5.4 Eigenvectors and Linear Transformations

based on out here cuz I'm tired from yesterday so we still got to wrap up that 5 point something 5 point 4 where did I put my 5 point for us and I hope this hasn't been too tedious for ya because I again I just decided to kind of take it step by step on some of these and black going here so this is the 5 point 4 exercise I won't write everything out it's on page 4 and now so we got this 2 by 2 matrix a 2 negative 1 1 3 and so we're thinking of the linear transformation that's just multiplication by a from r2 to r2 and I'm calling B this set of 2 vectors 1 1 & 3 0 and that's a basis for r2 and r2 I'm kind of picked a easy space for myself it's 2-dimensional so any set of two linearly independent vectors is going to be a basis and it because it's two-dimensional it's easy to tell if those two are linearly independent because they're not scalar multiples of each other right and so so we're gonna do this find the what's called B matrix for the linear transformation and two ways we're gonna approach this one is the slick way we'll do that second and you may already get the slick way it's based on similarity a matrices but so the first way you might call the long way is to just approach things very naively like we don't know anything that fancy I'm going to call these two vectors it's pretty cool color so let's call them b1 and b2 our two basis vectors and I want to write down what a b1 is so 2 1 negative 1 3 times 1 1 is 1 4 and a b2 is 2 negative 1 1 3 3 0 and that turned out to be 6 3 can check and see the the B matrix would be what you would multiply the B coordinates of a vector by to get the image vectors B coordinates the B matrix would take you from B coordinates to B coordinates so what we want to find out is and I'll put question mark here what are the B coordinates of those two vectors and I just chose some variables to use here x1 y1 x2 y2 so in the slick process there's like a just a fast way to pull this out but I'm not going to do it the fast way this oh okay so I want to take this guy lets me bring in smart color so let's take one four and if x1 and y1 are the B coordinates that just means one four is expressed as X 1 x 1 1 plus y 1 x 3 0 or turning that into a system of equations that's 1 equals x1 plus 3 y 1/4 equals x1 now I gave myself a real easy system to work with here but of course it could be any two-by-two system we could solve for x1 and y1 so I'm just thinking of this this algebraically x1 is for y1 is negative 1 and so that means the B coordinates of that 1/4 vector are 4 negative 1 and we can do the same thing with the 6 3 so 6 3 has to be equal to if I use this notation x2 times the 1 1 vector plus y 2 times 3 0 and that gives me a little system again I can solve really quickly and we what do we got we got X 2 is 3 and Y 2 if I just plug in there is gonna be 1 so 3 1 are the B coordinates of that image vector so if we say so B 1 which is 1 1 what are the B coordinates of B 1 because I danced around circles here it's just 1 0 right because it's my first why because B 1 is equal to 1 times B 1 plus 0 times B 2 right so the B coordinates of B 1 are just 1 0 I mean that these are coordinates designed for that basis so those basis vectors are going to look like kind of like standard vectors in their own coordinate system and so under this transformation 1 0 and B coordinates corresponds to or negative 1 can be coordinates B 2 which is 3 0 and B coordinates that's 0 1 because B 2 is 0 B 1 plus 1 B 2 and we just notice that corresponds to 3 1 and B coordinates so in the land of B coordinates this vector corresponds to that this vector corresponds to that so that means the B matrix is going to be for negative one three one it's just it's just gonna have these columns so you know so that's one way to get the answer and what it so you know what is this thing and this isn't part of the work but what we're saying is if you take the B coordinates of any vector they're going to correspond under this linear transformation to the B matrix times those B coordinates and this would give you the B coordinates of the image so nowhere along the line here do we see our standard coordinates like and what were used to looking at now what's a short way short and fancy actually that wasn't that long it just it was not that long because I gave myself an easy problem but it's just sort of tedious working through those those details call the B matrix of linear transformation I'm gonna call it C okay just because it matches some of the notation with you so I I just pulled a letter out of here to give it a name okay hay is gonna equal PC P inverse or to that matrix C that's what we were talking about last time and I want to add a little notation here the the P is going to be the change of basis matrix or the beat standard change and that means P inverse is that ye to be coordinate change right so we take a vector here and eat coordinates well put it on the right multiply by P inverse changes to be coordinates apply the transformation and B coordinates and then change the B coordinates back to standard and that really is what our original a matrix means now I can multiply on both sides here I can multiply on the left by P inverse on the right by P and I'm gonna get P inverse a P is equal to C again these I like to put this on here and let's say the P inverse was the e to be