Explore
For Enterprise
Join for Free
Log In

예술 & 인문학
비즈니스
컴퓨터 공학
데이터 과학
정보 기술
건강
수학 및 논리
자기개발
물리 과학 및 공학
사회 과학
언어 학습
학위 및 수료증

Explore all of Coursera

검색
포 엔터프라이즈 (For Enterprise)
로그인
무료 회원 가입

Calculating eigenvectors

Loading...

Mathematics for Machine Learning: Linear Algebra

임페리얼 칼리지 런던

4.7(2,958개의 평가) | 74K명의 학생이 등록함

Mathematics for Machine Learning 전문 분야에서 3의 강좌 1

강의 계획서 보기

In this course on Linear Algebra we look at what linear algebra is and how it relates to vectors and matrices. Then we look through what vectors and matrices are and how to work with them, including the knotty problem of eigenvalues and eigenvectors, and how to use these to solve problems. Finally we look at how to use these to do fun things with datasets - like how to rotate images of faces and how to extract eigenvectors to look at how the Pagerank algorithm works. Since we're aiming at data-driven applications, we'll be implementing some of these ideas in code, not just on pencil and paper. Towards the end of the course, you'll write code blocks and encounter Jupyter notebooks in Python, but don't worry, these will be quite short, focussed on the concepts, and will guide you through if you’ve not coded before. At the end of this course you will have an intuitive understanding of vectors and matrices that will help you bridge the gap into linear algebra problems, and how to apply these concepts to machine learning.

배우게 될 기술

Eigenvalues And Eigenvectors, Basis (Linear Algebra), Transformation Matrix, Linear Algebra

검토

4.7(2,958개의 평가)

5 stars
2,158 ratings
4 stars
638 ratings
3 stars
114 ratings
2 stars
31 ratings
1 star
17 ratings

NS

Dec 23, 2018

Professors teaches in so much friendly manner. This is beginner level course. Don't expect you will dive deep inside the Linear Algebra. But the foundation will become solid if you attend this course.

VC

Jul 11, 2018

The content of the course is very relevant, and the instructors are really fun and helpful.My only suggestion is to upload revisions for each assessment, so we can understand what we are doing wrong.

수업에서

Eigenvalues and Eigenvectors: Application to Data Problems

Eigenvectors are particular vectors that are unrotated by a transformation matrix, and eigenvalues are the amount by which the eigenvectors are stretched. These special 'eigen-things' are very useful in linear algebra and will let us examine Google's famous PageRank algorithm for presenting web search results. Then we'll apply this in code, which will wrap up the course.

Special eigen-cases3:32

Calculating eigenvectors10:07

강사:

David Dye
Professor of Metallurgy
Samuel J. Cooper
Lecturer
A. Freddie Page
Strategic Teaching Fellow

[MUSIC]

Hopefully, you will all now have a reasonable feeling for

what an eigen-problem looks like geometrically.

So in this video, we're going to formalise this concept into an algebraic expression,

which will allow us to calculate eigenvalues and

eigenvectors whenever they exist.

Once you've understood this method, we'll be in a good position to see

why you should be glad that computers can do this for you.

If we consider a transformation A,

what we have seen is that if it has eigenvectors at all, then these are simply

the vectors which stay on the same span following a transformation.

They can change length and even point in an opposite direction entirely.

But if they remain in the same span, they are eigenvectors.

If we call our eigenvector x, then we can say the following expression.

Ax = lambda x.

Where, on the left hand side, we're applying the transformation matrix A to a vector x.

And on the right-hand side,

we are simply stretching a factor x by some scalar factor lambda.

So lambda is just some number.

We're trying to find values of x that make the two sides equal.

Another way of saying this is that for our eigenvectors,

having A apply to them just scales their length or does nothing at all,

which is the same as scaling the length by a factor of 1.

So in this equation, A is an n dimensional transform,

meaning it must be an n by n square matrix.

The eigenvector x must therefore be an n-dimensional vector.

To help us find the solutions to this expression,

we can rewrite it by putting all the terms on one side and then factorizing.

So

(A - lambda I) x = 0.

If you're wondering where the I term came from,

it's just an n by n identity matrix, which means it's a matrix the same size as A but

with ones along the leading diagonal and zeros everywhere else.

We didn't need this in the first expression we wrote,

as multiplying vectors by scalars is defined.

However, subtracting scalars from matrices is not defined, so

the I just tidies up the maths, without changing the meaning.

Now that we have this expression we can see that for the left-hand side to

equal 0, either the contents of the brackets must be 0 or the vector x is 0.

So we're actually not interested in the case where the vector x is 0.

That's when it has no length or direction and is what we call a trivial solution.

Instead, we must find when the term in the brackets is 0.

Referring back to the material in the previous parts of the course, we can

test if a matrix operation will result in a 0 output by calculating its determinant.

So,

det (A - lambda I) = 0.

