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Current time:0:00Total duration:21:35

Video transcript

let's say I have a subspace V that is equal to all of the vectors all the vectors let me write it this way all of the X 1 X 2 X 3 so all the vectors like this that satisfy that satisfy X 1 plus X 2 plus X 3 is equal to 0 so if you think about it this is just a plane in r3 so this subspace is a plane in r3 and I'm interested in finding the transformation matrix for the projection for the projection of any vector X and r3 onto V so how can we do that so we can do it like we did in the last video we can find the basis for this the subspace right there and that's not too hard to do we could say x1 if we assume that let's say that X 2 and X 3 are kind of free variables and we could say X 1 is equal to minus X 2 minus X 3 and then let's just just so that we can write it in kind of our parametric form or if we can write our solution set as the combination of basis vectors we can say X 2 X 2 is equal to let's say it's equal to some arbitrary constant C 2 and let's say that X 3 is equal to some arbitrary constant C 3 then we can say that then we can say that V we can rewrite V we could say that V is I'll do it here V is equal to the set of all X 1 X 2 s and X 3 s that are equal to C 2 times so X 1 is equal to minus let me write let me rewrite this let me rewrite this with the SI toos this is equal to C 2 this is equal to C 3 so X 1 is equal to minus C 2 minus C 3 so X 1 is equal to minus 1 times C 2 plus C 3 times what plus C 3 times - one and then what is X 2 equal to X 2 is just equal to C 2 so it's 1 times C 2 plus 0 times C 3 X 3 is just equal to C 3 so it's 0 times C 2 plus 1 times C 3 and so this is another way of defining our subspace all of the vectors that satisfy this is equal to is equal to this definition here it's all the vectors whose components satisfy or that lie in this plane whose entries lie in that plane and that's for any real numbers right there or another way of writing that another way of writing this is that V is equal to the span V is equal to the span of the vectors minus 1 1 and 0 and the vector minus 1 0 and 1 just like that and we know that these are actually a basis from V because they're linearly independent there's no way I can take linear combinations of this guy and make the second entry be 1 here and likewise there's no way I can take linear combinations of this guy and make this third entry equal to 1 here so these are also a basis for V so given that just using the technique we did before we could set some vector we could some set some matrix a equal to minus 1 1 0 and then minus 1 0 and 1 and then we can figure out we can figure out that the projection of any vector X in r3 onto V is going to be equal to and we saw this it's going to be equal to a times the inverse of a transpose a all of that times a transpose and all of that times X and you can do it you can you have a here you can figure out what the transpose of a is very easy you can take at a transpose a then you can invert it and it'll be very similar to what we did in the last video it'll be a little less work because this is a 3 by 2 matrix instead of a 4 by 2 matrix but you saw it actually a lot of work it's very hairy and you might make some careless mistakes so let's think let's let's figure out if there's another way that we can come up there's another way that we can come up with this matrix right here now we know that if X is a member we know that if X is a member of r3 that X can be represented then X can be represented as a combination of some mem some vector V that is in our subspace plus some vector W that is in that is an R in the art in the orthogonal complement to the subspace where V V is a member of our subspace and W is a member of the orthogonal complement the orthogonal complement of our subspace now by definition this is that right there is the projection of X onto onto V and this is the projection of X onto the orthogonal complement of V so we can write that X X is equal to the projection onto V of X plus the projection onto V orthogonal complement or the orthogonal complement of V of X so this is by definition that any member of r3 can really represent it this way now if we want to write this as matrix vector products a little and look to videos ago I showed you that these are linear transformations so let me write that here so they're linear transformation so they can be written as matrix vector products you see that right there let me define this matrix I don't know let me call this let me call this I don't know let's just call this matrix T T let me just call it T and let me call let me call let me do another let me do a letter now let me do B and let's say that the projection the project of onto the orthogonal complement of V of X let's say that that's equal to some other vector sorry that's some other metric other matrix C times X we know this is a linear transformation so it can be represented as some matrix C times X so what are these going to be equal to well X if I want to write it as a linear transformation of X I could just write it as the 3 by 3 identity matrix times X right that's the same thing as X that's going to be equal to the projection of X onto V well that's just the same thing as B times X that's the same thing as B times X and then plus the projection of X onto V is orthogonal complement well that's just C times X plus C times X and if you want to factor