change because if you think of C by itself you start with B coordinates what what's C gonna do change the B coordinates to e multiplied by the matrix a and standard coordinates change the coordinates back to B that's a way of thinking what C is right so those are now what was our a our a was 2 1 negative 1 3 our P matrix somewhere in all our discussion we actually have a rule for writing down what that is the matrix that changes B coordinates to e coordinates is gonna have b1 and b2 as columns just think about that just plug it in and see and that'll remind you that that's what's happening so P is just our 1 1 3 0 and PN verse you would have to calculate so it is true maybe that's doesn't take that long to find the inverse of P but I'm not going to go through those steps yeah oh I guess I guess it it it let's just say it doesn't matter which one we call P or P n we just have a matrix in its inverse and I think we write our I just wrote it this way because this is how we define our similarity relationship so I don't think there's any law that says you you know you could just as well call this one p and this one p inverse the thing it's not so much the the p here is what changes B coordinates to B coordinates I think that's a thing to know so I don't okay so I'm not bothers me that you find that confusing but I don't I don't so that's just this guy and so again there's no reason why we couldn't call this P inverse and this P but then the thing is you've got to realize we're starting with this matrix and it's not so much which one we call PMPM verse you got to realize we're starting with the matrix whose columns are the B basis vectors is a matrix that's going to change B coordinates to standard coordinates and that that kind of makes sense because if you multiply by a two component vector I don't even want to put I don't know what letters to use here that wouldn't be confusing like Z 1 and Z 2 or something you're gonna take Z 1 times B 1 plus Z 2 times B 2 so those are exactly you know these guys stand for the B coordinates right the weights of the B vectors that we would get okay so calculate P inverse and I'm just gonna write this down zero a third one negative 1/3 so you got to go through the process of calculating and so if we write C as e inverse a P this is my zero 1 third one negative 1/3 a I'm just recapping 2 1 negative 1 3 P is 1 1 3 0 and it's so again it's not so much which one we call P which one P inverse I think the issue is this this is the the B to e standard basis matrix this is the standard e matrix for the transformation and this guy is gonna take our eco ordinates got to be that's the thing you want to think about and again you can plug the plug in a specific sector and watch what happens when you multiply all this out guess what we get it's on the board somewhere I mean it's just this guy okay the four negative one three so that's another way to obtain what we call the B matrix of linear transformation that's right here so I mean I would have to think about that like I wouldn't just and yet if you just work your way through those steps it's there's nothing that profound you just gotta keep once you see it you'll convince yourself that this similarity relationship gives us exactly what we want now I have I have a little bit more I wrote down here so this is just sort of some extra thoughts and I think this might help you guys to I just picked two five and B coordinates and this is sort of picked at random I just want to look at what happens to a particular particular vector here okay so two five and B coordinates what is that vector what is that vector is going to be the same thing as one one three zero two five which is the same thing as 2 times 1 1 plus 5 times 3 0 where this is just b1 and b2 colors mixed up here and so what's this guy turns out to be is 17 - and that's just in standard coordinates I mean you could put a little curly e there but we usually leave the e off right because that's our favorite way of describing a vector with the standard standard basis and all right so what am I gonna do now I want to take that to five vector and B coordinates and I want to multiply by the 4 negative 1 3 1 so the B basis vector I'm sorry the B basis matrix for the transformation and again that that matrix takes me from B coordinates to B coordinates so when I do what I got to do what is it that turns out to be 2033 ok and what are the standard coordinates question is standard coordinates I can do the same thing I did up above I can take one one three zero times twenty three three or take twenty three times 1 1 plus 3 times 3 0 I mean that's what it means for these to be B coordinates these are the weights of the B 1 and B 2 basis vectors and I get thirty two twenty three that's the standard coordinate way of thinking of that vector this is all beyond the end of the problem okay so I'm just sort of rambling on and on and on with my extra thoughts here but here's here's the picture I want to give you guys so remember I picked this this I ran I just picked one particular vector and B coordinates we can ask take our two five and beat coordinates apply the linear transformation the linear transformation in this example and we figured out right here we just mentioned it's gonna get assigned to twenty three three and B coordinates right there's the linear transformation this guy has standard basis coordinates