Calculating the determinants manually is a lot of work for high dimensional matrices.

So let's just try applying this to an arbitrary two by two transformation.

Let's say, A =

(a, b, c, d).

Substituting this into our eigen-finding expression gives the following.

Det (a b c d)-

lambda 0 0 lambda.

Evaluating this determinant, we get what is referred to as

the characteristic polynomial, which looks like this.

So lambda squared

-(a + d) lambda

+ ad - bc = 0.

Our eigenvalues are simply the solutions of this equation, and we can then plug

these eigenvalues back into the original expression to calculate our eigenvectors.

Rather than continuing with our generalized form, this is a good moment to

apply this to a simple transformation, for which we already know the eigensolution.

Let's take the case of a vertical scaling by a factor of two,

which is represented by the transformation matrix A = 1, 0, 0, 2.

We can then apply the method that we just described and

take the determinant of A minus lambda I and then set it to zero and solve.

So det (1- lambda,

0, 0, 2- lambda)

= 1- lambda, 2- lambda,

which is of course equal to 0.

This means that our equation must have solutions at lambda equals 1 and

lambda equals 2.

Thinking back to our original eigen-finding

formula, (A - lambda I) x = 0.

We can now sub these two solutions back in.

So thinking about the case where lambda = 1,

we can say (1- 1, 0, 0, 2- 1)

times this x vector,

x1 and x2, must equal to

(0, 0, 0, 1) x1, x2,

therefore we've got 0 and x2 must equal 0.

Now, thinking about the case where lamda equals 2,

at lamda = 2, you get 1- 2, and 2- 2,

And then you get of course minus 1, 0, 0, 0,

Which equals to minus x1, 0,

which equals zero.

So what do these two expressions tell us?

Well, in the case where our eigenvalue lambda equals one,

we've got an eigenvector where the x2 term must be zero.

But we don't really know anything about the x1 term.

Well, this is because, of course any vector that points along

the horizontal axis could be an eigenvector of this system.

So we write that by saying @ lambda = 1,

x, our eigenvector, can equal anything along the horizontal axis,

as long as it's 0 in the vertical direction.

So we put in an arbitrary parameter t.

Similarly for the lambda = 2 case,

We can say that our eigenvector must equal 0,t.

Because as long as it doesn't move at all in the horizontal direction,

any vector that's purely vertical would therefore also

be an eigenvector of this system, as they all would lie along the same span.

So now we have two eigenvalues, and their two corresponding eigenvectors.

Let's now try the case of a rotation by 90-degrees anti-clockwise, to ensure that

we get the result that we expect which, if you remember, is no eigenvectors at all.

The transformation matrix corresponding to a 90-degree rotation is as follows.

A = (0, -1), (1, 0).

So applying the formula once again we

get the det (0- lambda- 1),

(1, 0- lambda),

which if you calculate this through,

comes out to lambda squared + 1 = 0.

Which doesn't have any real numbered solutions at all.

Hence, no real eigenvectors.

We can still calculate some complex eigenvectors using imaginary numbers, but

this is beyond what we need for this particular course.

Despite all the fun that we've just been having, the truth is that you will almost

certainly never have to perform this calculation by hand.

Furthermore, we saw that our approach required finding the roots of

a polynomial of order n, i.e., the dimension of your matrix.

Which means that the problem will very quickly stop being

possible by analytical methods alone.

So when a computer finds the eigensolutions of

a 100 dimensional problem it's forced to employ iterative numerical methods.

However, I can assure you that developing a strong conceptual understanding of

eigen problems will be much more useful than being really good at calculating them

by hand.

In this video, we translated our geometrical understanding of eigenvectors

into a robust mathematical expression, and validated it on a few test cases.

But I hope that I've also convinced you that working through lots of eigen-problems,

as is often done in engineering undergraduate degrees, is not a good

investment of your time if you already understand the underlying concepts.

This is what computers are for.

Next video, we'll be referring back to the concept of basis change to

see what magic happens when you use eigenvectors as your basis.

See you then.

[MUSIC]

Coursera 카탈로그 살펴보기

무료로 참여해 맞춤화된 추천, 업데이트 및 제안을 받아보세요.

Coursera는 세계 최고의 교육에 대한 보편적인 접근 기회를 제공하며, 최고의 대학 및 기관들과 제휴하여 온라인 강좌를 제공합니다.

© 2019 Coursera Inc. All rights reserved.

앱스토어에서 다운로드하기

구글 플레이에서 받기

Coursera
소개
리더십
직업
카탈로그
수료증
MasterTrack™ 수료증
학위
포 엔터프라이즈 (For Enterprise)
정부용

커뮤니티
Learners
Partners
Developers
Beta Testers
Translators

연결
블로그
Facebook
Linkedin
Twitter
Instagram
YouTube
기술 블로그

이상
조건
개인정보 보호
도움말
접근성
보도 자료
문의하기
디렉토리
계열사