Placeholder Image

字幕表 動画を再生する

  • The following content is provided under a Creative

  • Commons license.

  • Your support will help MIT OpenCourseWare

  • continue to offer high quality educational resources for free.

  • To make a donation, or to view additional materials

  • from hundreds of MIT courses, visit MIT OpenCourseWare

  • at ocw.mit.edu.

  • GILBERT STRANG: So this is a pretty key lecture.

  • This lecture is about principal component analysis, PCA--

  • which is a major tool in understanding a matrix of data.

  • So what is PCA about?

  • Well first of all, let me remember what

  • was the whole point of last--

  • yesterday's lecture-- the singular value decomposition,

  • that any matrix A could be broken into r rank 1 pieces--

  • r being the rank of the matrix.

  • And each piece has a U times a V transpose.

  • And the good-- special thing is, the U's are orthonormal,

  • and also, the V's are orthonormal.

  • OK.

  • So that's the whole matrix.

  • But we have a big matrix, and we want

  • to get the important information out of it--

  • not all the information.

  • And people say, in machine learning,

  • if you've learned all the training data, you haven't

  • learned anything, really.

  • You've just copied it all in.

  • The whole point of neural nets and the process

  • of machine learning is to learn important facts about the data.

  • And now, here we're at the most basic stage of that.

  • And I claim that the important facts about the matrix

  • are in its largest k singular values--

  • the largest k pieces.

  • We can take-- k equal 1 would tell us

  • the largest single piece.

  • But maybe we have space and computing power

  • to handle a hundred pieces.

  • So I would take k equal 100.

  • The matrix might have ranked thousands.

  • So I claim that Ak is the best.

  • Now here's the one theorem for today, that Ak--

  • using the first k pieces of the SVD--

  • is the best approximation to A of rank k.

  • So I'll write that down.

  • So that really says why the SVD is perfect.

  • OK.

  • So that statement says, that if B--

  • another matrix-- has rank k, then the distance from A to B--

  • the error you're making in just using B--

  • that error is greater than or equal to the error

  • you make for the best guy.

  • Now that's a pretty straightforward,

  • beautiful fact.

  • And it goes back to people who discovered

  • the SVD in the first place.

  • But then a couple of psychologists

  • gave a proof in a later paper--

  • and it's often called the Eckart-Young Theorem.

  • There is the theorem.

  • Isn't that straightforward?

  • And the hypothesis is straightforward.

  • That's pretty nice.

  • But of course, we have to think, why is it true?

  • Why is it true?

  • And to give meaning to the theorem,

  • we have to say what these double bars are.

  • Do you know the right name for this?

  • So that double bar around a matrix is called the--

  • the norm of the matrix, the norm.

  • So I have to say something about matrix norms.

  • How big is-- that's a measure of how big it is.

  • And what I have to say is, there are many different measures

  • of a matrix--

  • how large that matrix is.

  • Let me tell you, for today, three possible measures

  • of a matrix.

  • So different ways to measure--

  • I'll call the matrix just A, maybe.

  • But then I'm going to apply the measure to A minus B,

  • and to A minus AK, and show that that is smaller.

  • OK.

  • So I want to tell you about the norm of A--

  • about some possible norms of A. And actually, the norms I'm

  • going to take today will be--

  • will have the special feature that they can be found--

  • computed by their singular values.

  • So let me mention the L2 norm.

  • That is the largest singular value.

  • So that's an important measure of the--

  • sort of the size of a matrix.

  • I'm talking here about a general m by n matrix

  • A. Sigma 1 is an important norm--

  • often called the L2 norm.

  • And that's where that index 2 goes.

  • Oh.

  • I should really start with vectors--

  • norms of vectors-- and then build to the norms of matrices.

  • Let me do norms of vectors over on this side.

  • The L2 norm of a vector--

  • do we know what that is?

  • That's the regular length of the vector that we all expect--

  • the square root of v1 squared up to vn squared.

  • The hypotenuse-- the length of the hypotenuse

  • in n dimensional space.

  • That's the L2 norm, because of that 2.

  • The L1 norm of a vector is just add up those pieces

  • without squaring and square rooting them.

  • Just add them.

  • That's the L1 norm.

  • And you might say, why do we want two norms?