seventeen to the same linear transformation is going to assign this 17 to vector 232 23 in accordance which are the e coordinates of that guy so same linear transformation you can think of it in B coordinates or in a coordinate what's done so what process this is this is multiplication by is that what I called C didn't we call that C along the way today tell me if that doesn't match up with my seat I think that's multiplication by C right and this is the same linear transformation this is multiplication by so kind of my point is if I were a student it would really bother me that two different matrices defined the same linear transformation somehow like they seem to be doing very different things but this is the way we look at it it's the same linear transformation in one coordinate basis C is the matrix you use the same transformation and the standard basis is a multiplication by a and what we haven't done yet is if I take 17 - x a now you remember what a was somewhere what was a a was 2 1 negative 1 3 right 2 negative 1 right make sure I wrote that down right that should give me this guy and we haven't done that calculation on the board here right so I am I am purposely going around in circles and what do I get I get 34 minus 2 is 32 I get 17 plus 6 23 Oh doesn't that make you feel good so I hope you wrote this down because I know if I were sitting there I wouldn't just be saying oh yeah I have to think about a little bit but it's not the math isn't complicated but it's confusing these changes and I so this was just my attempt than these extra thoughts too it's this really this diagram that helps me understand what's going on and remember to five I picked at random so I just picked by random vector applied the transformation to it converted to standard coordinates applied the transformation to it so it was all driven by that random choice of 2/5 that's that's not part of the exercise I just it just helps me to look at what happens to a very specific vector so I if you want you don't have to shout it out now and embarrass me but like if this is like the most obvious thing in the world to you and you feel like I'm just somebody come and tell me it's like why does he keep because I felt like it wasn't obvious and it was worth spending a little bit all right now we got we got one more exercise on that sheet and it has to do with the diagonalization and our really the textbook focuses on diagonalization in this section so this was sort of a journey and more abstraction even though it's a very specific concrete example and our - I thought it was worth thinking about that that general situation but if we get back to diagonalization so this is the page five exercise [Music] so we're talking about T is the transformation that takes a vector to a time's the vector right and here what we want to do is find a basis for r2 with the property so this is the the matrix of that transformation in the and the basis B is diagonals sometimes you can do this sometimes you can't so the way we're going to do this is so we will find diagonal matrix is similar to a and how to do it right I mean that's what it means to diagonalize what do we got to start out with we need to find the eigenvalues we went through this process so start with a minus lambda I which is 1 minus lambda 1 6 2 minus lambda now this is not a hard calculation where to buy to imagine a ten by ten or a hundred by a hundred matrix where someone would be trying to diagonalize right kind of an enormous computing challenge and so the determinant of a minus lambda I is gonna be 1 minus lambda 2 minus lambda minus 6 or 1 minus 3 lambda right plus lambda squared minus 6 lambda squared minus 3 lambda minus 5 so there's our characteristic polynomial so we're gonna set that equal to zero and I think this guy factors right no it doesn't factor what did i write down incorrectly 2-3 oh you guys are so 2 minus 3 lambda plus lambda squared minus 6 and so I have lambda squared minus 3 lambda minus 4 oh I almost I would have got perfect and so we got lambda minus 4 and it's written right here on my sheet you know if I would just look at my piece of paper lambda minus 4 times lambda plus 1 so so lambda equals 4 and negative 1 are the eigenvalues what's our diagonal matrix going to be we already know what the diagonally occurs it's gonna be right we're gonna use the eigenvalues in either order doesn't there's no rule about which order on the diagonal zeros elsewhere but now those are not the basis vectors the basis vectors are going to be the columns of either the P or the P inverse matrix right that we're trying to find so we're not done yet so for so for each eigenvalue we got to describe the eigenspace right we got to find a basis vector find the eigenspace a minus 4i it's gonna be negative 3 1 6 negative 2 so solve negative 3 6 1 negative 2 x equals 0 I'll solve the homogeneous equation we're looking for eigen vectors right and you're gonna get I'm just gonna jump to the conclusion X 2 is free and X is going to have the form X 2 times one third one we got to do the same for the other eigenvalue that was negative one right so we got to take a minus one I plug in the lambda value of it's actually gonna be a plus one all right right minus a negative one I and so we get two one six three and again so a homogeneous equation 2 6 1 3 X 0 and that's easy and you get here again you get X 