# Least squares approximation | Linear Algebra | Khan Academy

Let’s say I have
some matrix A. Let’s say it’s an n-by-k
matrix, and I have the equation Ax is equal to b. So in this case, x would have to
be a member of Rk, because we have k columns here, and
b is a member of Rn. Now, let’s say that it just so
happens that there is no solution to Ax is equal to b. What does that mean? Let’s just expand out A. I think you already know
what that means. If I write a like this, a1, a2,
if I just write it as its columns vectors right there,
all the way through ak, and then I multiply it times x1,
x2, all the way through xk, this is the same thing as
that equation there. I just kind of wrote out
the two matrices. Now, this is the same thing as
x1 times a1 plus x2 times a2, all the way to plus xk times ak
is equal to the vector b. Now, if this has no solution,
then that means that there’s no set of weights here on the
column vectors of a, where we can get to b. Or another way to say it is, no
linear combinations of the column vectors of a will
be equal to b. Or an even further way of saying
it is that b is not in the column space of a. No linear combination of these
guys can equal to that. So let’s see if we can
visualize it a bit. So let me draw the column
space of a. So maybe the column space of
a looks something like this right here. I’ll just assume it’s
a plane in Rn. It doesn’t have to be a plane. Things can be very general, but
let’s say that this is the column space. This is the column space of a. Now, if that’s the column space
and b is not in the column space, maybe we
can draw b like this. Maybe b, let’s say this is the
origin right there, and b just pops out right there. So this is the 0 vector. This is my vector b, clearly
not in my column spaces, clearly not in this plane. Now, up until now, we would
get an equation like that. We would make an augmented
matrix, put in reduced row echelon form, and get a line
that said 0 equals 1, and we’d say, no solution, nothing
we can do here. But what if we can do better? You know, we clearly can’t
find a solution to this. But what if we can find
a solution that gets us close to this? So what if I want to find some
x, I’ll call it x-star for now, where– so I want to find
some x-star, where A times x-star is– and this is
a vector– as close as possible– let me write this–
as close to b as possible. Or another way to view it, when
I say close, I’m talking about length, so I want to
minimize the length of– let me write this down. I want to minimize the length
of b minus A times x-star. Now, some of you all
might already know where this is going. But when you take the difference
between 2 and then take its length, what
does that look like? Let me just call Ax. Ax is going to be a member
of my column space. Let me just call that v. Ax is equal to v. You multiply any vector in Rk
times your matrix A, you’re going to get a member of
your column space. So any Ax is going to be
in your column space. And maybe that is the vector v
is equal to A times x-star. And we want this vector to get
as close as possible to this as long as it stays–
I mean, it has to be in my column space. But we want the distance between
this vector and this vector to be minimized. Now, I just want to show you
where the terminology for this will come from. I haven’t given it its
proper title yet. If you were to take this
vector– let just call this vector v for simplicity– that
this is equivalent to the length of the vector. You take the difference between
each of the elements. So b1 minus v1, b2 minus v2,
all the way to bn minus vn. And if you take the length of
this vector, this is the same thing as this. This is going to be equal
to the square root. Let me take the length
squared, actually. The length squared of this is
just going to be b1 minus v1 squared plus b2 minus v2 squared
plus all the way to bn minus vn squared. And I want to minimize this. So I want to make this value the
least value that it can be possible, or I want to get the
least squares estimate here. And that’s why, this last minute
or two when I was just explaining this, that was just
to give you the motivation for why this right here is called
the least squares estimate, or the least squares solution,
or the least squares approximation for the equation
Ax equals b. There is no solution to this,
but maybe we can find some x-star, where if I multiply A
times x-star, this is clearly going to be in my column space
and I want to get this vector to be as close to
b as possible. Now, we’ve already seen in
several videos, what is the closest vector in any
subspace to a vector that’s not in my subspace? Well, the closest vector to
it is the projection. The closest vector to b, that’s
in my subspace, is going to be the projection of
b onto my column space. That is the closest
vector there. So if I want to minimize this,
I want to figure out my x-star, where Ax-star is equal
to the projection of my vector b onto my subspace or onto
the column space of A. Remember what we’re
doing here. We said Axb has no solution, but
maybe we can find some x that gets us as close
as possible. So I’m calling that my least
squares solution or my least squares approximation. And this guy right here is
clearly going to be in my column space, because you take
some vector x times A, that’s going to be a linear combination
of these column vectors, so it’s going to
be in the column space. And I want this guy to be as
close as possible to this guy. Well, the closest vector in my
column space to that guy is the projection. So Ax needs to be equal
to the projection of b on my column space. It needs to be equal to that. But this is still pretty
