 In the previous video, we introduced the notion of a coordinate vector for a vector given a basis, but it was very important we emphasize that the coordinate vector was relative to a basis. Now, given a vector space v, one could actually have numerous, numerous different basis for that same vector space. So in this example, let's consider two different bases for this vector space v. We're going to take the first basis b, which will consist of vectors b1 and b2, and we'll have another basis c, which will consist of vectors c1, c2. Now, I'm not going to specify what the vectors b1, b2, c1, c2 are. The only thing we can know for right now is that because one basis contains two vectors, all bases for this vector space will contain two vectors because the base or the vector space is two dimensional. What we will say about the basis is the following. Since b is a basis for v, we know that every vector in v can be uniquely expressed as a linear combination of the vectors in b. Now, because c is likewise a basis for v, these c1 and c2 are vectors inside of the vector space v, excuse me. Therefore, we can uniquely express the vector c1 as a linear combination of the b's. Hypothetically, let's say that c1 equals 2b1 minus 3b2. Likewise, the vector c2, which is in v, we can express as a unique linear combination with respect to the vectors b1 and b2, and let's say that hypothetically that linear combination is negative 3b1 plus 5b2. That's because b is a basis, so every vector, including the other basis, can be uniquely expressed in terms of b's. Now, if we have this linear combination of c1, this gives us the c1 coordinate vector relative to b. We'll just grab the coefficients 2 and negative 3, and that gives us our coordinate vector. Likewise, we have a coordinate vector for c2 with respect to b coordinates. We would just grab the coefficients in this linear combination negative 3 and 5, so we can express the c basis in terms of b coordinates. Now, let's take another vector inside of v. This time, let's call it x. Now, x can be uniquely expressed in terms of the c basis, because again, c is a basis, so every vector v can be uniquely expressed in terms of the vector c1, c2. In this situation, let's say that the vector x is written as 1c1 plus 3c2. Much like the vector c, then, we can write a coordinate vector for x, but this would be the coordinate vector of x relative to c, and its coordinates would be 1 and 3 that we see right here. Because we have these two different bases, and since we have a coordinate vector for x right here, c1, we don't actually know what x is. And since we don't actually know what the b's and the c's are, we can't give you a specific vector to represent what's going on right here. But we do have enough information to determine what would the b coordinate vector be for x. And it follows from the following idea here. If we look at the coordinate vector, if we look at the coordinate vector for x relative to b, well, since x can be written as a combination of the c's, we can make a substitution and say that the coordinate vector of x will be the coordinate vector of c1 plus 3c2 relative to b. Now, as we mentioned in the previous video, the coordinate map is, well, first of all, it's a one to one and on to map, but most importantly here, it's a linear map. That is, linear transformations are preserved by the coordinate map. So a linear transformation, a linear combination inside of a coordinate vector becomes a linear combination of coordinate vectors. c1 plus 3c2 inside of this coordinate vector relative to b becomes a combination of coordinate vectors 1c1b plus 3c2b. Now the c1b, the coordinate vector of c1 relative to b, we know it's 2 negative 3. And the coordinate vector of c2 relative to b, we also know negative 3, 5. And so now this right here just becomes a calculation in f2. And therefore, we can simplify this, 2 plus negative 9 is negative 7, negative 3 plus 15 is 12. And so this right here captures then the coordinate vector of x relative to b. So since we knew x and c coordinates and we knew the vector c's in b coordinates, we were able to transition from c coordinates to b coordinates for our vector x. And I want to look at this above equation from a different light. So what we observed earlier is that the b coordinates of x can be written as the combination of the c coordinate or the b coordinates of the c's with these coefficients 1 and 3. Now this linear combination, in fact, every linear combination can be factored as a matrix vector product where the matrix will be the, the column vectors of the matrix will be the vectors in the linear combination. So we get a, we get a matrix whose column vectors are c1b and c2b. And then the vector in this factorization, it's will be the scalars associated to this linear combination. So you're going to get these scalars 1 and 3. You're going to get these column vectors 2, negative 3 and negative 3, 5. But the first column is just c1b. The second column is just c2b. And then the vector 1, 3 right here, this was the vector x with c coordinates. And so what we see here is that there was a matrix. We could times this, we could times the c coordinate vector by a certain matrix. And this would give us the b coordinate, the b coordinate vector. And so the significance of this matrix right here and right here is that it changed from c coordinates to b coordinates. We want to generalize this idea. So again, let's take a vector space v, which has two bases. B is the basis consistent of vectors will be 1b2 up to bn. And the basis c will be a basis consistent of vectors c1, c2 up to cn. Notice they both have the same number of vectors because that's the dimension of v. It's n right here. Then there will be a unique in by in matrix, which we will denote p over c2b. So notice that there's a little arrow here. The arrow goes from c to b. And this will be called the change of basis matrix to be from c. Oops, my little c there looks like the complex numbers. That's just supposed to be the little c we had above. We want to take the change of basis matrix from b to c. And this matrix will have the property that if you multiply the c coordinates of a vector x by this change of basis matrix, it'll give you the b coordinates of that matrix. And so this actually indicates the mnemonic device we're using right here on the bottom