Orthogonal projections
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Projections onto Subspaces
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Visualizing a projection onto a plane
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A Projection onto a Subspace is a Linear Transforma
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Subspace Projection Matrix Example
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Another Example of a Projection Matrix
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Projection is closest vector in subspace
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Least Squares Approximation
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Least Squares Examples
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Another Least Squares Example
Another Least Squares Example Using least squares approximation to fit a line to points
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- So I've got four Cartesian coordinates here.
- This first one is minus 1, 0.
- I tried to draw them ahead of time.
- So minus 1, 0 is this point right there.
- Doing this in these new colors.
- The next point is a 0, 1, which is
- that point right there.
- Then the next point is 1, 2, which is that
- point right up there.
- And then the last point is 2, 1, which is that point there.
- Now my goal in this video is to find some line, y equals mx
- plus v, that goes through these points.
- Now the first thing I'd say is, hey Sal, there is not
- going to be any line that goes through these points, and you
- can see that immediately.
- You could find a line that maybe goes through these
- points, but it's not going to go through
- this point over here.
- If you try to make a line to goes through these two points,
- it's not going to go through those points there.
- So you're not going to be able to find a solution that goes
- through those points.
- Let's set up the equation that we know we can't find the
- solution to and maybe we can use our least squares
- approximation to find a line that almost goes through all
- these points.
- Or it's at least the best approximation for a line that
- goes through those points.
- So this first one, I can express my line, y
- equals mx plus b.
- Let me just express it as f of x is equal to mx plus b, or y
- is equal to f of x.
- We can write it that way.
- So our first point right there -- let me do it in that color,
- that orange -- that tells us that f of minus 1, which is
- equal to m times -- let me just write this way -- minus 1
- times m, it's minus m plus b, that that is going
- to be equal to 0.
- That's what that first equation tells us.
- The second equation tells us that f of 0, which is equal to
- 0 times m, which is just 0 plus b is equal to 1.
- f of 0 is 1.
- This is f of x.
- The next one -- let me do it in this yellow color -- tells
- us that f of 1, which is equal to 1 times m, or just m, plus
- b, is going to be equal to 2.
- And then this last one down here tells us that f of 2,
- which is of course 2 times m plus b, that that is going to
- be equal to 1.
- These are the constraints.
- If we assume that our line can go through all of these
- points, then all of these things must be true.
- Now you could immediately, if you wish, try to solve this
- equation, but you'll find that you won't find a solution.
- We want to find some m's and b's that satisfy
- all of these equations.
- Or another way of writing this -- We want to write it as a
- matrix vector or a matrix equation .
- We could write it like this.
- Minus 1, 1, 0, 1, 1, 1, 2, 1, times the vector mv has got to
- be equal to the vector 0, 1, 2, 1.
- These two systems, this system and this system right here,
- are equivalent statements, right?
- Minus 1 times m plus 1 times b has got to be equal to that 0.
- 0 times m plus 1 times b has got to be equal to that 1
- That's equivalent to that statement right here.
- And this isn't going to have a solution.
- The solution would have to go through all of those points.
- So let's at least try to find a least squares solution.
- So if we call this a, if we call that x, and let's call
- this b, there is no solution to ax is equal to b.
- Now maybe we can find a least -- Well, we can definitely
- find a least squares solution.
- So let's find our least squares solution such that a
- transpose a times our least squares solution is equal to a
- transpose times b.
- Our least squares solution is the one that
- satisfies this equation.
- We proved it two videos ago.
- So let's figure out what a transpose a is and what a
- transpose b is, and then we can solve.
- So a transpose will look like this.
- b minus 1, 1, 0, 1, 1, 1, and then 2, 1.
- This first column becomes this first row; this second column
- becomes this second row.
- So we're going to take the product of a transpose and
- then a-- a is that thing right there --minus 1, 0, 1, 2, and
- we just get a bunch of 1's.
- So what does this equal to?
- We have a 2 by 4 times a 4 by 2.
- So we're going to have a 2 by 2 matrix.
- So this is going to be -- Let's do it this way.
- Well, we're going to have minus 1 times minus 1, which
- is 1, plus 0 times 0, which is 0 -- so we're at 1 right now
- --plus 1 times 1.
- So that's 1 plus the other 1 up there, so that's 2,
- plus 2 times 2.
- 2 times 2 is 4, so we get 6.
- That's that row, dotted with that column, was equal to 6.
- Now let's take this row dotted with this column.
- So it's going to be negative 1 times 1, plus 0 times 1, so
- all of these guys times 1 plus each other.
- So minus 1 plus 0 plus 1 -- that's all 0's --plus 2.
- So it's going to get a 2.
- I just dotted that guy with that guy.
- Now I need to take the dot of this guy with this column.
- So it's just going to be 1 times minus 1 plus 1 times 0
- plus 1 times 1 plus 1 times 2.
- Well, these are all 1 times everything, so it's minus 1
- plus 0 plus 1, which is 0 plus 2.
