Gram-Schmidt example with 3 basis vectors Gram-Schmidt example with 3 basis vectors
Gram-Schmidt example with 3 basis vectors
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- Let's do one more Gram-Schmidt example.
- So let's say I have the subspace V that is spanned by
- the vectors-- let's say we're dealing in R4, so the first
- vector is 0, 0, 1, 1.
- The second vector is 0, 1, 1, 0.
- And then a third vector-- so it's a three-dimensional
- subspace of R4-- it's 1, 1, 0, 0, just like that,
- three-dimensional subspace of R4.
- And what we want to do, we want to find an orthonormal
- basis for V.
- So we want to substitute these guys with three other vectors
- that are orthogonal with respect to each other
- and have length 1.
- So we do the same drill we've done before.
- We can say-- let's call this guy v1, this guy is v2, and
- let's call this guy v3.
- So the first thing we want to do is replace v1-- and I'm
- just picking this guy at random because he was the
- first guy on the left-hand side.
- I want to replace v1 with an orthogonal version of v1.
- So let me call u1 is equal to-- well, let me just find
- out the length the v1.
- I don't think I have to explain too much of the theory
- at this point.
- I just want to show another example.
- So the length of v1 is equal to the square root of 0
- squared plus 0 squared plus 1 squared plus 1 squared, which
- equals the square root of 2.
- So let me define my new vector u1 to be equal to 1 over the
- length of v1, 1 over the square root of 2, times v1,
- times 0, 0, 1, 1.
- And just like that, the span of v1, v2, v3, is the same
- thing is the span of u1, v2, and v3.
- So this is my first thing that I've normalized.
- So I can say that V is now equal to the span of the
- vectors u1, v2, and v3.
- Because I can replace v1 with this guy, because this guy is
- just a scaled-up version of this guy.
- So I can definitely represent him with him, so I can
- represent any linear combination of these guys with
- any linear combination of those guys right there.
- Now, we just did our first vector.
- We just normalized this one.
- But we need to replace these other vectors with vectors
- that are orthogonal to this guy right here.
- So let's do v2 first. So let's replace-- let's call it y2 is
- equal to v2 minus the projection of v2 onto the
- space spanned by u1 or onto-- you know, I could call it c
- times u1, or in the past videos, we called that
- subspace V1, but the space spanned by u1.
- And that's just going to be equal to y2 is equal to v2,
- which is 0, 1, 1, 0, minus-- v2 projected onto that space
- is just a dot product of v2, 0, 1, 1, 0, with the spanning
- vector of that space.
- And there's only one of them, so we're only going to have
- one term like this with u1, so dotted with 1 over the square
- root of 2 times 0, 0, 1, 1, and then all of that times u1.
- So 1 over the square root of 2 times the vector 0, 0, 1, 1.
- And so this is going to be equal to v2,
- which is 0, 1, 1, 0.
- The square root of 2, let's factor them out.
- So then you just get-- or kind of reassociate them out.
- So then you get this is 1 over the square root of 2 times 1
- over the square root of 2 is minus 1/2.
- You times-- what's the dot product of these two guys?
- You get 0 times 0 plus 1 times 0, which is still 0, plus 1
- times 1 plus 0 times 0.
- So you're just going to have times 1 times this out here:
- 0, 0, 1, 1.
- I'll write that a little bit neater.
- I'm getting careless.
- 1, 1.
- So this is just going to be equal to 0, 1, 1, 0 minus--
- 1/2 times 0 is 0.
- 1/2 times 0 is 0.
- Then I have two halves here.
- So y2 is equal to-- let's see, 0 minus 0 is 0, 1 minus 0 is
- 1, 1 minus 1/2 is 1/2, and then 0 minus 1/2 is minus 1/2.
- So V, we can now write as the span of u1, y2, and v3.
- And this is progress.
- u1 is orthogonal, y2-- sorry, u1 is normalized.
- It has length 1.
- Y2 is orthogonal to it or they're orthogonal with
- respect to each other, but y2 still has not been normalized.
- So let me replace y2 with a normalized version of it.
- The length of y2 is equal to the square root of 0 plus 1
- squared, which is 1, plus 1/2 squared, which is 1/4, plus
- minus 1/2 squared, which is also 1/4, so plus 1/4.
- So this is 1 and 1/2.
- So it's equal to the square root of 3/2.
- So let me define another vector here.
- u2, which is equal to 1 over the square root of 3/2, or we
- could say is the square root of 2/3, I'm just inverting it.
- It's 1 over the length of y2.
- So I'll just find the reciprocal, so it's the square
- root 2 over 3 times y2, times this guy right here, times 0,
- 1, 1/2, and minus 1/2.
- And so this span is going to be the same thing as the span
- of u1, u2, and v3.
- And there's our second basis vector.
- And we're making a lot of progress.
- These guys are orthogonal with respect to each other.
- They both have length 1.
- We just have to do something about v3.
- And we do it the same way.
- Let's find a vector that is orthogonal to these guys, and
- if I sum that vector to some linear combination of these
- guys, I'm going to get v3, and I'm going to
- call that vector y3.