of this matrix. Notice how it's pointing from c to b. So it's clear with the direction we're going. We're changing from c coordinates to b coordinates. But why are we pointing it from right to left as opposed to maybe left to right? The reason here is that in terms of matrix multiplication, you're going to times a by a, if you times a matrix a by a vector x, the vector x goes on the right hand side. And so what we're trying to say is that the matrix, the change of basis matrix p right here, it's going to go on the left of the matrix. And so we want the arrow pointing to the left to show you the transition. So we go from c coordinates right here to b coordinates over here. And so this is the change of basis matrix to b from c. And so we're doing that, of course, intentionally here. Now the other thing I want to mention about this change of basis matrix is not that just what it does, but we actually have a formula from it, which we discovered from this example, generalizing the principles and play, of course, if we want to change to b coordinates from c coordinates, we're going to take the old basis, which is C, we take all those C1, C2, CNs, and we're going to write the b coordinates of the c vectors. Like you can see right here. And so multiplication by this change of basis matrix p will convert from c coordinates to b coordinates. And to change between the two matrices, we need the coordinates of the old basis in terms of the new basis. Let's look at an example of such a thing. So let's again, let's take some, well, actually this time, let's let's just take R2 right here. Neither one of these is going to be the standard basis per se. But let's take two bases for R2. We're going to take the b basis will consist of B1, B2, which is illustrated here B1 is 32 and B2 is 43. Then we have a second basis C, which will consist of C1 and C2, C1 being the vector 1, 2, and C2 being the vector 5, 1. I should mention that the the dimension of R2, of course, is two dimensional. So every basis should have two elements. And let's can look at these ones right here. If you look at the B's, these vectors are not linear combinations of each other. There's no multiple of B1 that'll give you B2 and vice versa. So this is an independent set of two vectors and a two dimensional vector space. So it'll be a basis. Same thing for the C's, no multiple of C1 will give you C2 and vice versa. So the set C will consist of two independent vectors in a two dimensional vector space. So it has to be a spanning set because again, the magic number in R2 is its dimension two. So let's consider how would we compute the change of basis matrix to be from C? Well, we need to compute the coordinate vectors C1, B and C2, B. And so how would you do that to find the coordinate vector of C1, B? What you would do is you would take the basis B. So you take the column vectors B1, B2, and you'd augment that with the C1 here. So it's the first sort of consideration is like, well, how, how would you write C1 as a linear combination of the B's? So you take this matrix whose coefficient matrix is essentially the basis B and whose augmented column is C1 right here. If we row reduce that kind of skipping over the steps, the row reduced echelon form here will be this one zero zero one. Notice that in terms of bases, the first column is just E1. And then the second column would be E2. That is these E vectors or the vectors which has a one in the in the I spot and zeros everywhere else. That's what's going to happen when you row do row reduce a matrix, which is a basis, it'll always turn into the standard basis of fn right there. In this case, it turns into the base standard basis for R2. And then what we see here is what happened to the C1. So what this tells us is that C1 is equal to negative five times B1 plus four times B2. And the negative five and the four give us the the B coordinates here. That's a typo it should say C1 right there C1 of B. That's going to be this negative negative four negative five comma four. And so if we do the same operations, we have to basically do the same problem here that if we want to figure out how to write C2 and B coordinates, we're going to take the matrix whose coefficient matrix is the new basis B, and whose augmented column is the old basis element C2. So the exact same coefficient matrix, but now we have a different augmented column this time. If we do row operations, we have to row reduce the coefficient matrix. And although we didn't see what those operations were, I can promise you they're going to be the exact same row operations that we had when we did the row reduction right here. And so the basis B will row reduce to this to the basis E. And then C2 will transition to this vector 11 negative seven, which would then be the coordinate vector for C2 with respect to B coordinates which we see right here. And so by the calculation we've done here, we now see that the change of basis matrix to be from C would then be the putting together of these two coordinate vectors negative five four and 11 negative seven. So this is the change of basis matrix. We can change from C coordinates to B coordinates using this matrix. But I also want to illustrate that in this example, we kind of solve two two linear systems, which essentially were the same linear system. What we could have done is we could have done the following. We could have taken B1, B2, augmented with C1, but we could also augment it with C2 here because when it comes to row reduction, we only care about what's happening on the left hand side. That is to say in terms of how the choices we make, should we do a replacement? Should we do interchange? Should we do scaling? It'll depend on the left hand side of this augmented matrix. The right hand side is just along for the journey. We want to see what happens to it. So we could actually row reduce these things simultaneously. That is we could have more than one augmented column. Why not? And if we had done that, this would have row reduced to be one zero zero one. You get the standard basis for R2. And then you get negative five, 11, four, and negative seven. And so we could have actually found the change of basis, the change of basis matrix by actually solving one linear system simultaneously. And this is actually how we are going to generally solve for the change of basis matrix.