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0:00
let's do a tour de force with this linear system and learn as much as we can about it in every way we've been thinking about these things all at once so you really want to concentrate we're going to put all of our ideas together including the ideas related to change of coordinates which I haven't really emphasized in full depth in the lectures yet
0:21
so it's a system of linear equations both in scalar form and in Matrix form
0:27
to find the general solution you want to find the eigenvalues and the eigenvectors
0:32
you create the Matrix a minus Lambda I and take its determinant
0:38
and find the roots of that polynomial why you might ask well that's discussed in the lectures the motivation for this is related to the straight line Solutions that's where this these ideas come from you're looking for points in the plane where the output of the vector field is parallel to the position doctor for the point in other words like if you imagine a point right there say
1:11
is that points position vector and the output of the vector field is parallel sometimes it is okay and in the special case when it is that's going to correspond to straight line Solutions so that arrow is supposed to represent the position Vector of a point and if you attach a vector in the vector field to it that happens to be parallel pointing in the same direction not necessarily the same length
1:38
and not necessarily pointing in the same direction it could point in the opposite direction but it's parallel that that would be a situation where the vector field represented as a times V say
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is a scalar multiple of V Lambda times V a times V is the output of the vector field Lambda times V is a scalar times v a scalar multiple of B this equation is saying the output of the vector field is parallel to the position Vector that's it an algebraic way of describing a geometric reality
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and then essentially what you do is you rearrange this equation a v equals Lambda V which is the definition of what it means for V to be an eigenvector and Lambda to be a corresponding eigenvalue and rearrange it as a v minus Lambda V equals the zero vector
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and then factor out the V like this
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and you're looking for non-zero solutions for that you want to find vectors V that are non-zero that solve this equation well the only way that's going to happen is if this Matrix is what's called singular which means non-invertible which means it's determined it is zero
2:56
that's why you set the determinant of this Matrix equal to zero ultimately set this equal to zero and solve for Lambda to find the possible values of Lambda that give you non-zero Solutions of this equation non-trivial null spaces of this Matrix non-trivial kernels of the corresponding linear transformation it's all connected
3:19
when you solve such a system by plugging in those values of Lambda in this case negative three and negative two you will get a system that has non-zero Solutions you will get free variables I'm not even bothering to go to rref here I can see these two rows our scalar multiples of each other they both represent the equation essentially x minus y equals zero in other words x equals y if you want to think of Y as a free variable then x equals y so any
3:50
scalar multiple of this Vector 1 1 would be a solution it's always tricky to remember you got to bring these things over to the other side of the equation so even though you see a minus sign there you don't see any minus signs there you gotta imagine bringing it over to the other side of the equation so you should have gotten an eigenvector Lambda negative 3 corresponding to an eigen a nugget value of negative 3 corresponding to an eigenvector of 1 1. for the other eigenvalue
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negative two the corresponding eigenvector is 2 1. replace Lambda with negative two
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both here and here you get a matrix a coefficient Matrix where the two rows or multiples of multiples of each other in fact they're the exact same Row in this case and they correspond to the equation x minus two y equals zero so X is 2y if Y is a free variable and you say let it equal one then X is forced to be 2. but these eigenvectors are not unique eigenvalues are unique okay these are the only two eigenvalues
4:54
but eigenvectors are not unique you could use any non-zero scalar multiples of these things and get another eigenvector okay but I just always try to pick the simplest ones I can think of don't pick zero zero zero the zero Vector is not an eigenvector
5:15
um these do then help you find the general solution and also help you draw the phase plane let's talk about that next so the general solution
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you take a linear combination of e to the Lambda 1 times T times its eigenvector
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and e to the Lambda 2 times T times its eigenvector K1 and K2 are arbitrary constants I started making a c there that's why it looks a little funny
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this is related this is related still to Vector space Theory
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ideas of basis and dimension
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these two functions
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are both individually Solutions of the system and they're linearly independent they're not scalar multiple multiples of each other being a two-dimensional system it's possible to show that the solution space is a two-dimensional Vector space and therefore these two functions functions emphasis on the word function form a basis for the solution space and so if I take a linear combination of them that's guaranteed to be the general solution
6:40
all that Vector space theory is behind the scenes here
6:45
in writing this General solution the dimension of the solution space is two-dimensional so any two linearly independent Solutions will be a basis and therefore an arbitrary linear combination of them will be a general solution but again the v's themselves are not unique
7:09
let's draw the Base plane started drawing it here with null clients to help
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the XML client has set the x DT equal to zero that's a length of the origin with a slope of negative one-half the Y no client I set the dydt equation equal to zero that's a line through the origin with with a slope of positive 1 4. these no clients can be used with linear systems the textbook doesn't do it because it waits till chapter five I think it's good to introduce early
7:46
in chapter five we still apply null clients to non-linear systems it's even more useful there
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are the no clients this one's the X and alkaline so you cross it vertically
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this one's the Y no client C cross Solutions cross horizontally
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we can also I haven't done so yet draw on the straight line Solutions
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based on this being your general solution the straight line Solutions are parallel to the eigenvectors one one and two one and they both have negative eigenvalues so as T goes to infinity solutions head towards zero the solution curves head toward the origin it's going to be a sink in the phase plane parallel to 1 1 and parallel to 2 1.
