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like we've done in the last several videos let's assume that we have some set of basis vectors B and let's say so our basis is going to be v1 v2 all the way to VK so this will span a subspace of dimension K and let's assume that each of these guys each of these guys are members of RN so v1 v2 all the way to VK they're all members of RN now in the last video we saw that we can define a change of basis matrix and it's a fancy word but all it means is a matrix that has these basis vectors as its columns so v1 v2 all the way to VK as its columns so we're going to have K columns we're gonna have K columns and we're going to have n rows because each of these guys are members of RN so they're going to have n entries so we're going to have n rows so it's going to be an N by K matrix and we saw in the last video that if I have some matrix a we have a sorry if I have some vector a that is a member of RN then and assuming that a is in the span of B I can represent a I could say that a is equal to is equal to the change of basis matrix times the coordinates of a with respect to our basis this is what we saw in the last video if I have the coordinates of a with respect to B I can multiply it by the change of basis matrix and I'll get my vector a in standard coordinates or if I have my vector a in standard coordinates then I can solve for my vector a in coordinates with respect to B we saw that in the last video now let's take the special case let's take the special case let's assume that let's assume let's assume that C is invertible let's assume that C is invertible what does that mean or what does that tell us about C well if c is invertible it's two things it means that c is a square matrix or it has the same number of rows and columns and then it's rows or columns you can pick either of them have to be linearly independent so linearly independent let's just pick columns now the second statement is a bit redundant we know that C has linearly independent columns because its columns are basis for a subspace so basis by definition all of the vectors have to be linearly independent so we know this is this is a bit redundant redundant but what's interesting is if we know that c is invertible c has to be square and if all of these vectors are members of RN then k has to be equal to then K has to be equal to n so C is square means that K is equal to n or that we have or that we have n basis vectors and basis vectors now if that's the case what is the span of B think about it we have K with sorry we have n linearly independent vectors n linearly independent vectors independent vectors in RN in RN so anytime you have n linearly independent vectors in RN those guys are a basis for RN because any basis that has n entries and they're all linearly independent is going to be a basis for RN so then B B is a basis for RN so if we know that c is invertible we also know that you can get to any vector in RN by some linear combination of your of your basis vectors right there so we don't you know in the last video we had to make sure that this guy that this guy was in the span of these vectors but now we don't have to make sure because if c is invertible if c is invertible invertible then then the span of B is going to be equal to RN or another way you could say it is if the span of B is equal to RN if we have n vectors here if K was equal to and then we know that the span of B would be equal to RN and so we'd have n vectors here and linearly independent columns here and it would be an N by n matrix with all the columns linearly independent so then C would be invertible so we could write if and only if so we can write it the other way if the span of B is RN then C is invertible and that's useful because if either of these things are true then we can rewrite the same equation so let's say if we know if we know this and we're looking for that we can just multiply C times that let's say where we know this and we're looking for that before we have to do that augmented matrix and solve for it what not but if we know C is invertible if we know C is invertible then 1 we know that any vector here can be represented but it's in the span of our basis so any vector here can be represented as linear combinations of these guys so you know that this is that any vector can be represented in these coordinates or with coordinates with respect to our basis and so we can just take we can multiply both sides of this equation times C inverse and what do you get if you multiply so it becomes C inverse C times our coordinates of a with respect to B is equal to C inverse times a this is just the identity matrix right there so we just get another way of writing this is that the coordinates of a with respect to our basis B which spans all of our N is equal to C inverse times our vector a let's apply this a little bit let's apply this let's use this information what we've done in this video let's do some concrete examples so let's say I have some basis let's say my basis is equal to let me define two vectors