Projection is closest vector in subspace Showing that the projection of x onto a subspace is the closest vector in the subspace to x
Projection is closest vector in subspace
- Let's say I've got some subspace, v,
- that's a plane in R3.
- So let me see if I can draw it respectably well.
- Let me draw some plane in R3.
- Maybe it looks something like that.
- That is my subspace.
- I think that's good enough.
- Let me see if I can draw it a little bit better than that.
- There you go.
- So that is my plane in R3.
- This is v.
- This is a subspace.
- And let's say that I have some other vector, x,
- any vector in R3.
- So my vector x looks like this.
- That is my vector x.
- Now what I want to show you in this video is that the
- projection of x onto our subspace-- and let's say that
- this is our 0 vector right there.
- I want to show you that the projection of x onto our
- subspace is the closest vector in our subspace to x.
- So let me draw that out, and maybe it'll make
- a little more sense.
- So the projection of x on to the subspace will look
- something like this.
- It will look something like that.
- That right there, that green vector right there, is the
- projection of the vector x onto our subspace v.
- That's our vector x.
- Now, let's take some arbitrary other vector in our subspace.
- Let's just take this one.
- This is just some other arbitrary
- vector in our subspace.
- Let me draw a little bit differently.
- We draw it like that.
- Let's call that vector v.
- That's clearly another vector on our subspace.
- It lies on that plane.
- What I want to show you is that the distance from x to
- our projection of x on to v is shorter than the distance from
- x to any other vector.
- From the distance from x to any other vector.
- And obviously, the way I've drawn it, it looks pretty
- clear that this line is shorter than that line.
- But that was just a particular choice that I picked.
- Let's prove it that's general.
- So what I want to prove is that the distance between x
- and its projection onto the subspace-- and the way we can
- get that is essentially just take the length of the vector
- of x minus the projection of x onto my subspace.
- This length right here is this length right here, is this
- length right here.
- So x minus the projection of x onto v, that's going to be
- this vector right there.
- Let me do that in a different color.
- Don't want to reuse colors too often.
- That's going to be that vector right there.
- We could call that vector a.
- It's clearly in the orthogonal complement of v, because it's
- orthogonal to this guy.
- And that's the definition of a projection, actually, so this
- is equal to a.
- My claim, what I want to show you, is that this distance a
- is shorter than any distance here, is less than or equal to
- the distance between x and v, where v is any member.
- So that's this distance, right here.
- This vector right here, that distance right-- let me draw
- this vector.
- The vector x minus v looks like this.
- It looks like that.
- That is the vector x minus v, right?
- If you take v plus x minus v, you're going to go to x.
- So what I want to show is that this distance, the length of
- a, of the difference between x and its projection, is always
- going to be less than the distance between x and any
- other vector in the subspace.
- So that is x minus v.
- So let's see if we can prove that.
- So let's take the square of this distance.
- So the square of x-- actually, let me do that.
- Yeah, let me write that way.
- So we want to concern ourselves with the square of
- the distance of x minus v, where x is some vector in R3,
- and v is some vector in R3 that's also a
- member of our subspace.
- It sits on this plane.
- So what's the square of this going to be?
- Well, x minus v is equal to this vector.
- Let me draw a new vector here.
- It is equal to this-- wait, let me draw
- it in in this yellow.
- It's equal to this vector.
- It's equal to this yellow vector plus a.
- Right? x minus v is-- this magenta vector that starts
- here and goes there-- clearly equal to this yellow vector
- plus this orange vector.
- So let me call that yellow vector b.
- Now, what is b equal to?
- Well, b is going to be equal to this vector, this green
- vector, which is the projection of x onto v, minus
- this purple vector.
- Minus this mauve vector, I guess.
- Minus v.
- That's what b is.
- So we could write x minus v as being equal to the sum of the
- vector b plus the vector a.
- So x minus v is equal to b plus a.
- And if we're taking the length of x minus v squared, that's
- the same thing as the length of b plus a squared.
- And that's just equal to b plus a dotted with b plus a,
- which is the same thing as b dot b.
- Let me write that a little bit neater.
- Let me write it down here.
- This is going to be equal to, I'll switch colors, b dot b.
- Right, b dot b plus b dot a plus a dot b.
- So plus 2 times a dot b, plus a dot a.
- Now, a and b are clearly orthogonal.
- b is the difference between two vectors in our subspace.
- The subspace is closed under addition and subraction.
- So b is a member of our subspace.
- a is orthogonal to everything in our subspace, by
- So since a is clearly orthogonal to b, a is-- by
- definition-- going to be in the orthogonal compliment of
- the subspace.
- This is going to be 0.
- And then this right here will simplify to
- the length of b squared.
- And then this right here is going to be plus
- the length of a squared.
- So we get the distance between x and some arbitrary member of
- our subspace squared is equal to the length of b, right
- here, plus the length of a squared.
- Now, a was a distance between our vector x and our
- projection, right?
- That's what the definition of a was. a was the distance
- between our vector x and our projection.
- Now, this number right here is going to be
- at least 0 or positive.
- So this right here is definitely going to be greater
- than or equal to a squared.
- Or another way to say it is that the distance between x
- minus v squared is definitely going to be greater than or
- equal to the distance of a squared.
- Or the distance between x minus v-- this is still going
- to be a positive quantity, length is always going to be
- positive-- is greater than or equal to the
- length of vector a.
- Or what's that length of vector a? a is just this thing
- right here.
- So let's write our result.
- The length of the vector x minus v, or the distance
- between x and some arbitrary member of our subspace, is
- always going to be greater than or equal to the length of
- a, which is just the distance between x and the projection
- of x onto our subspace.
- So there you have it.
- We've shown, and the original graph kind of hinted at it,
- that the projection of x onto v is the closest vector in our
- subspace to x.
- It's closer than any other vector in v to our arbitrary
- vector in R3, x.
- And we've proven it right there.
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