the columns of matrix . Then:
Choosing a proper basis can simplfy many calculations. For example, computing projection of vector space . That is:
If is an orthonormal basis, then:
For any projection :
is the inner product between the basis and the vector to be projected. Also you may find:
Exactly the generalized Pythagoras theorem but not necessary to choose the special basis .
Gram-Schmidt Process
Now we need to find a way to create orthonormal vectors. Gram-Schmidt process is one possible way.
Suppose a basis . (Of course they are linearly independent.) Start from and . I want to separate from to make them orthogonal as well as linearly independent. A straight way is to make projection from to :
Here . By fundamental theorem 2, should be orthogonal to . What does this decomposition tell us? The first part can be present by , while the second pary can only be produced by . Therefore, the only important components of is . In short:
You can check spans the same subspace that does.
Again, deplete from the subspace of by decomposition in projection:
Therefore, the "efficient" part of is:
That is:
NOTICE: You should use not to calculate because the part you deplete from subspace should also be orthogonal to .
In principle, you can repeatedly do this process until :
Once you get all :
unify the basis. is an orthonormal basis of space .
Such a construction is to:
Distribute into sub-orthonormal subspace created byProjection in each directions: .
It finally gives: , or:
In matrix form:
Gram-Schmit: From independent basis , columns of , it constructs orthonromal basis , columns of . The matrices with these columns satisfy , is upper-triangular.
4. Exercise
Exercise 1.1: For , which is a unit vector, prove:
for every vector .
(Hint: ).
Exercise 2.1: Prove
Exercise 3.2: In least square approximation, if has no solution, we can use the solution to to approximate. If has independent columns, then it can be factorized into , where is a orthonormal basis. Proof:
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Basis is not unique.Given a space S: every basis for the space has the same number of vectors.The number is the dimension of the space,called dim S.基不是唯一的,當給定一個空間S時,它的每一組基都含有相同數量的向量。
The lecture will focus on his recent work with Columbia University's Center for Resilient Cities & Landscapes. Please join us, either in-person or online.
Professor of Department of Philosophy, Peking UniversityAbstract: In October 1917, in the middle of the war, when its outcome was still uncertain, Max Weber delivered a public lecture
But I slowly came to feel that abandoning the concept of a lecture in favor of more 「modern」 teaching techniques was not the way to go, at least for me.然而,經過嘗試,這些教學方法似乎都不是適合我的完美方案。我喜歡做解釋。