  • Or there are more norms.

  • Let me just tell you one more.

  • The infinity norm-- and there is a reason for the 1 and the 2

  • and the infinity--

  • is the largest of the v's.

  • OK.

  • Have you met norms before?

  • I don't know.

  • These are vector norms, but maybe you have met.

  • Then we're going to have matrix norms, that maybe will be new.

  • So this is the norm that we usually think of.

  • But this one has become really, really important,

  • and let me tell you just why.

  • And then we'll-- later section of the notes and a later

  • lecture in this course will develop that--

  • develop this.

  • This is the L1 norm.

  • So this is L2, L1, and L infinity--

  • [INAUDIBLE]

  • So what's special about this one?

  • Well, it just turned out-- and it was only discovered

  • in our lifetimes--

  • that when you minimize some function using the L1 norm,

  • you minimize some, let's say, signal the noise,

  • or whatever you minimize--

  • some function.

  • If you use L1, the winning vector--

  • the minimizing vector-- turns out to be sparse.

  • And what does sparse mean?

  • Sparse means mostly zero components.

  • Somehow, when I minimize in L2--

  • which historically goes back to Gauss,

  • the greatest mathematician of all time.

  • When you minimize something in L2, you do the least squares.

  • And you find that the guy that gives you the minimum

  • has a lot of little numbers--

  • lot of little components.

  • Because when you're square those little ones,

  • they don't hurt much.

  • But Gauss-- so Gauss didn't do least L1 norm.

  • That has different names-- basis pursuit.

  • And it comes into signal processing and sensing.

  • Right.

  • And then it was discovered that if you minimize--

  • as we'll see in that norm--

  • you amazingly get-- the winning vector has--

  • is mostly zeros.

  • And the advantage of that is that you can understand

  • what its components are.

  • The one with many small components,

  • you have no interpretation for that answer.

  • But for an answer that just has a few non-zero components,

  • you really see what's happening.

  • And then this is a important one, too.

  • OK.

  • Now I'm going to turn just to-- so

  • what's the property of a norm?

  • Well, you can see that the norm of C times a vector is--

  • just multiplying by 6, or 11, or minus

  • pi, or whatever-- is the size of C. Norms

  • have that nice property.

  • They're homogeneous, or whatever word.

  • If you double the vector, you should double the norm--

  • double the length.

  • That makes sense.

  • And then the important property is that--

  • is the famous triangle in equality--

  • that if v and w are two sides of a triangle,

  • and you take the norm of v and add to the norm of w--

  • the two sides--

  • you get more than the straight norm along the hypotenuse.

  • Yeah.

  • So those are properties that we require,

  • and the fact that the norm is positive, which is--

  • I won't write down.

  • But it's important too.

  • OK.

  • So those are norms, and those will apply also

  • to matrix norms.

  • So if I double the matrix, I want to double its norm.

  • And of course, that works for that 2 norm.

  • And actually, probably-- so the triangle in equality for this

  • norm is saying that the largest singular value of A plus B--

  • two matrices-- is less or equal to the larger

  • the singular value of A plus the largest singular value of B.

  • And that's-- we won't take class time to check minor,

  • straightforward things like that.

  • So now I'm going to continue with the three norms

  • that I want to tell you about.

  • That's a very important one.

  • Then there is another norm that's named--

  • has an F. And it's named after Frobenius.

  • Sorry about that.

  • And what is that norm?

  • That norm looks at all the entries in the matrix--

  • just like it was a long vector--

  • and squares them all, and adds them up.

  • So in a way, it's like the 2 norm for a vector.

  • It's-- so the squared--

  • or shall I put square root?

  • Maybe I should.

  • It's the square root of all the little people in the matrix.

  • So a1, n squared, plus the next a2, 1 squared, and so on.

  • You finally get to a-m-n squared.

  • You just treat the matrix like a long vector.

  • And take this square root just like so.

  • That's the Frobenius norm.

  • And then finally, not so well known,

  • is something that's more like L1.

  • It's called the nuclear norm.

  • And not all the faculty would know about this nuclear norm.

  • So it is the sum of the sigma of the singular values.

  • I guess there are r of them.

  • So that's where we would stop.

  • Oh, OK.

  • So those are three norms.