hard to find. You saw how, you know, you took
A times the inverse of A transpose A times A transpose. That’s hard to find that
transformation matrix. So let’s see if we can find an
easier way to figure out the least squares solution, or kind
of our best solution. It’s not THE solution. It’s our BEST solution
to this right here. That’s why we call it the least
squares solution or approximation. Let’s just subtract b from
both sides of this and we might get something
interesting. So what happens if we take Ax
minus the vector b on both sides of this equation? I’ll do it up here
on the right. On the left-hand side we
get A times x-star. It’s hard write the x and
then the star because they’re very similar. And we subtract b from it. We subtract our vector b. That’s going to be equal to the
projection of b onto our column space minus b. All I did is I subtracted
b from both sides of this equation. Now, what is the projection
of b minus our vector b? If we draw it right here, it’s
going to be this vector right– let me do it in
this orange color. It’s going to be this
right here. It’s going to be that vector
right there, right? If I take the projection of b,
which is that, minus b, I’m going to get this vector. you
we could say b plus this vector is equal to
my projection of b onto my subspace. So this vector right
here is orthogonal. It’s actually part of the
definition of a projection that this guy is going to be
orthogonal to my subspace or to my column space. And so this guy is orthogonal
to my column space. So I can write Ax-star minus
b, it’s orthogonal to my column space, or we could
say it’s a member of the orthogonal complement
of my column space. The orthogonal complement is
just the set of everything, all of the vectors that are
space right here. So this vector right here
that’s kind of pointing straight down onto my plane
is clearly a member of the orthogonal complement
of my column space. Now, this might look familiar
to you already. What is the orthogonal
complement of my column space? The orthogonal complement of
my column space is equal to the null space of a transpose,
or the left null space of A. We’ve done this in many,
many videos. So we can say that A times my
least squares estimate of the equation Ax is equal to
b– I wrote that. So x-star is my least squares
solution to Ax is equal to b. So A times that minus
b is a member of the null space of A transpose. Now, what does that mean? Well, that means that if I
multiply A transpose times this guy right here, times
Ax-star– and let me, no I don’t want to lose the vector
signs there on the x. This is a vector. I don’t want to forget that. Ax-star minus b. So if I multiply A transpose
times this right there, that is the same thing is that,
what am I going to get? Well, this is a member of the
null space of A transpose, so this times A transpose has
got to be equal to 0. It is a solution to A transpose
times something is equal to the 0 vector. Now. Let’s see if we can simplify
this a little bit. We get A transpose A times
x-star minus A transpose b is equal to 0, and then if we add
this term to both sides of the equation, we are left with A
transpose A times the least squares solution to Ax
equal to b is equal to A transpose b. That’s what we get. Now, why did we do
all of this work? Remember what we started with. We said we’re trying to find a
solution to Ax is equal to b, but there was no solution. So we said, well, let’s find
at least an x-star that minimizes b, that minimizes
the distance between b and Ax-star. And we call this the least
squares solution. We call it the least squares
solution because, when you actually take the length, or
when you’re minimizing the length, you’re minimizing the
squares of the differences right there. So it’s the least squares
solution. Now, to find this, we know
that this has to be the closest vector in our
subspace to b. And we know that the closest
vector in our subspace to b is the projection of b onto our
subspace, onto our column space of A. And so, we know that A–
let me switch colors. We know that A times our least
squares solution should be equal to the projection of b
onto the column space of A. If we can find some x in Rk that
satisfies this, that is our least squares solution. But we’ve seen before that
the projection b is easier said than done. You know, there’s a
lot of work to it. So maybe we can do
it a simpler way. And this is our simpler way. If we’re looking for this,
alternately, we can just find a solution to this equation. So you give me an Ax equal to
b, there is no solution. Well, what I’m going to do is
I’m just going to multiply both sides of this equation
times A transpose. If I multiply both sides of this
equation by A transpose, I get A transpose times Ax is
equal to A transpose– and I want to do that in the same
blue– A– no, that’s not the same blue– A transpose b. All I did is I multiplied
both sides of this. Now, the solution to this
equation will not be the same as the solution to
this equation. This right here will always
have a solution, and this right here is our least
squares solution. So this right here is our
least squares solution. And notice, this is some matrix,
and then this right here is some vector. This right here is
some vector. So long as we can find a
solution here, we’ve given our best shot at finding a solution
to Ax equal to b. We’ve minimized the error. We’re going to get Ax-star,
and the difference between Ax-star and b is going
to be minimized. It’s going to be our least
squares solution. It’s all a little bit abstract
right now in this video, but hopefully, in the next video,
we’ll realize that it’s actually a very, very
useful concept.

