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Gram-Schmidt example with 3 basis vectors. Created by Sal Khan.
Video transcript
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.