  • Now why do I pick on those three norms?

  • And here's the point--

  • that for those three norms, this statement is true.

  • I could cook up other matrix norms

  • for which this wouldn't work.

  • But for these three highly important norms,

  • this Eckart-Young statement, that the closest rank k

  • approximation is found from the first k pieces.

  • You see, that's a good thing, because this is

  • what we compute from the SVD.

  • So now we've solved an approximation problem.

  • We found the best B is Ak.

  • And the point is, it could use all the-- any of those norms.

  • So there would be a--

  • well, somebody finally came up with a proof that

  • does all three norms at once.

  • In the notes, I do that one separately from Frobenius.

  • And actually, I found--

  • in an MIT thesis--

  • I was just reading a course 6 PhD thesis--

  • and the author-- who is speaking tomorrow, or Friday in IDSS--

  • Dr. [? Cerebro ?] found a nice new proof of Frobenius.

  • And it's in the notes, as well as an older proof.

  • OK.

  • You know, as I talk here, I'm not too sure

  • whether it is essential for me to go through the proof,

  • either in the L2 norm--

  • which takes half a page in then notes--

  • or in the Frobenius norm, which takes more.

  • I'd rather you see the point.

  • The point is that, in these norms-- and now,

  • what is special about these norms of a matrix?

  • These depend only on the sigmas--

  • only on the-- oh.

  • Oh.

  • I'll finish that sentence, because it was true.

  • These norms depend only on the singular values.

  • Right?

  • That one, at least, depends only on the singular value.

  • It's the largest one.

  • This one is the sum of them all.

  • This one comes into the Netflix competition, by the way.

  • This was the right norm to win a zillion dollars in the Netflix

  • competition.

  • So what did Netflix put-- it did a math competition.

  • It had movie preferences from many, many Netflix subscribers.

  • They gave their ranking to a bunch of movies.

  • But of course, they hadn't seen--

  • none of them had seen all the movies.

  • So the matrix of rankings--

  • where you had the ranker and the matrix--

  • is a very big matrix.

  • But it's got missing entries.

  • If the ranker didn't see the movie, he isn't--

  • he or she isn't ranking it.

  • So what's the idea about Netflix?

  • So they offered like a million dollar prize.

  • And a lot of math and computer science people

  • fought for that prize.

  • And over the years, they got like higher 92, 93, 94% right.

  • But it turned out that this was--

  • well, you had to--

  • in the end, you had to use a little psychology

  • of how people voted.

  • So it was partly about human psychology.

  • But it was also a very large matrix problem

  • with an incomplete matrix--

  • an incomplete matrix.

  • And so it had to be completed.

  • You had to figure out what would the ranker have

  • said about the post if he hadn't seen it,

  • but had ranked several other movies, like All

  • the President's Men, or whatever--

  • given a ranking to those?

  • You have to-- and that's a recommender system, of course.

  • That's how you get recommendations from Amazon.

  • They've got a big matrix calculation here.

  • And if you've bought a couple of math books,

  • they're going to tell you about more math books--

  • more than you want to know.

  • Right.

  • OK.

  • So anyway, it just turned out that this norm

  • was the right one to minimize.

  • I can't give you all the details of the Netflix competition,

  • but this turned out to be the right norm

  • to do a minimum problem, a best not least squares.

  • These squares would look at some other norm,

  • but a best nuclear norm completion of the matrix.

  • And that-- and now it's--

  • so now it's being put to much more serious uses for MRI--

  • magnetic resonance stuff, when you go in and get--

  • it's a noisy system, but you get--

  • it gives a excellent picture of what's going on.

  • So I'll just write Netflix here.

  • So it gets in the--

  • and then MRIs.

  • So what's the point about MRIs?

  • So if you don't--

  • if you stay in long enough, you get all the numbers.

  • There isn't missing data.

  • But if you-- as with a child--

  • you might want to just have the child

  • in for a few minutes, then that's

  • not enough to get a complete picture.

  • And you have, again, missing data

  • in your matrix in the image from the MRI.

  • So then, of course, you've got to complete that matrix.

  • You have to fill in, what would the MRI have

  • seen in those positions where it didn't look long enough?

  • And again, a nuclear norm is a good one for that.

  • OK.