- y3 is equal to v3 minus the projection of v3 onto the
- subspace spanned by u1 and u2.
- So I could call that subspace-- let me
- just write it here.
- The span of u1 and u2, just for notation, I'm going to
- call it v2.
- So it's v3, and actually, I don't even have
- to write that .
- Minus the projection of v3 onto that.
- And what's that going to be?
- That's going to be v3 dot u1 times u1, times the vector u1.
- And actually let me just-- plus v3 dot u2
- times the vector u2.
- Since this is an orthonormal basis, the projection onto it,
- you just take the dot product of v2 with each of their
- orthonormal basis vectors and multiply them times the
- orthonormal basis vectors.
- We saw that several videos ago.
- That's one of the neat things about orthonormal bases.
- So what is this going to be equal to?
- A little bit more computation here.
- y3 is equal to v3, which was up here.
- That's v3.
- v3 looks like this.
- It's 1, 1, 0, 0 minus v3 dot u1.
- So this is minus v3, 1, 1, 0, 0, dot u1.
- So it's dot 1 over the square root of 2 times 0, 0, 1, 1.
- That's u1-- so that's this part right here-- times u1, so
- times 1 over the square root of 2 times 0, 0, 1, 1.
- This piece right there is this piece right there.
- And then we can distribute this minus sign, so it's going
- to be plus.
- You know, we have a plus, but there's this minus over here
- so we put a minus v3.
- Let me switch colors .
- Minus v3 , which is 1, 1 0, 0 dotted with u2, dotted with
- the square root of 2/3 times 0, 1, 1/2, minus 1/2 times u2,
- times the vector u2, times the square root of 2/3, times the
- vector 0, 1, 1/2, minus 1/2.
- And what do we get?
- Let's calculate this.
- So we could take the-- so this is going to be equal to the
- vector 1, 1, 0, 0, minus-- so the 1 over the square root of
- 2 and the 1 over the square root of 2, multiply them.
- You're going to get a 1/2.
- And then when you take the dot product of these two, 1 times
- 0-- let's see, this is actually all going to be, if
- you take the dot product of all of these, then it actually
- gets 0, right?
- So this guy, v3, was actually already orthogonal to u1.
- This will just go straight to 0, which is nice.
- We don't have to have a term right there.
- I took the dot product 1 times 0 plus 1 times 0 plus 0 times
- 1 plus 0 times 1, all gets zeroed.
- So this whole term drops out.
- We can ignore it, which makes our computation simpler.
- And then over here we have minus the square root of 2/3
- times the square root of 2/3 is just 2/3 times the dot
- product of these two guys.
- So that's 1 times 0, which is 0, plus 1 times 1, which is 1,
- plus 0 times 1/2, which is 0, plus 0 times minus 1/2, which
- is 0, so we just get a 1 there, times the vector 0, 1,
- 1/2, minus 1/2.
- And then what do we get?
- We get-- this is the home stretch-- 1, 1, 0, 0 minus 2/3
- times all of these guys.
- So 2/3 time 0 is 0.
- 2/3 times 1 is 2/3.
- 2/3 times 1/2 is 1/3.
- And then 2/3 times minus 1/2 is minus 1/3.
- So then this is going to be equal to 1 minus 0 is 1, 1
- minus 2/3 is 1/3, 0 minus 1/3 is minus 1/3, and then 0 minus
- minus 1/3 is positive 1/3.
- So this vector y3 is orthogonal to these two other
- vectors, which is nice, but it still hasn't been normalized.
- So we finally have to normalize this guy, and then
- we're done.
- Then we have an orthonormal basis.
- We'll have u1, u2, and now we'll find u3.
- So the length of my vector y-- actually, let's do something
- even better.
- It'll simplify things a little bit.
- Instead of a writing y this way, I
- could scale up y, right?
- All I want is a vector that's orthogonal to the other two
- that still spans the same space.
- So I can scale this guy up.
- So I could say, I don't know, let me call it y3-- let me
- call it y3 prime.
- And I'm just doing this to ease the computation.
- I could just scale this guy up, multiply him by 3.
- So what do I get?
- I probably should have done it some of the other ones.
- 3, 1, minus 1, and 1.
- And so I can replace y3 with this guy, and then I can just
- normalize this guy.
- It'll be a little bit easier.
- So the length of y3 prime that I just defined is equal to the
- square root of 3 squared, which is 9, plus 1 squared
- plus minus 1 squared plus 1 squared, which is equal to the
- square root of 12, which is what?
- That's two square roots of 3.
- That is equal to 2 square roots of 3, right?
- Square root of 4 times the square root of 3, which is two
- square roots of 3.
- So now I can to find u3 as equal to y3 times 1 over the
- length of y3, so it's equal to 1 over two square roots of 3
- times the vector 3, 1, minus 1, and 1.
- And then we're done.
- If we have a basis, an orthonormal basis would be
- this guy-- let me take the other ones down here-- and
- these guys.
- All of these form-- let me bring it all the way down.
- If I have a collection of these three vectors, I now
- have an orthonormal basis for V, these three right there.
- That set is an orthonormal basis for my original subspace
- V that I started off with.
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