out the X on this side we know that the matrix vector products exhibit the distributive property so we could write that the identity matrix times X is equal to B plus C times X or another way to view this equation is that this matrix must be equal to these two matrices so we've get that the identity matrix the identity matrix in r3 is equal to the projection matrix onto V plus the projection matrix onto these orthogonal complement so if we're trying to remember the whole point of this problem is to figure out this thing right here is to solve for B and we know a technique for doing it you take a transpose you do you can do this whole thing but that might be pretty hairy but maybe it's easy to find this guy maybe I don't know it actually turns out in this video this one will be easy so if it's easy to find this guy we can just solve for B we get if we subtract C from both sides we get that B is equal to is equal to I is equal to the identity matrix minus the transformation matrix for the transformation onto these orthogonal complement so let's see what this is let's see if we can figure out what C is right there so let's go back to our original so remember let me rewrite the problem actually remember that V remember that V was equal to it's equal to all of the X 1 X 2 s X 3s that satisfy that satisfy X 1 plus X 2 plus X 3 is equal to 0 or another way to say it is all the X 1 X 2 s and X 3s that satisfy the equation 1 1 1 times X 1 X 2 X 3 is equal to the 0 vector or in this case it'll just be 0 we could write the zero vector like that just like that so 1 times X 1 plus 1 times X 2 plus 1 times X 3 is going to equal the 0 vector this is another way to write V now all of the X's that satisfy this right here what is that this this is saying that V is equal to the null space of this matrix right there right the null space of this matrix is all of the vectors satisfy this equation so V is equal to the null space the null space of let me write it this way the null space of 1 1 1 just like that up here we kind of figured out V in the kind of traditional way we figured out that V is the span of these things but now we know that's the same thing as the null space of 1 1 1 these two statements are equivalent now we at least at a hunch that maybe you know we could figure out straight up the this B here by doing all of this a transpose and you know by doing all of this silliness here but our hunch is maybe if we can figure out the transformation matrix for the orthogonal complement of V for the authoritative V right there then that we can just apply this kind of that we can this solve for B given that the identity matrix minus this guy is going to be equal to so let's see if we can figure out let's see if we can figure out the projection matrix if we can figure out the the transformation matrix for the orthogonal projection for X onto the orthogonal projection of V so this this is V what is what is V complement the complement is going to be equal to the orthogonal complement or V perp is going to be equal to the orthogonal complement of the null-space of this of this matrix right here which is equal to what remember the null space it's orthogonal complement then a null space is orthogonal complement is equivalent to is equivalent to the row space or the column space of a transpose we saw that multiple times or you could say the the orthogonal complement of the row space is the null space we've seen this many many times before so the orthogonal complement of this guy is going to be the column space of his transpose so the column space of the transpose of this guy right so it's 1 1 1 just like that or we can write or we can write that V these orthogonal complement is equal to the span the span of 1 1 1 right the column space of this matrix we only have one column in it so it's column space is going to be the span of that one column so just to visualize what we're doing here that original equation for V that satisfies that that's just going to be some plane in r3 that's going to be some plane in r3 that is V right there and now we just figured out what these orthogonal complement is it's going to be aligned in r3 right is going to be all of the linear combinations of this guy so it's going to be some line in r3 I haven't drawn it you know this is going to be tilted more and so is this but it's going to be some line so this is the orthogonal complement of V so let's see if we can figure out so remember the projection the projection let me do it this way so let's create so this is the basis for these orthogonal complement so let's construct some matrix let's construct some matrix on a limb use a new letter that I haven't used before let me construct some matrix D whose columns are the basis vectors for the orthogonal complement of V well there's only one basis vector so it's going to be that and we learned the last video in the video before that that the projection the projection of any vector in r3 on two V's orthogonal complement is going to be equal to D times D transpose D inverse times D transpose times DS transpose times X or another way to view it is that this thing right here that thing right there is the transformation matrix for this projection that is the transformation matrix transformation matrix so let's see if this is easier to solve this thing then this business up here where we had a three by two matrix that was a whole motivation for doing this problem to figure out the projection matrix