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let's say I've got some basis B and it's made up of K vectors let's say it's v1 v2 all the way to VK and let's say I have some vector a and I know what a is coordinate SAR with respect to B so this is the coordinates of a with respect to B are c1 c2 and I'm going to have K coordinates because we have K basis vectors or if this describes a subspace this is a K dimensional subspace so I'm going to have K of these guys right there and all this means by our definition of coordinates with respect to a basis this literally means that I can represent my vector a I can represent my vector a as a linear combination of these guys where these coordinates are the weights so a would be equal to c1 times v1 plus c2 times v2 plus all the way keep adding them up all the way to CK times VK now another way to write this another way to write this is that let me write it this way if I had a if I had a matrix where the column vectors were the basis vectors of B so let me write it just like that so let me say I have some matrix C that looks like this where its column vectors are just these basis vectors so we have v1 v2 all the way to VK if we assume that these are let's say that all of these are a member of RN then each of these are going to have n entries or going to be it's going to be an N N by K matrix each of these guys have n entries so we're gonna have n rows and we have K columns so this let's imagine this matrix right there another way to write this expression right there is to say that a is equal to a is equal to the vector c1 c2 all the way to CK multiple by this matrix right there this would be equal to a this would be equal to a this statement over here and this expression over here are completely identical if I take this matrix vector product what do I get I get c1 times v1 c1 times v1 plus c2 times v2 plus c2 times v2 plus c3 times v3 all the way to CK times VK is equal to a we've seen this multiple times in multiple different contexts but what's interesting here is this expression is the same thing that's really I'm just applying new words to things that we've seen probably a hundred times by now we can rewrite this expression this is C and remember C is just a matrix with our basis vectors as columns C is equal to this guy this is just the coordinates of a with respect to the basis B so C times the vector that has the coordinates of the vector a with respect to the basis B that is going to be equal to a that is equal to a now why did I go through the trouble of doing this because now you have a fairly straightforward way if if I were to give you this if I were to give you that right there and say hey what is what what is a if I wanted to write it in in the standard in standard coordinates or with respect to the standard basis which is just kind of the way we've been writing vectors all along then you just multiply it times this matrix C this matrix that has the basis vectors as columns on it the other way if you have some matrix a let's or sorry if you have some vector a that you know can be represented as a linear combination of B or it's in the span of this basis vectors then you could solve for this guy right here to figure out a coordinate with respect to V and so this little matrix right here what does it do it helps us change bases if you multiply it times this guy you're going from the vector represented by coordinates with respect to some basis and you multiply it times this guy you're going to get to the vector and just the standard just with standard coordinates so we call this matrix right here chain basis matrix change of basis change of basis matrix which sounds very fancy but all it literally is is a matrix with the basis vectors as columns so let's just apply this a little bit to see if we can do anything vaguely constructive with it let's say that I have some basis let's say B for basis let's say let's say I have two vectors I'll define the vectors up here let's say vector one is we're dealing with r3 so vector 1 is 1 2 3 and let's say that vector 2 vector 2 is vector 2 is 1 0 1 and let's say I'm going to define some basis B as being the set of the vectors V 1 and V 2 and you can leave it to you to verify that these are not linear combinations of each other so this is a valid basis that you know these aren't in any way linearly dependent now let's say that I know some vector that's in the span of these guys let's say that I know some vector that's in the span of these and all I know is how it happens to be represented in coordinates with respect to this basis so let's say is I have some vector a and when I represent the coordinates of a with respect to this basis with respect to this basis it's equal to I don't know tate's seven seven minus four so how can we represent this guy in its standard coordinates or you know how would we what is a equal to what is a equal to well you could just you know you say oh well a is equal to seven times v1 minus four times v2 and you would be completely correct but let's actually use this change of basis matrix that I've introduced you to in this video so the change of basis matrix here is going to be is going to be just a matrix with v1 and v2 as its columns so 1 2 3 and then 1 0 1 and then if we multiply a change of beta basis matrix times the vector representation with respect to that basis so times 7 minus 4 7 7 minus 4 we're going to get the vector represented in standard coordinates so what is this going to be equal to what is this we have a 3 by 2 matrix times a 2 by 1 we're going to get a 3 by 1 matrix which makes completely which makes complete sense because we're dealing in r3 we're dealing in r3 this vector right here a is going to be a member of r3 a is a member of r3 so when we write it with standard coordinates we should have three coordinates right there now when we represented a with respect to the basis we only had two coordinates because a was in the plane spanned by these two guys actually this is good this is a good excuse to draw this so let me draw it in three dimensions let's say the span of v1 and v2 looks like this the span of v1 v2 looks like this and this is the let's say this is the zero vector right there so this right here is the span span of v1 and v2 or another way this is the subspace that B is the basis for and so we know that a is in this guy so let's say v1 looks like this let's say v1 looks like this and that v2 I'm not even looking at the numbers I'm just doing it fairly abstract let's say v2 looks like let's say v2 looks like this right here now the fact that a can be represented as a linear combination of v1 and v2 tells us that a is also going to be in this plane in r3 and in fact it's seven times v1 so seven times v1 so 2 1 V 1 to V one's three ones 4 5 6 7 so it's 7 in that direction and then it's a minus 4 in the v2 direction so that's one in the v2 direction this is minus one in the v2 direction minus 2 minus 3 minus 4 or we could do it here 1 2 3 4 so our vector a is going to look like