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Changing basis

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Mathematics for Machine Learning: Linear Algebra

Imperial College London

4.7 (3,033 notes) | 75K étudiants inscrits

Cours 1 de 3 dans Mathematics for Machine Learning Spécialisation

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.

Visualiser le programme de cours

Compétences que vous apprendrez

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

Avis

4.7 (3,033 notes)

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PL

Aug 26, 2018

Great way to learn about applied Linear Algebra. Should be fairly easy if you have any background with linear algebra, but looks at concepts through the scope of geometric application, which is fresh.

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.

À partir de la leçon

Vectors are objects that move around space

In this module, we look at operations we can do with vectors - finding the modulus (size), angle between vectors (dot or inner product) and projections of one vector onto another. We can then examine how the entries describing a vector will depend on what vectors we use to define the axes - the basis. That will then let us determine whether a proposed set of basis vectors are what's called 'linearly independent.' This will complete our examination of vectors, allowing us to move on to matrices in module 3 and then start to solve linear algebra problems.

Changing basis11:24

Basis, vector space, and linear independence4:13

Applications of changing basis3:28

Enseigné par

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

Transcription

Now, so far, we haven't really

talked about the coordinate system of our vector space,

the coordinates in which all our vectors exist.

But it turns out that in doing this thing of projecting,

of taking the dot-product,

we're projecting our vector onto one.

Which we might use as

part of a new definition of the coordinate system.

So in this video, we'll look at

what we mean by a coordinate system,

and then we'll do a few case of changing from

one coordinate system to another.

So remember that a vector like this guy r here is

just an object that takes us from

the origin into some point in space.

Which could be some physical space

or could be some space of data,

like bedrooms, and thousands of years

for the house, or something like that.

What we haven't talked about so far really,

is the coordinate system that we use to describe space.

So we could use

a coordinate system defined by

these two vectors here. I'm going to give them names.

We call them i and j before,

I'm going to give names e1 and

e2 and found to be of unit length.

So I'm going to give them a little hat

meaning they have unit length.

I'm going to define them to be the vectors one,

zero and zero one.

If I have more dimensions in my space,

I could have e3 hat,

e4 hat, e5 hat,

e million hat, whatever.

Here the instruction then is that r is going to

be equal to doing a vector sum of two e1s or three e1s,

and then some number of e2s.

So we'll call it during 3 e1s hats plus 4 e2 hats,

and some will write it down as a little list, three four.

So always, the three four,

here's the instruction to do 3 e1 hats plus 4 e2 hats.

But if you think about it, my choice of

e1 hat and e2 hat is arbitrary.

It depends entirely on

the way I set up the coordinate system.

There's no reason I couldn't have set up

some quota system, at some angle to that.

Or even use vectors to find the axis

that weren't even at 90 degrees to each other,

were of different lengths.

I could still have described r as being some sum of

some vectors I used to define the space.

So I could have another set of vectors b,

I'll call b1 here in the vector two one,

and I could have another vector here

b2 as the vector minus two four,

and I've defined it in terms of the coordinates

e. I can then describe r in terms of,

your using those vectors b1 and b2,

is just the numbers in r would be different.

So we call the vectors we use to define the space,

these guys e or these guys b,

we call them basis vectors.

So the numbers I've used to define r,

only have any meaning when I

know about the basis vectors.

So r refer to these basis vectors e is three four,

but r referred to the basis vectors b also exists.

We start out with the numbers earlier.

So this should be amazing,

the vector r has some existence

in a deep mathematical sets,

completely independently of the coordinate system

we use to describe the numbers in the list,

describing r. All the vector

that takes us from there from the origin to that,

still exist, independently of

the numbers we used in r. This is neat.

Right. So the fundamentally idea.

Now, if the new basis vectors, these guys b,

are at 90 degrees to each other,

then it turns out that projection products

has a nice application.

We can use the projection or dot-product to find out

the numbers for r in the new basis b,

so long as we know what the bs are in

terms of e. So here I've described b1 as being two one,

as being e1 plus e2,

twice e1 plus e2.

I've described b2 as being minus two e1s plus four e2s.

If I know b in terms of e,

I'm going to be able to do,

use the projection product to

find r described in terms of the bs.