  • So there will be a whole section on norms, maybe just about--

  • in stellar by now.

  • OK.

  • So I'm not going to--

  • let me just say, what does this say?

  • What does this tell us?

  • I'll just give an example.

  • Maybe I'll take-- start with the example that's in the notes.

  • Suppose k is 2.

  • So I'm looking among all rank 2 matrices.

  • And suppose my matrix is 4, 3, 2, 1, and all the rest 0's.

  • Diagonal.

  • And it's rank 4 matrix.

  • I can see its singular values.

  • They're sitting there.

  • Those would be the singular values, and the eigenvalues,

  • and everything, of course.

  • Now, what would be A2?

  • What would be the best approximation

  • of rank 2 to that matrix, in this sense to be completed?

  • What would A2 do?

  • Yeah.

  • It would be 4 and 3.

  • It would pick the two largest.

  • So I'm looking at Ak.

  • This is k to the 2, so it has to have rank 2.

  • This has got rank 4.

  • The biggest pieces are those.

  • OK.

  • So this thing says that if I had any other matrix B,

  • it would be further away from A than this guy.

  • It says that this is the closest.

  • And I just-- could you think of a matrix that could possibly

  • be closer, and be rank 2?

  • Rank two 2 the tricky thing.

  • The matrices of rank 2 form a kind of crazy set.

  • If I add a rank 2 matrix to a rank 2 matrix,

  • probably the rank is up to 4.

  • So the rank 2 matrices are all kind of floating around

  • in their own little corners.

  • This looks like the best one.

  • But in the notes I suggest, well, you could get a rank 2--

  • well, what about B?

  • What about this B?

  • For this guy, I could get closer--

  • maybe not exact-- but closer, maybe by taking 3.5, 3.5.

  • But I only want to use rank--

  • I've only got two rank 2 to play with.

  • So I better make this into a rank--

  • I have to make this into a rank 1 piece, and then

  • the 2 and the 1.

  • So you see what I--

  • what I thought of?

  • I thought, man, maybe that's better--

  • like on the diagonal, I'm coming closer.

  • Well, I'm not getting it exactly here.

  • But then I've got one left to play with.

  • And I'll put, maybe, 1.5 down here.

  • OK.

  • So that's a rank 2 matrix--

  • two little rank 1s.

  • And on the diagonal, it's better.

  • 3.5s-- I'm only missing by a half.

  • 1.5s-- I'm missing by half.

  • So I'm only missing by a half on the diagonal

  • where this guy was missing by 2.

  • So maybe I've found something better.

  • But I had to pay a price of these things off the diagonal

  • to keep the rank low.

  • And they kill me.

  • So that B will be further away from A.

  • The error, if I computed A minus B, and computed its norm,

  • I would see bigger than A minus A2.

  • Yeah.

  • So, you see the point of the theorem?

  • That's really what I'm trying to say, that it's not obvious.

  • You may feel, well, it's totally obvious.

  • Pick 4 and 3.

  • What else could do it?

  • But it depends on the norm and so on.

  • So it's not--

  • Eckart-Young had to think of a proof, and other people, too.

  • OK.

  • So that's-- now, but you could say-- also say--

  • object that I started with a diagonal matrix here.

  • That's so special.

  • But what I want to say is the diagonal matrix

  • is not that special, because I could take A--

  • so let me now just call this diagonal matrix D--

  • or let me call it sigma to give it

  • another sort of appropriate name.

  • So if I thought of matrices, what I want to say

  • is, this could be the sigma matrix.

  • And there could be a U on the left of it,

  • and a sigma on the right of it.

  • So A is U sigma V transpose.

  • So this is my sigma.

  • And this is like any orthogonal matrix U.

  • And this is like any V transpose.

  • Right?

  • I'm just saying, here's a whole lot more matrices.

  • There is just one matrix.

  • But now, I have all these matrices

  • with Us multiplying on the left, and V transpose ones

  • on the right.

  • And I ask you this question, what

  • are the singular values of that matrix, A?

  • Here the singular values were clear--

  • 4, 3, 2, and 1.

  • What are the singular values of this matrix A,

  • when I've multiplied by a orthogonal guy on both sides?

  • That's a key question.

  • What are the singular values of that one?

  • AUDIENCE: 4, 3, 2, 1.