this it's going to look like this going to sit on the plane it's going to sit on the plane so this is our vector a it's going to sit on the plane and when we represent it with respect to this basis when we represent as coordinates with respect to our basis B we say ok it's 7 of this guy I didn't even I'm just doing this abstract li don't pay attention to the numbers just now I just want to understand the idea we say it's 7 of this guy minus 4 of this guy so it takes you back here and you get this vector which is in this plane so we only needed two coordinates to specify it within this plane because this subspace was two-dimensional but we're dealing in r3 we're dealing in r3 and if we just want the general version of a in standard coordinates we'll have to essentially get three coordinates three coordinates and I want you understand that a is sitting on this plane this plane just keeps going on and on and on in all of these directions a actually sits on that plane it's a linear combination of that guy in that guy but let's figure out what a looks like in standard coordinates in standard coordinates we get the first term is going to be 1 times 7 plus 1 times minus 4 so that's going to be 3 then you get 2 times 7 plus 0 times minus 4 that is 14 you get 3 times 7 plus 1 times minus 4 so 3 times 7 is 21 minus 4 is 17 so a is the vector 314 17 that is equal to a now let's say we wanted to go the other way let's say we wanted to go the other way let's say we have some vector and let me pick a letter I haven't used recently let's say I have some vector D let's say so I have some vector D which is 8 8 minus 6 2 and let's say we know we know that D D is a member of the span of the span of our basis vectors the span of v1 and v2 v1 and v2 which tells us that D can be represented as a linear combination of these guys or that D is in this subspace or that D can be represented as coordinates with respect to the basis B remember the basis B the basis B was just equal to the set of v1 and v2 that's all the basis B was now we know we know that if we have our change of basis matrix our change of basis matrix x times the vector made up of the coordinates of D with respect to B so let me write that down D with respect to B is equal to D we know that we know if we have this guy's coordinates and we multiply by the change of basis matrix we'll just get the the regular standard coordinate representation representation of D now in this case we have D we're given this we of course know what the change of basis matrix is so if we wanted to represent D in coordinates with respect to B we're going to have to solve this equation so let's do that so our change of basis matrix is 1 1 2 0 3 1 and we're going to have to multiply it some times some coordinates this thing right here we can represent it as do it in yellow we're going to represent it we're going to need to cord it's going to be some multiple of of v1 times plus some multiple of V 2 so it's C 1 C 2 and we know it has to be two coordinates because this matrix vector product is only well-defined if this is a member of r2 right because this is a three by two matrix we have two columns here so we have to have two entries here and then that's going to be equal to D that's going to be equal to D so if 8-6 two and so if we figure out what this vector is we've figured out what the representation or the coordinates of D with respect to B are so let's solve this let's solve this so to solve this we can just set up an Augmented matrix we can just set up an Augmented matrix that's just our traditional way of solving a linear equation so what we have we have 1 1 2 0 3 1 we augment it with this side right there so we have 8 minus 6 and 2 and let's keep my first row the same 1 1 augmented it with 8 and let's replace my second row let's replace my second row with the second row minus 2 times the first row so I'm going to get 2 minus 2 times 1 is actually let me do it the other way let me replace my second row with 2 times my first row minus my second row so 2 times 1 minus 2 is 0 2 times 1 minus 0 is 2 2 times 8 is 16 minus 6 is 10 now let's replace let's replace the third row with 3 times the first row minus the third row so 3 times 1 minus 3 is 0 3 times 1 minus 1 is 2 and then 3 times 8 3 times 8 is 24 minus 2 3 times 8 is 24 minus 2 is going to be 22 see it looks like I must have made a mistake someplace because I have these two would lead to no solution so let me verify what I did make sure I didn't make any strange errors so the second row I replace it with 2 times the first row minus the second row so 2 times 1 minus 2 is 0 2 times 1 minus 0 is 2 2 times 8 minus minus 6 so there's my error that's equal to 22 that was my error so these two things are equivalent now let me let me let me do I'll do one step at a time let me divide let me let me replace my third row with my third row minus my second row just get it out of the way so I'll keep this one one eight zero two twenty two and then the third row is going to be my third row I'm going to replace it with my third row minus my second row so it's going to be zero zero zero so that just gets zeroed out now let me divide my second row let me divide my second row by two so I get 1 1 & 8 and then this one becomes 0 1 and 11 and then of course the third row is just a bunch of 0 and then let me keep my middle row the same so it's 0 1 and 11 and then let me replace my first row with my first row - my middle row so 1 minus 0 is 1 1 minus 1 is 0 8 minus 11 is minus 3 and I'll keep my last row the same so I've put the left hand side in reduced row echelon form so this right here is essentially telling me my solution I know that C 1 the solution so I can write it this way I could write that 1 0 0 1 0 0 times C 1 C 2 is equal to is equal to minus 3 11 0 or another way of writing this is that 1 times C 1 plus 0 times C 2 or C 1 is equal to minus 3 right 1 times C 1 plus 0 times C 2 is equal to minus 3 and then we could have and then we have 0 times C 1 plus 1 times C 2 plus 1 times C 2 is going to be equal to is going to be equal to 11 so our solution to this equation our solution to this equation is is minus 311 or another way of saying this is that if I wanted to write my vector D in coordinates with respect to my basis B it would be the coordinates through minus 311 so if I want to write my vector D in coordinates with respect with respect to my basis B it's going to be equal to minus 311 which implies which implies we write it this way which implies that D is equal to minus 3 times vector 1 plus 11 times vector 2 and I'll leave you I'll leave that for you to verify but just like that using this change of basis matrix we can go back and forth if you have this representation it's very easy to take the product and get the standard representation for D if you have the standard representation or the coordinates with respect to the standard basis it's easy well it's a little more involved but then you just solve for your coordinates with respect to the with respect to be