2 is free so one free variable means a one dimensional solution space here X 2 is free the solution X it's gonna look like it's two times now you really can pick any nonzero scalar multiple of this vector any nonzero scalar multiple of that vector two uses your two basis vectors alright so you know you don't have to actually choose those two but because they're vectors that correspond to distinct eigen values they got to be linearly independent okay here we're 2 by 2 and so we only have two vectors and we see that they're not scalar multiples of each other that's another reason so my P matrix here is one third one negative 1/2 one and my B basis it's gonna consist of those two columns oh and did I I got to be careful here I got to make sure my column 1 3 1 here which correspond to lambda equals 4 so you got to make sure that matches my diagonal matrix right matches the order of eigenvalues in deep so my two basis vectors are 1/3 1 and negative 1/2 1 be basis means if we take the B coordinates of a vector times P that will convert to a coordinate and then we can apply the matrix a right and then I'm gonna stop all right I didn't know today the answer here by the way is just you know this is what we were trying to get this basis that diagonalizes the matrix yes no okay yes aha okay III had more I mean I think this is just what you're asking right let's just write this down if I so I take B coordinates of a vector so this this is just six times one-third 1 plus 8 times negative 1/2 one so that's negative 214 and standard coordinates right okay is this what you're getting at and we can look at a specific example how these matrices act so under the linear transformation this 6 8 and B coordinates is gonna correspond to something x 6 8 actually that would be a good question for my test like after we do all this what goes in there and we've got all these that's actually gonna be the diagonal one right right that was the whole point the B coordinates in this linear transformation yeah have the 4 0 0 negative 1 matrix just keep up on that blue and that's gonna give me 24 negative 8 and B coordinates now 24 negative 8 and B coordinates is what that's 24 times 1 3 1 minus 8 times negative 1/2 1 so these these are just my basis vectors right from the two eigenspaces and what did that come out to be that comes out to be 12 16 and standard coordinates all right so what we're saying is okay this guy's standard coordinates this is really what we're used to thinking of as the vector negative 12 14 and this is what we're used to thinking of as the vector 12 16 and if all this is working I should be able to take 1214 and multiplied by the original matrix the original a which was one six one two and that better give me twelve sixteen or I'm out of here so negative two plus 14 is 12 yes negative 12 plus 28 song sounds like 16 to me okay so we're getting this this is the linear transformation which was originally just multiplication by the matrix a this is the original a this is the diagonal little ization of a so it's a same diagram I drew a moment ago and with a more general situation but here it's just being applied to this diagonalization situation alright was that so here question no no right so that the P the both the P and the P inverse would change the diagonally tricks would not change oh yes yeah yes yeah so there there are really four that p.m. p inverse business there are infinitely many choices of p and infinitely many choices of p inverse but they are paired once you choose a P that sets the P inverse so so I mean should we say a B basis is right so it's isn't like there's just one one choice yeah that's a good point I shouldn't use the definite article I wanted to say six my one that's whatever I didn't finish to just leave this let this go because we're you're gonna have to just think about it and it but it's not hurt I don't think so and I do think it was valuable just to for me at least I hope it was for you Felix if if I makes no sense to multiply this vector by a by a peek should we let's just write that down yeah yeah so changing from B to e that should be what our change of coordinate matrix is right so if we take 24 negative 8 times p-1 3 negative 1/2 1 1 there's our P matrix okay we get 8 plus 4 is 12 we get 24 minus 8 is 16 this is B coordinates this is e coordinates this is a B to e coordinate change again that so the magic isn't so much why you call PP inverse it's like this is the B to e change its inverse is the e to be change and this arrow notation is just meant to help you see that the B vector goes on the right-hand side and the e vector is you know what you end up with after you do the multiplication well no I mean if we gave this a vector named V that's V that's the same vector they are equal now we're used to like if we're looking for the point in the XY plane we're used to looking at it at 12 16 in the XY plane so this is that same point in the XY plane go back to that was at the end of the previous chapter where I had that graph grid where we looked at the same vector under different bases this is but these are the weights of the B basis vectors that would give you that point so it's those are equal absolutely equal all right sorry to keep you and so we're a day off the schedule will

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Length: 49min 31sec (2971 seconds)

Published: Sun Nov 17 2019