  • GILBERT STRANG: 4, 3, 2, 1.

  • Didn't change.

  • Why is that?

  • Because the singular values are the--

  • because this has a SVD form--

  • orthogonal times diagonal times orthogonal.

  • And that diagonal contains the singular values.

  • What I'm saying is, that my--

  • and our-- trivial little example here, actually

  • was all 4 by 4's that have these singular values.

  • I could-- my whole problem is orthogonally invariant,

  • a math guy would say.

  • When I multiply by U or a V transpose, or both--

  • the problem doesn't change.

  • Norms don't change.

  • Yeah, that's a point.

  • Yeah.

  • I realize it now.

  • This is the point.

  • If I multiply the matrix A by an orthogonal matrix U,

  • it has all the same norms--

  • doesn't change the norm.

  • Actually, that was true way back for vectors with this length--

  • with this length.

  • What's the deal about vectors?

  • Suppose I have a vector V, and I've computed

  • its hypotenuse and the norm.

  • And now I look at Q times V in that same 2 norm.

  • What's special about that?

  • So I took any vector V and I know what its length is--

  • hypotenuse.

  • Now I multiply by Q.

  • What happens to the length?

  • Doesn't change.

  • Doesn't change.

  • Orthogonal matrix-- you could think

  • of it as just like rotating the triangle in space.

  • The hypotenuse doesn't change.

  • And we've checked that, because we could--

  • the check is to square it.

  • And then you're doing QV, transpose QV.

  • And you simplify it the usual way.

  • And then you have Q transpose Q equal the identity.

  • And you're golden.

  • Yeah.

  • So the result is you get the same answer as V.

  • So let me put it in a sentence now, pause.

  • Multiplying that norm is not changed by orthogonal matrix.

  • And these norms are not changed by orthogonal matrices,

  • because if I multiply the A here by an orthogonal matrix,

  • I have--

  • this is my A. If i multiply by a Q,

  • then I have QU sigma V transpose.

  • And what is really the underlying point?

  • That QU is an orthogonal matrix just as good as U. So if I--

  • let me put this down.

  • QA would be QU sigma V transpose.

  • And now I'm asking you, what's the singular value

  • decomposition for QA?

  • And I hope I may actually-- seeing it.

  • What's the singular value decomposition of QA?

  • What are the singular values?

  • What's the diagonal matrix?

  • Just look there for it.

  • The diagram matrix is sigma.

  • What goes on the right of it?

  • The V transpose.

  • And what goes on the left of it is QU.

  • Plus, that's orthogonal times orthogonal.

  • Everybody in this room has to know

  • that if I multiply two orthogonal matrices,

  • the result is, again, orthogonal.

  • So I can multiply by Q, and it only affects the U part, not

  • the sigma part.

  • And so it doesn't change any of those norms.

  • OK.

  • So that's fine.

  • That's what I wanted to say about the Eckart-Young

  • Theorem--

  • not proving it, but hopefully giving you

  • an example there of what it means--

  • that this is the best rank to approximate that one.

  • OK.

  • So that's the key math behind PCA.

  • So now I have to--

  • want to, not just have to-- but want to tell you about PCA.

  • So what's that about?

  • So we have a bunch of data, and we want to see--

  • so let me take a bunch of data--

  • bunch of data points--

  • say, points in the plane.

  • So I have a bunch of data points in the plane.

  • So here's my data vector.

  • First, vector x1-- well, x.

  • Is at a good--

  • maybe v1.

  • These are just two component guys.

  • v2.

  • They're just columns with two components.

  • So I'm just measuring height and age,

  • and I want to find the relationship between height

  • and age.

  • So the first row is meant-- is the height of my data.

  • And the second row is the ages.

  • So these are-- so I've got say a lot of people,

  • and these are the heights and these are the ages.

  • And I've got n points in 2D.

  • And I want to make sense out of that.

  • I want to look for the relationship between height

  • and age.

  • I'm actually going to look for a linear row relation

  • between height and age.

  • So first of all, these are all over the place.

  • So the first step that a statistician does,

  • is to get mean 0.

  • Get the average to be 0.

  • So what is-- so all these points are all over the place.

  • From row 1, the height, I subtract the average height.