### 43 Replies to “Least squares approximation | Linear Algebra | Khan Academy”

1. fascist27 says:

2. fascist27 says:

Respond to this video…

3. rbfreitas says:

Good video!!!! And nice work! Good luck with the KhanAcademy 🙂

4. elmiramb says:

Thanks a lot, very comprehensive ! great job!

5. Steve Tran says:

thaks

6. Vahagn Nahapetyan says:

Excelent video.
Thanks much :))))))))

Vahag

7. priestofrhythm says:

I am the 60th guy liking it !! 😛 😀
Great vid, thank you. 🙂

8. Liam McArdle says:

can you teach me cubic expressions and cubic equations 🙂
eg. solve the equation x(3X3X3) – 2x(2X2) – x + 2 = 0
by using the factor theorem formula 🙂

9. itrustedyou says:

I was just watching some stanford video lectures, and they were talking about some linear algebra concepts, and I thought to myself ' I don't remember linear algebra being so hard when I did it with Sal', and then I came back to these videos, and it turns out I was right. Linear Algebra is so so so intuitive the way Sal explains it, and I had no trouble just picking up from this video without watching all the previous ones. Stupid non-Khan teachers. make me feel retarded.

10. spechtbert says:

n1

11. BlackfireGippal says:

I wish to know how to solve this: x has values of : -2 0 1 2 3 and y : 17 5 2 1 2 and i'm asked to use the least squares method, but i've been absent and i don't know exactly what my teacher ment by that or what that method consists of. Can anyone help me solve this ?

12. Neil Jan says:

what happens when AT*A is singular. How do we solve for the least square solution?

very helpful! Thanks a lot! you are doing great things! I also listened to your other videos, all very wonderful!

14. Aida Waller says:

You are like a billion times better than my professor… and my professor isn't even bad. On the contrary he's my favorite! You're just even better at explaining things.

Plus it's impossible for me to lose focus with the pretty colors and your beautiful handwriting. lol

I have my Linear Algebra final tomorrow (technically today) and I owe the A that I'm sure to get to you and all your helpful videos!

15. Rajagopal Jayaram says:

Your videos are just great !!! The concepts with geometrical examples make very good sense !!! Thanks a lot

16. utte12 says:

nice vid, but why did you take the length squared? i understand that the length of the vector would be sqrt(b1^2 + b2^2…bn^2) but why did you square even that?

17. J. Charles says:

Very useful! In my lecture slides I had this term Hx=z for the same problem and I couldn't make sense of how we could get to this as the best solution: x = (Ht*H)^-1 * Ht * z.
Now I understand:-)

18. Sam Ma says:

Very useful man you are doing an amazing job this literally saved me hours of searching and reading can't thank you enough 🙂

19. Han Nguyen says:

Love the diagram. It always give the big picture.

20. Zhiqi Guo says:

love this guy

21. Deathangle says:

I have a question..
does least sequare approximation has always solution..

22. Michal Fašánek says:

god dang it I knew I should have chosen other bachelor thesis..

It would be great having links when says "I explained (whatever) in a different video" to access that explanation. In this case I wanted to know why C(A)transpose=N(Atranspose).

Thanks¡

24. Kavish Doshi says:

accha hai

25. Jo L says:

26. xesan555 says:

Thanks so much Khan…wonderful explanation in two videos that explains everything…great. You are wonderful

27. 墨石 says:

Comes handy while studying machine learning.

28. Vrushabh says:

Khaini thuk k aa pehle

29. Winnie Shi says:

how did you know that it was a projection to the Col(A) and not anything else like the Range(A)?

30. johnfy.k hikc says:

Best approach to the problem. No gradient, no multivariable calculus. you're master!

31. Kartar Singh says:

Super clarity……

32. Hatta Gomuzuki says:

2018? Im alone 🙁

33. Otto Omen says:

Excellent explanation of a valuable technique.

34. S. Wang says:

This guy is good………..

35. yark park says:

Should have used n instead of k its usually mxn in R^n

36. Insert here says:

great geometric intuition of linear regression

37. biwom Abu says:

Thank you so much. You just simplified long boring hours of confusing lecture

38. yoMeela says:

Helpful exploration of least square properties

39. Sanwa Official says:

I have one question, whether the LSS always consistent? if yes, how can I prove it? please answer

40. jack sjsjs says:

Why is this method chosen over the derivative method with minimize the error. Is this method more useful for multi linear or just multiple variable problems

41. Kin Man says:

Great video, thanks for your explaination, my I ask is Least – squares inversion the same thing as this with different name?

42. Float Circuit says:

I was doing an online machine learning course and got lost when the lecturer introduced the normal equation (which this is, with a different name). Needless to say, I'm finna binge-watch your linear algebra lectures now because I get insecure about using equations I don't understand. Thanks for the playlist, I really wanna put ML in my toolset so we're doing this!

43. Sara Benavidez says:

"Some of you might already know where this is going.."

Me: Nope