But this is a big if,

the b1 and b2 have to be at 90 degrees to each other.

If they're not we end up being in big trouble and need

matrices to do what's

called a transformation of axis from

the e to the b based on basis vectors.

We'll look at matrices later,

but this will help us out a lot for now.

Using dot-products in this special case where

the new basis vectors are orthogonal to each other,

is computation a lot faster and easier,

it's just less generic.

But if you can arrange the new axis

to be orthogonal, you should,

because it makes the calculations much

faster and easier to do.

So you can see that if I project r down onto b1,

so I look down from here,

and project down at 90 degrees,

I get a length here for scalar product,

and that's scalar projection is the shadow of r to b1.

A number of the scalar projection describes

how much of this factor I need.

The vector projection is going

to actually give me a vector in the direction of b1,

of length equal to that projection.

Now, if I take the vector projection of

r onto b2 going this way,

I'm going to get a vector in the direction of

b2 of length equal to that projection.

If I do a vector sum of that vector projection,

plus this guy's vector projection,

I'll just get r. So if I

can do those two vector projections,

and add that their vector sum,

I'll then have our b

being the numbers in those two vector projections.

So I found how to get from r

in the e set of basis vectors,

to the b set of basis vectors.

Now, how do I check that

these two new basis vectors

are at 90 degrees to each other?

I just take the dot-product.

So we said before the dot-product cos Theta was

equal to the dot of two vectors together,

so b1 and b2,

divided by their lengths.

So if b1.b2 is zero,

then cos Theta zero,

if they're at 90 degrees to

each other, they are orthogonal.

So I don't even need calculate length.

So I'll just calculate the dot-product.

So b1.b2 here, I take two times minus two,

and I add it to one times four,

which is minus four plus four which is zero.

So these two vectors are at 90 degrees to each other.

So it's going to be safe to do the projection.

So having thought through it,

let's now do it numerically.

So if I want to know what r described in the basis e,

and I'll use pink to write.

If I take r in the basis e,

and I'm going to dot him with b1.

The vector projection divides

by the length of b1 squared.

So r in e dotted with b1 is going to be three times two,

plus four times one,

divided by the length of b1 squared.

So that's the sum of the squares of the components of b,

that's two squared plus one squared.

So that gives me, six plus four is 10,

divided by five which is two.

So this projection here is of length two times b1.

So that projection there,

that vector is going to be two times b1.

So that is, in terms of the original set of

vectors b, original vectors e,

re.b1 over b1 squared,

times b1 is two times the vector two one,

is the vector four two.

I can do, then now this projection onto e2.

So I can do re dotted with b2 I divided by b.

The length of b2 squared,

and re.b2 is three times minus two,

plus four times four,

divided by the length of b2 squared,

which is minus two squared plus four squared.

So that's, three times minus two is minus six,

plus four times four is 16.

So that's a minus six plus 16 is 10,

divided by this length here is,

four squared is 16, two squared is four so 20.

So that's equal to a half.

So this vector projection here is that guy times b2.

So that's re.b2 over the modulus of b2 squared,

that's my half, times the vector b2.

So that's a half,

times the vector b2,

which is minus two four.

Now, if I add those two together,

four two, this bit,

that vector projection, plus this vector projection.

So this guy is going to be a half b2,

plus half minus two four is minus one two.

If I add those together,

I've got three four,

which is just my original vector r three four in

the basis e. So in the basis of b1 and b2,

rb is going to be two a half.

Very nice. Two a half.

So actually in the basis b it's going to

be two a half there.

So rb is 2 times b1 plus a half times b2. Very nice.

So I've converted from the e set of

basis vectors to the b set of

basis vectors, which is very neat.

Just using a couple of dot-products.

So this is really handy, this is really cool.

We've seen that a vector describing our data,

isn't tied to the axes that we

originally used to describe it at all.

We can re-describe it using some other axis,

some other basis vectors.

So the basis vectors were use to

describe the space of data,

and choosing them carefully to help us solve a problem

will be a very important thing in

any algebra, and in general.

What we've seen is we can move the numbers in the vector,

we use to describe a data item from one basis to another.

We can do that change just by

taking the dots or projection product,

in the case where

the new basis vectors are orthogonal to each other.

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