Changing basis

Loading...

En provenance du cours de Imperial College London

Mathematics for Machine Learning: Linear Algebra

1973 notes

Imperial College London

Mathematics for Machine Learning: Linear Algebra

1973 notes

Cours 1 sur 3 dans Specialization Mathematics for Machine Learning

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.

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 independence4:18

Applications of changing basis3:28

Rencontrer les enseignants

David Dye
Professor of Metallurgy
Department of Materials
Samuel J. Cooper
Lecturer
Dyson School of Design Engineering
A. Freddie Page
Strategic Teaching Fellow
Dyson School of Design Engineering

[MUSIC]

Now, so far we haven't really talked about the coordinate system of our vector space,

the coordinates in which all of our vectors exist.

But it turns out in doing this thing of projecting,

of taking the dot product, we are 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 coordinate systems and

we'll do a few cases of changing from one coordinate system to another.

So, remember that a vector, like this guy r here if just an object that takes us

from the origin to some point in space.

Which could be some physical space or it could be some data space,

like bedrooms and thousands of Euros 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 called them i and j before,

I'm going to give them names e1 and e2.

I'm going to define them to be of unit lengths, so

I'm going to give them a little hat, meaning they're of unit length, and

I'm going to define them to be the vectors 1,0, and 0,1.

And if I had more dimension in my space, I could have e3 hat, e4 hat,

e5 hat, e 1 million hat, whatever.

Here the instruction then is that r is going to be equal

to doing a vector sum of 2e1 or 3e1 and then some number of e2.

So we'll call it to 3e1 hats plus 4e2 hats.

And so we'll write it down as a little list 3,4.

So 3,4 here is the instructions to do 3e1 hats plus 4e2 hats.

If you think about it, my choice of e1 hat and e2 hat here is kind of arbitrary.

It depends entirely on the way I set up the coordinates.

There's no reason I couldn't have set up some co ordinate system

at some angle to that, or you can use vectors to find the axis

that weren't even at 90 degrees to each other or 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, say, another set of vectors b

I'll call b1 here in the vector 2,1, and

I could have another vector here, b2 as the vector -2,4.

And I've defined it in terms of the coordinates e.

And I could then describe r in terms of you, you're using, those vectors b1 and b2.

It's just the numbers in r would be different.

So, we call the vectors we used to define this 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 referred to these basis vectors e is 3,4.

But r referred to the basis vectors b also exists.

We just don't know what the numbers are in there.

So this should be kind of amazing.

r, the vector r, has some existence in a deep sort of mathematical sense completely

independently of the coordinate system we use to describe, the numbers in

the list describing r.

r, the vector takes us from there,

from the origin to there still exists, independently of the numbers used in her.

Which is kind of neat, right?

Sort of fundamentally, sort of idea.

Now, if the new basis vectors, these guys b are at 90 degrees to each other,

then it turns out the projection product 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 b's are in terms of e.

So here I've described b1 as being 2,1,

as being e1 plus e2 twice e1 plus 1 e2.

And I've described b2 as being minus 2e1's plus 4e2's.

And if I know b in terms of e, I'm going to be able to

use the projection product to find r described in terms of the b's.

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 axes, from the e to the b set of 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 computationally 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 computations 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 scalar projection is the shadow of r1 to b1.

And the number of the scalar projection describes how much of this vector I need.

And 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 protection 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.

And 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 up their

vector sum, I'll then have our b being the numbers in those two vector projections.

And 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?

Well, 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 0, then cosine theta is 0, and

cosine theta is 0 if they're 90 degrees to each other, if they're orthogonal.

So I don't even need to calculate a thing, so I just calculate the dot product.

So b1.b2 here, I take 2 times minus 2 and

I add it to 1 times 4, which is, minus 4 plus 4 which is 0.

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

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

So having talked through it, let's now do it numerically.

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

and r is pink right, if I take r in the basis e,

and I'm going to dot him with b1,

and the vector projection divides by the length of b1 squared.

So r in e dotted with b1 is going to be 3

times 2 plus 4 times 1, 4 times 1,

divided by the length of b1 squared.

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

So that's 2 squared plus 1 squared.

So that gives me 6 plus 4 is 10, divided by 5 which is 2.

So this projection here is of length 2 times b1.

So that projection there,

that vector is going to be 2 times b1.

So that is in terms of the original set of vectors e,

r_e.b1 over b1 squared times b1 is 2 times

the vector 2,1, is the vector 4,2.

And I can do then now this projection onto e2.

So I can do r_e dotted with b2 and

divide by the length of b2 squared,

and r_e.b2 is 3 times minus 2,

plus 4 times 4.

Divided by the length of b2 squared, which is -2 squared plus 4 squared.

So that's 3 times minus 2 is minus 6 plus 4 times 4 is 16,

so that's minus 6, plus 16 is 10.

Divided by this length here is 4 squared is 16, 2 squared is 4, so

20, so that's equal to a half.

So this vector projection here is that guy times b2,

so that's r_e.b2 over the modulus of b2 squared, that's my half,

times the vector b2, so that's a half times the vector b2 -2,4.

Now if I add those two together,

4,2 this bit, that vector projection,

plus this vector projection, so

this guy is going to be half b2 plus half -2,4 is -1,2.

If I add those together, I've got 3,4,

which is just my original vector r, 3,4 in the basis e.

So in the basis of b1 and b2, r_b is going to be 2

one-half, very nice, 2,1/2.

So actually in the basis b, it's going to be 2,1/2, there.

So r_b is 2 times b1 plus one-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 our vector describing our data isn't tied

to the axis that we originally used to describe it at all.

We can redescribe it using some other axis, some other basis vectors.

So the basis vectors we use to describe the space of data, and

choosing them carefully to help us solve our problem,

will be a very important thing in intermediate algebra, and in general.

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

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

We can do that change just by taking the dot or projection product in

the case where the new basis factors are orthogonal to each other.

Explorer notre catalogue

Rejoignez-nous gratuitement et obtenez des recommendations, des mises à jour et des offres personnalisées.

Coursera propose un accès universel à la meilleure formation au monde, en partenariat avec des universités et des organisations du plus haut niveau, pour proposer des cours en ligne.

© 2019 Coursera Inc. Tous droits réservés.

Télécharger dans l'App Store

Disponible sur Google Play