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Expressing a projection on to a line as a matrix vector prod

Expressing a Projection on to a line as a Matrix Vector prod. Created by Sal Khan.

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Video transcript

In the last video we saw that if we had some line that was defined as all of the scalar multiples of some vector-- I'll just write it like this. The scalar multiples, obviously, are any real number. We defined a transformation, and I didn't speak of it much in terms of transformations, but it was a transformation. We defined a projection onto that line L as a transformation. In the video, we drew it as transformations within R2, but it could be, in general, a transformation from Rn to Rn. We defined it as, the projection of x onto L was equal to the dot product of x, with this defining vector. x dot this defining vector, divided by that defining vector dotted with itself. All of that times the defining vector of the line. This was our definition. A couple of things might have popped out at you right when we first saw this. When you dot a vector with itself, what's that equal to? We know that if I take some vector, and I dot it with itself, that is equivalent to the length of the vector squared. We can rewrite this as being equal to x dot v, over the length of v squared, all of that times v. Wouldn't it be nice if the length of v was 1. The length of v was equal to 1. If the length of v was 1, or this is another way of saying that, v is a unit vector. Our formula for our projection would just simplify to x dot v. All of that times, this will just be some scalar number, that times v. You say, hey Sal, how do we know if this is a unit vector or not. What you can realize is that any-- Let me draw it this way. When I drew it in the previous video, I just picked a line, like that. The line can be really defined to this vector v in the line. It can be any of the vectors that's contained in the line. The vector v could be like that. Let's say someone gives you a vector v that isn't a unit vector. Let's say that the length of v is not equal to 1. How can you define a line using sum unit vector. You can just normalize v. You can define sum unit vector right here. You could define sum vector right there. Let's call it u, and I'll say it's a unit vector. That is equal to 1 over 1 over the length of v times v. I showed you this in the unit vector video. You can construct a unit vector that goes in the same direction as any vector, essentially just by dividing, or I guess multiplying, that vector times 1 over its length. In general, we can always redefine the line. All of the possible scalar multiples of v are going to be the same thing as all of the scalar multiples of our unit vector, u, which is just a scalar multiple of v. We can redefine our line. If we redefine our line, L, as being equal to all the possible scalar multiples of our unit vector, where the scalars are any members of the real numbers. Our projection definition simplifies a good bit. The projection of x onto L becomes x dot our unit vector, times the unit vector, times the unit vector itself. That case that I did in the previous video, where I had those two vectors. Where I said the vector v that defined the line, I think it was vector 2, 1. Our vector x was equal to 2, 3. If you want to do this definition, we just have to turn this guy into a unit vector first. The way you turn him into a unit vector, you figure out the magnitude. In this case the magnitude of v is equal to what. 2 squared plus 1 squared is 1. You take the square root of that. Let me just write. It's equal to square root of 2 squared plus 1 squared, which is equal to the square root of 5. You can define your u-- your unit vector could just be 1 over this, times that guy. 1 over the square root of 5 times 2, 1. You could multiply it out, or not. You could just leave it in this form. You can always, for any vector v, you can always find a unit vector that goes in the same direction, assuming that we're dealing with non-zero vectors. You can always reduce anything like this, to some other definition, like this. Where this is a unit vector version of your vector v up there. I just said that, look, this is a transformation from Rn to Rn. The one thing that we're not sure of, is this a linear transformation. We can always write it like this. Let's see if this is always going to be a linear transformation. There's two conditions for it to be a linear transformation. Let's see what happens if I take the projection onto L of two vectors. Let's say the vector a plus the vector v. If I take the sum of their vectors. If this is a linear transformation, this should be equivalent to taking each of their projections individually, and then summing. Let's see if this is the case. This is equal to, by our definition, we'll use the unit vector version, because it's simpler. This is equal to a plus b, that's our x, dot u. And then, all of that times our unit vector. We know that the dot product has a distributive property, so that this is equal to a dot u plus b dot u. These are unit vectors. All of that times the vector u. These are just scalar numbers. So scalar multiplication has distributive properties. This is equal to a dot u, times our vector u. Remember, this is just going to be some scalar. Plus b dot u times our unit vector u. What is this equal to. This right here is equal to the projection of a. This is equal to the projection of a onto L, by definition, right here. By this definition. If we assume that we're dealing with the unit vector definition for the line. This is equal to, this whole thing right here, is equal to plus the projection onto L of the vector b. We see our first condition for this being a linear transformation holds. A projection of the sum of the vectors is equal to the sum of the projections of the vectors. Our second condition is that the projection of a scalar multiple should be equal to a scalar multiple of the projection. Let me write that down. What is the projection onto L of some scalar multiple of some vector a. That is equal to ca dot our unit vector u times unit vector u. This one is a little bit more straightforward. This is the scalar multiple. We see it in our dot product properties, this is equal to c times a dot u, times the vector u. This is just equal to c times, this right here, is a projection