- It's going to be 2.
- And then finally -- Well.
- I mean, I think you see some symmetry here.
- We're going to have to take the dot of this guy and this
- guy over here.
- So what is that?
- That's 1 times 1, which is 1, plus 1 times 1, which is 2,
- plus 1 times 1.
- So we're going to have 1 plus itself four times.
- So we're going to get that it's equal to 4.
- So this is a transpose a.
- And let's figure out what a transpose b looks like.
- Scroll down a little bit.
- So a transpose is this matrix again-- let me switch colors
- --minus 1, 0, 1, 2.
- We get all of our 1's just like that.
- And then the matrix b is 0, 1, 2, 1.
- We have a 2 by 4 times a 4 by 1, so we're just going to get
- a 2 by 1 matrix.
- So this is going to be equal to a 2 by 1 matrix.
- We have here, let's see, minus 1 times 0 is 0, plus 0 times 1
- is still 0.
- Plus 1 times 2, which is 2, plus 2 times
- 1, which is 4, right?
- This is 2 plus 2, so it's going to be 4 right there.
- And then we have 1 times 0, plus 1 times 2, plus-- So 1
- times all of these guys added up.
- So 0 plus 1 is 1, 1 plus 2 is 3, 3 plus 1 is 4.
- So this right here is a transpose b.
- So just like that, we know that the least squares
- solution will be the solution to this system.
- 6, 2, 2, 4, times our least squares solution, is going to
- be equal to 4, 4.
- Or we could write it this way.
- We could write it 6, 2, 2, 4, times our least squares
- solution, which I'll write-- Remember, the
- first entry was m .
- I'll write it as m star.
- That's our least square m, and this is our least square b, is
- equal to 4, 4.
- And I can do this as an augmented matrix or I could
- just write this as a system of two unknowns, which is
- actually probably easier.
- So let's do it that way.
- So this, if I were to write it as a system of equations, is 6
- times m star plus 2 times b star, is equal to 4.
- And then I get 2 times m star plus 4 times b star is
- equal to this 4.
- So let me solve for my m stars and my b stars.
- So let's multiply this second equation, actually let's
- multiply that top equation by 2.
- This is just straight Algebra 1.
- So times 2, what do we get?
- We get 12m star plus 4b star is equal to 8.
- We just multiplied that top guy by 2.
- Now let's multiply this magenta 1 by negative 1.
- So this becomes a minus, this becomes a minus, that becomes
- a minus, and now we can add these two equations.
- So we get minus 2 plus 12m star, that's 10m star.
- And then the minus 4b and the 4b cancel out, is equal to 4,
- or m star is equal to 4 over 10, which is equal to 2/5.
- Now we can just go and back-substitute into this.
- We can say 6 times m star-- This is just
- straight Algebra 1.
- So 6 times our m star, so 6 times 2 over 5, plus 2 times
- our b star is equal to 4.
- Enough white, let me use yellow.
- So we get 12 over 5 plus 2b star is equal to 4, or we
- could say 2b star-- let me scroll down a little bit --2b
- star is equal to 4.
- Which is the same thing as 20 over 5, minus 12 over 5, which
- is equal to-- I'm just subtracting the 12 over 5 from
- both sides --which is equal to 8 over 5.
- And you divide both sides of the equation by 2, you get b
- star is equal to 4/5.
- And just like that, we got our m star and our b star.
- Our least squares solution is equal to 2/5 and 4/5.
- So m is equal to 2/5 and b is equal to 4/5.
- And remember, the whole point of this was to find an
- equation of the line.
- y is equal to mx plus b.
- Now we can't find a line that went through all of those
- points up there, but this is going to be our
- least squares solution.
- This is the one that minimizes the distance between a times
- our vector and b.
- No vector, when you multiply times that matrix a-- that's
- not a, that's transpose a --no other solution is going to
- give us a closer solution to b than when we put our
- newly-found x star into this equation.
- This is going to give us our best solution.
- It's going to minimize the distance to b.
- So let's write it out.
- y is equal to mx plus b.
- So y is equal to 2/5 x plus 2/5.
- Let's graph that out.
- y is equal to 2/5 x plus 2/5.
- So its y-intercept is 2/5, which is about there .
- This is at 1.
- 2/5 is right about there.
- And then its slope is 2/5.
- Let's think of it this way: for every 2 and 1/2 you go to
- the right, you're going to go up 1.
- So if you go 1, 2 and 1/2, we're going to go up 1.
- We're going to go up 1 like that.
- So our line-- and obviously this isn't precise --but our
- line is going to look something like this.
- I want to do my best shot at drawing it because
- this is the fun part.
- It's going to look something like that.
- And that right there is my least squares estimate for a
- line that goes through all of those points.
- And you're not going to find a line that minimizes the error
- in a better way, at least when you measure the error as the
- distance between this vector and the vector a times our
- least squares estimate.
- Anyway, thought you would find that neat.
Be specific, and indicate a time in the video:
At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?
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