I'll do it this way so let's say I have v1 is equal to the vector 1 3 and let's say v2 is equal to the vector 2 1 and I have a Asus that is equal to the set of v1 and v2 now I'll leave it for you to verify that these guys are linearly independent but if I have two linearly independent vectors in r2 then B is a basis for r2 and if we draw it that if we make the chain if we write the change of basis matrix if we say C is equal to 1 3 2 1 we know that c is invertible and actually to show that c is invertible we can actually just calculate its inverse so what's the determinant of C the determinant of C is equal to 1 times 1 1 times 1 minus 2 times 3 so it's equal minus 5 that's the determinant of C and so C inverse we figured out a general formula for doing this for 2 by 2 matrices is equal to 1 over the determinant of C so 1 over 5 or so 1 over minus 5 times you switch these two guys so you switches the ones and you make these two guys negative so minus 2 and then minus 3 and the very fact that this guy was the determinant of C was nonzero told us that this was invertible but anyway this is C inverse so let's say that I have some vector let's say that I have some vector a that is a member of r2 let's say that a I'm just going to pick some random numbers a is equal to 7/2 and I want to find out what the coordinates of a are with respect to my basis B with my respect with respect to the basis B well we go to this situation we know what we know what a is so we just multiply a times C inverse to get this guy right here to get the coordinates of a with respect to B so let me write that down so what is C so C is that C inverse is that so we could write the coordinates of a with respect to B is equal to C inverse times the standard coordinates of a or this is the same thing let me put the actual numbers here the coordinates of a with respect to B are going to be equal to C inverse which is minus 1/5 times 1 minus 3 minus 2 1 times times a times a times 7/2 and what is this equal to this is equal to minus 1/5 and then we're going to get 1 times 7 plus 2 minus 2 times 2 so it's minus 4 so 7 minus 4 is 3 and then we're gonna get minus 3 times 7 which is minus 21 plus 1 times 2 so minus 21 plus 2 is minus 19 so the coordinates of a with respect to the basis B so the coordinates of a with respect to B are going to be equal to let me just multiply the negative 1/5 you get minus 3/5 and then you get plus 19 fifth so the 19 over 5 just like that and let's verify that let's take this means that a is equal to minus 3/5 times our first basis vectors plus 19 fifths times our second basis vector let's verify that that's the case so let's see minus 3/5 times our 1 3 plus 19 5th times 2 1 let's see what this is going to be equal to this is equal to this is equal to let me write the two vectors this is minus 3/5 minus 3/5 this is minus 3/5 times 3 is minus 9 fifths minus 9 fifths and then we're going to add it to this guy so this guy is 2 times 19 is 38 fifths 38 fifths right and then 1951 is 19 this and then if you add these two vectors together what do we get we get minus 3/5 plus 38 fifths that's thirty five fifths thirty-five fifths is seven minus nine fifths minus nine fifths plus 1950s ten fifths or two and there you have it that was our original a so we see that a can definitely be represented as minus three-fifths times our first basis vector plus nineteen fifth times our second basis vector now that was the case where we had some vector a and we wanted to represent it in coordinates with respect to B what if we had the other way what if we said that some vector let's say some vector w's coordinates with respect to B R I'll do something simple r11 are one one then what is W in standard coordinates well there we can just multiply we can just multiply the change of remember W is just equal to the change of basis matrix times W coordinates with respect to the basis B so W is going to be equal to the change of basis matrix which is just 1 3 2 1 it's 1 3 2 1 times the coordinates of W with respect to B times 1 1 which is equal to 1 times 1 plus 2 times 1 is 3 and then 3 times 1 3 times 1 plus 1 plus 1 so 3 times 1 is 3 plus 1 is 4 so W is just equal to the vector 3/4 so there you see if our change of basis matrix is invertible which is really just another way of saying that our basis spans are n in this example it was R 2 then you can easily go back and forth between coordinate representations in our standard coordinates in our standard coordinates and coordinate representations and coordinate representations in not with respect to our basis right this is with respect to the basis this is in standard this was with respect to the basis this is in standard coordinates and you can do that just simply by either using this information or just saying oh the coordinates with respect to the base is equal to C inverse times a are the inverse of our change of basis matrix times a or saying our coordinates in standard with respect to the standard basis is just equal to the change of basis matrix times the coordinates with respect to the basis