of a onto L. We met both of our conditions for linear transformations. We know that our projection onto a line L in Rn is a linear transformation. That tells us that we can represent it as a matrix transformation. We know that projection of x onto L, we already know this definition, it can be rewritten. It doesn't hurt to rewrite it. As x dot some unit vector that defines our line. Let me draw it with a little hat to show that it is a unit vector. Times the unit vector itself, so that we actually get a vector. How can I write this as some matrix product. Some matrix vector product. I want to write it as a product of some matrix times x. To simplify things, since we're actually dealing with a matrix, let's limit ourselves to the case of R2. I'm assuming that my projection onto L is going to be a mapping from R2 to R2. You could do what I'm doing here with an arbitrary dimension. If we're doing it in R2, then our matrix A, right there, is going to be a 2 by 2 matrix. We've seen in multiple videos that to figure out the matrix A, we just take the identity matrix that has the standard basis vectors as columns. 0, 1. Or 1, 0, and then 0, 1. And we apply the transformation to each of these columns. We could say that A is going to be equal to-- its first column is going to be equal to the projection onto L of this thing right here. We'll do it in this orange color, right here. What is that going to be. That is going to be this dot u. Let me write my u. My unit vector, let's just assume that u can be rewritten as my unit vector is equal to sum u1 and u2. Just like that. I need to take this dot my unit vector, let me write this down. Let me write this on the side. The first thing I want to do is figure out what the projection-- the projection onto L, let me write it this way. We know the projection is just equal to this dot this times that vector. Let me write that. The vector 1, 0 dot the unit vector u, which is just u1, u2. We're going to have that times my unit vector. I'll write it like this. Times the vector u1, u2. This is going to be my first column in my transformation matrix. My second column is going to be the same thing, but I'm not ready to take the projection of this guy. The definition of our projection is you dot this guy with our unit vector. So we dot it. We're taking the dot product of 0, 1. 0, 1 dot my unit vector dot u1, u2. I'm going to multiply that times my unit vector, times u1, u2. This seems very complicated, but it should simplify when we actually try to work out our transformation matrix. Let's do it. When I dot these two guys, what do I get. Let me write it here. My matrix A will become 1 times u1, plus 0 times u2. That's just u1. This whole thing just simplifies to u1, when I take the dot product of these two things. Times u1, u2. That's going to be my first column. My second column, if I dot these two guys, I get 0 times u1 plus 1 times u2. So I'm going to get u2 times my unit vector, u1, u2. If I multiply that out, this will be equal to what. I can just write them as columns. u1 times u1 is u1 squared. u1 times u2 is u1, u2. u2 times u1 is u2 times u1. Then, u2 times u2 is u2 squared. You give me any unit vector and I will give you the transformation that gives you any projection of some other vector onto the line defined by that. That was a very long way of saying that. Let's go back to what I did before. Let's say we want to find any projection onto the line, onto the vector, I'll draw it here. We'll do the same example that we did in the last video. If I have some vector v that looks like that. We said the vector v was equal to the vector 2, 1. That was my vector v. How can we find sum transformation for the projection onto the line defined by v? Onto this line right here. The line defined by v. What we can first do is convert v into a unit vector. We can convert v into a unit vector that goes in the same direction. Some unit vector u. We did that already up here. Where we essentially just divided [? bv ?] by it's length. Let's take v and divide by it's length. The unit vector is this, 1 over the square root of 5 times our vector v. It was 1 over the square root of 5 times our vector v, right there. You start with a unit vector there. You just create this matrix, and then we will have our transformation matrix. If this is our u, what will our matrix be equal to. This is u. Then our matrix would be equal to u1 squared. What is u1 squared. Let me rewrite our u a little bit, not at angles. Our vector u, our unit vector that defines this line, is equal to the vector 2 over the square root of 5 and 1 over the square root of 5. I just multiplied out this scalar. If we want to construct this matrix, we get A is equal to u1 squared. What is this squared? It becomes 2 squared 4 over the square root of 5 squared, which is just 5. Equals 4 over 5. What is u1 times u2? 2 times 1 over square root of 5 times square root of 5. So, 2/5. I just multiplied these two. What is u2 times u1. Same thing. Order doesn't matter when you multiply. This will also be 2/5. What is u2 squared. 1 squared over the square root of 5 squared is just 1/5. Now we can say-- and that's the neat thing about creating these matrices, that the projection-- Let's say we have some, let's say this is the origin right here, we have some other vector x, right here. We can now define our transformation. The projection onto L where L is equal to any scalar multiple of our unit vector u. It's right here. Is a member of the reals. That is our line L. The projection onto L of any vector x is equal to this matrix. Is equal to the matrix 4, 5, 2/5, 2/5, 1/5 times x. Which is a pretty neat result, at least for me. We once again reduced everything to just a matrix multiplication. You take this x and you multiply it by this matrix, you're going to get its projection onto the L, onto the line. If you take this vector, let's say a, and you multiply it times this matrix right there, you're going to get its projection. Its projection onto the line. If you could take this vector-- No, it should go through the origin. I want to draw it in standard position. If you take this vector, right there, and multiply it times this matrix, you're going to get this vector, right here, that is contained in the line. When you subtract it from this, it's orthogonal. We know the definition. It's kind of the shadow of that vector. Anyway, I think this is pretty neat.