  • So this is A-- the matrix I'm really going to work on is

  • my matrix A-- minus the average height--

  • well, in all components.

  • So this is a, a, a, a--

  • I'm subtracting the mean, so average height and average age.

  • Oh, that was a brilliant notation,

  • a sub a can't be a sub a.

  • You see what the matrix has done--

  • this matrix 2 means?

  • It's just made each row of A. Now adds to row.

  • Now add to what?

  • If I have a bunch of things, and I've

  • subtracted off their mean--

  • so the mean, or the average is now 0--

  • then those things add up to--

  • AUDIENCE: Zero.

  • GILBERT STRANG: Zero.

  • Right.

  • I've just brought these points into something like here.

  • This is age, and this is height.

  • And let's see.

  • And by subtracting, it no longer is

  • unreasonable to have negative age and negative height,

  • because--

  • so, right.

  • The little kids, when I subtract it off the average age,

  • they ended up with a negative age.

  • The older ones ended up still positive.

  • And somehow, I've got a whole lot of points,

  • but hopefully, their mean is now zero.

  • Do you see that I've centered the data at 0, 0?

  • And I'm looking for-- what am I looking for here?

  • I'm looking for the best line.

  • That's what I want to find.

  • And that would be a problem in PCA.

  • What's the best linear relation?

  • Because PCA is limited.

  • PCA isn't all of deep learning by any means.

  • The whole success of deep learning

  • was the final realization, after a bunch of years,

  • that they had to have a nonlinear function in there

  • to get to model serious data.

  • But here's PCA as a linear business.

  • And I'm looking for the best line.

  • And you will say, wait a minute.

  • I know how to find the best line, just use least squares.

  • Gauss did it.

  • Can't be all bad.

  • But PCA-- and I was giving a talk

  • in New York when I was just learning about it.

  • And somebody said, what you're doing

  • with PCA has to be the same as least squares--

  • it's finding the best line.

  • And I knew it wasn't, but I didn't know how

  • to answer that question best.

  • And now, at least, I know better.

  • So the best line in least squares--

  • can I remind you about least squares?

  • Because this is not least squares.

  • The best line of least squares--

  • so I have some data points.

  • And I have a best line that goes through them.

  • And least squares, I don't always center the data

  • to mean zero, but I could.

  • But what do you minimize in least squares--

  • least squares?

  • If you remember the picture in linear algebra

  • books of least squares, you measure the errors--

  • the three errors.

  • And it's how much you're wrong at those three points.

  • Those are the three errors.

  • A-- difference between Ax and B--

  • the B minus Ax that you square.

  • And you add up those three errors.

  • And what's different over here?

  • I mean, there's more points, but that's not the point.

  • That's not the difference.

  • The difference is, in PCA, you're measuring perpendicular

  • to the line.

  • You're adding up all these little guys, squaring them.

  • So you're adding up their squares and minimizing.

  • So the points-- you see it's a different problem?

  • And therefore it has a different answer.

  • And this answer turns out to involve the SVD, the sigmas.

  • Where this answer, you remember from ordinary linear algebra,

  • just when you minimize that, you got

  • to an equation that leads to what equation for the best x?

  • So do you remember?

  • AUDIENCE: [INAUDIBLE]

  • GILBERT STRANG: Yeah.

  • What is it now?

  • Everybody should know.

  • And we will actually see it in this course,

  • because we're doing the heart of linear algebra here.

  • We haven't done it yet, though.

  • And tell me again, what equation do I solve for that problem?

  • AUDIENCE: A transpose A.

  • GILBERT STRANG: A transpose A x hat equal A transpose b.

  • Called the normal equations.

  • It's sort part of--

  • it's this regression in statistics language.

  • That's a regression problem.

  • This is a different problem.

  • OK.

  • Just so now you see the answer.

  • So that involves-- well, they both

  • involve A transpose A. That's sort of interesting,

  • because you have a rectangular matrix A,

  • and then sooner or later, A transpose A is coming.

  • But this involves solving a linear system of equations.

  • So it's fast.

  • And we will do it.

  • And it's very important.

  • It's probably the most important application in 18.06.

  • But it's not the same as this one.

  • So this is now in 18.06, maybe the last day is PCA.