for the subspace we'd have to do this with a three by two matrix it seems pretty difficult instead let's find the projection matrix to get to the projection onto these orthogonal complement which is this so what is D transpose so D transpose is just going to be equal to one one one what does D transpose times D well that's D transpose this is D just like that so what is this going to be equal to this is just the dot product of that in that right one times one plus one times one plus one times one it equals three so this thing right here is equal to a one by one matrix three so let's write down so this is equal to D D which is this matrix 1 1 1 times D transpose D inverse so d transpose D is just a 1 by 1 matrix we have to invert it actually I've never defined the inverse of a 1 by 1 matrix for you just now so this is mildly exciting times D transpose so D transpose looks like this 1 1 1 and then all of that's times X but this is the transformation matrix right there now what is the inverse of a 1 by 1 matrix now you just have to remember you just have to remember that a inverse times a is equal to the identity matrix if we're dealing with a 1 by 1 matrix then we're just I'm just trying to figure out what let's say what matrix times 3 is going to be equal to the 1 by 1 identity matrix so if I say let's say you know I don't know let's say that 3 inverse 3 inverse times 3 has to be equal to has to be equal to the identity matrix 1 by 1 identity matrix well the only matrix that's going to make this work out to get the center up to take this guy's entry times that guy's entry is going to be this guy right here the inverse of this 1 by 1 matrix has to be the matrix 1/3 1/3 times 3 is equal to 1 this is almost trivially simple but this is the inverse that right there is the inverse matrix for the 1 by 1 matrix 3 so this right here is just 1/3 and we could actually just take that out it's a 1 by 1 matrix which is essentially equivalent to a scalar so this is going to be equal to this is going to be equal to let me just draw a line here this thing is equal to is equal to 1/3 actually I don't want to confuse it let me rewrite it so we get the projection the projection of any vector in r3 onto the orthogonal complement of a V is equal to 1/3 that's 1/3 times the vector 1 1 1 times sorry or way it is a vector the matrix one on one times that matrix transpose 1 1 1 and then all of that times X and you can see this is a lot simpler than when we did it with when if we had to do all of this business if we did all of this business with this matrix that's a harder matrix to deal with this 1 1 1 matrix is very easy now what is this going to be equal to what is this going to be equal to this is going to be equal to 1/3 times we have a 3 by 1 times a 1 by 3 matrix so it's going to result in the 3 by 3 matrix 3 by 3 matrix and what do we get this first entry is going to be 1 times 1 which is 1 second entries going to be 1 times 1 which is 1 second third entries gonna be 1 times 1 which is 1 I think you see the pattern the this guy the second row first column 1 times 1 is 1 so this is going to be a 3 by 3 matrix of ones so just like that we were able to get that was a pretty straightforward situation we were able to get the projection matrix for any matrix any vector in r3 on two V's orthogonal complement now we just we know that this is our this thing right here this thing right here is our original see that we said now we said that we said that the identity matrix we did it wrote it up here let me refer back to what I wrote way up here we said look the identity matrix is equal to the transformation matrix for the product for the projection onto V plus the transformation matrix for the projection onto B's orthogonal complement or we can write that the transformation matrix for the projection onto V is equal to the identity matrix minus the the transformation matrix for the projection onto V is orthogonal complement so we can write so B is our transformation matrix on to so if we say that the projection onto V of X is equal to B times X we know that B is equal to the 3 by 3 identity matrix minus C and this is C right there so B is equal to the identity matrix so that's just 1 0 0 0 1 0 0 0 1 ty minus C minus 1/3 times 1 1 1 1 1 1 1 1 1 just like that what is this going to be equal to this is going to be equal to let's see whenever this is one list in our heads multiply this out this all of these entries are going to be one third essentially if we multiply this out like that so if we have 1 minus 1/3 right I could write it out like that's 1/3 1/3 1/3 everything is one-third one-third one-third one-third one-third one-third one-third and this just becomes a 1 so 1 minus 1/3 is 2/3 and all of the ones minus 1/3 are going to be 2/3 so we could just go down the diagonal and then the 0 is minus 1/3 are going to be minus 1/3 minus 1/3 minus one third minus 1 third you have minus 1/3 minus 1/3 and minus 1/3 and just like that we've been able to figure out our projection or transformation matrix for the projection of any vector X onto V by essentially finding this guy first for finding the transformation matrix for the projection of any X onto V S orthogonal complement anyway that would I thought that was pretty neat and you could rewrite this you could rewrite this as be equal to 1/3 times 2 to 2 to s along the diagonals and then you have minus ones everywhere else minus ones everywhere else anyway see you in the next video