  • So I didn't put those letters--

  • Principal Component Analysis-- PCA.

  • Which statisticians have been doing for a long time.

  • We're not doing something brand new here.

  • But the result is that we--

  • so how does a statistician think about this problem,

  • or that data matrix?

  • What-- if you have a matrix of data--

  • 2 by 2 rows and many columns-- so many, many samples--

  • what-- and we've made the mean zero.

  • So that's a first step a statistician

  • takes to check on the mean.

  • What's the next step?

  • What else does a statistician do with data

  • to measure how-- its size?

  • There's another number.

  • There's a number that goes with the mean,

  • and it's the variance--

  • the mean and the variance.

  • So somehow we're going to do variances.

  • And it will really be involved, because we

  • have two sets of data-- heights and ages.

  • We're really going to have a covariance--

  • covariance matrix-- and it will be 2 by 2.

  • Because it will tell us not only the variance in the heights--

  • that's the first thing a statistician

  • would think about--

  • some small people, some big people--

  • and variation in ages--

  • but also the link between them.

  • How are the height, age pairs--

  • does more height-- does more age go with more height?

  • And of course, it does.

  • That's the whole point here.

  • So it's this covariance matrix.

  • And that covariance matrix--

  • or the sample covariance matrix, to give it its full name--

  • what's the-- so just touching on statistics for a moment here.

  • What's the-- when we see that word sample in the name, what

  • is that telling us?

  • It's telling us that this matrix is computed from the samples,

  • not from a theoretical probability distribution.

  • We might have a proposed distribution

  • that the height follows the age--

  • height follows the age by some formula.

  • And that would give us theoretical variances.

  • We're doing sample variances, also called

  • empirical covariance made.

  • Empirical says-- empirical-- that word

  • means, from the information, from the data.

  • So that's what we do.

  • And it is exactly--

  • it's AA transpose.

  • You have to normalize it by the number of data points, N.

  • And then, for some reason--

  • best known to statisticians--

  • it's N minus 1.

  • And of course, they've got to be right.

  • They've been around a long time and it should be N minus 1,

  • because somehow 1 degree of freedom

  • was accounted for when we took away--

  • when we made the mean 0.

  • So we-- anyway, no problem.

  • But the N minus 1 is not going to affect our computation here.

  • This is the matrix that tells us that's

  • what we've got to work with.

  • That's what we've got to work with-- the matrix AA transpose.

  • And then the-- so we have this problem.

  • So we have a--

  • yeah.

  • I guess we really have a minimum problem.

  • We want to find--

  • yeah.

  • What problem are we solving?

  • And it's-- yeah.

  • So our problem was not least squares--

  • not the same as least squares.

  • Similar, but not the same.

  • We want to minimize.

  • So we're looking for that best line

  • where age equals some number, c, times the height, times the--

  • yeah-- or height.

  • Maybe it would have been better to age here and height here.

  • No.

  • No, because there are two unknowns.

  • So I'm looking for c.

  • I'm looking for the number c--

  • looking for the number c.

  • And with just two minutes in class left, what is that number

  • c going to be, when I finally get the problem stated

  • properly, and then solve it?

  • I'm going to learn that the best ratio of age to height

  • is sigma 1.

  • Sigma 1.

  • That's the one that tells us how those two are connected,

  • and the orthogonal-- and what will be the best--

  • yeah.

  • No.

  • Maybe I didn't answer that the right--

  • maybe I didn't get that right.

  • Because I'm looking for--

  • I'm looking for the vector that points in the right way.

  • Yeah.

  • I'm sorry.

  • I think the answer is, it's got to be there in the SVD.

  • I think it's the vector you want.

  • It's the principal component you want.

  • Let's do that properly on Friday.

  • I hope you see--

  • because this was a first step away

  • from the highlights of linear algebra

  • to problem solve by linear algebra,

  • and practical problems, and my point

  • is that the SVD solves these.

The following content is provided under a Creative

字幕と単語

ワンタップで英和辞典検索 単語をクリックすると、意味が表示されます

B1 中級

7.エッカート・ヤングAに最も近いランクkの行列 (7. Eckart-Young: The Closest Rank k Matrix to A)

  • 1 0
    林宜悉 に公開 2021 年 01 月 14 日
動画の中の単語