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one straight line solution parallel to one one is going to be along that line that's the line Y equals X it's not a nullcline it's a straight line solution draw arrows going toward the origin
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another one is parallel to the vector two one which has a slope of one-half well like this
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once again the eigenvalues are negative so the solutions head toward the origin
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now you take this information and do your best to draw the phase plane the face portrait all the other Solutions you can also think about the signs where these quantities positive and where are they negative I mean if we start down here
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over here where X is positive and Y is zero say we're on the x axis to start with say
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X is positive and Y is zero DX DT is negative and d y t is positive
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Solutions have to go to the Northwest DX DT is negative d y d t is positive
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so they gotta head this way they got to cross this Y no Cline with a horizontal tangent and then they're in this region up here where I bet it's going to go looking at the picture here to the Southwest and head toward the origin I want to double check do Solutions in this region go to the Southwest are both DX DT and dydt negative pick a point up here where X is zero and Y is positive
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both of these are going to be negative when x is 0 and Y is positive all Solutions in this region up here between these null clients got to go to the South West I mean I can see the straight line Solutions as well go to the Southwest
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if I started with an initial condition down here
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for consistency sake I hope it's pretty intuitive it's got to go up this way to the Northeast cross this axon alkaline with a vertical tangent then go to the northwest
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is it going to cross that Y in Oakland before it heads toward the origin yes it will it will not head toward the origin without Crossing that wine outline I mean it's going to be very close to the origin before it crosses I know it can't I know it's got to cross that other nullcline
11:10
because if it didn't if it just stayed in this region here there would have to be another straight line solution in there and there's not because these Solutions end up being tangent to straight line Solutions question I have a question um or so are you talking about with initial conditions in hearsay yeah yeah okay let's let's let's be careful that's a good point let's let me draw some other ones first and then we'll come back to your
11:47
point okay um probably the easiest ones to draw are similar sort of the opposite across the origin from these uh like you could have one here going this way and tangent it seems to the straight line solution if you start up here it's going to go down this way and yes I know it's got to cross this green line here before it goes toward the origin because there are no other straight line Solutions in there and yeah what happens if you have an initial condition here or here
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I claim all solutions except along the line Y equals X have to be tangent to this straight line solution so I claim if you start up here you got to go this way and be tangent to this one
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you start down here you got to go this way and be tangent to the same line actually on the other side of the origin the only solutions that are not tangent to this red line with a slope of one half
12:47
the solutions on the straight line solution with a slope of one
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those are the only solutions that are not tangent to that what's the intuition behind that
12:59
this is a very subtle point but important this straight line solution Y equals X by the way that's not the equation of the straight line solution itself that's just the line it lies on
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the equation of the straight line solution itself
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is essentially just that expression as a parametric curve
13:26
that corresponds to the eigenvalue negative three this was Lambda 1 equals negative three the other straight line solution corresponds to the eigenvalue negative two Lambda 2 equals negative two here's the quick way to know which line of eigenvalue vectors most solution curves are tangent to as T goes to Infinity it's which whichever one corresponds the eigenvalue closer to zero
13:55
negative 2 is closer to zero than negative three is
14:00
and so I'm just telling you based on that that's why this line is the line that most solutions are tangent to as T goes to Infinity but why what's the intuition the intuition there is because negative 3 being further away from zero corresponds to an exponential decay that's faster
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you look at these two functions here which decays to zero faster as T goes to Infinity that one does
14:30
with the eigenvalue further away from zero
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so the Decay to zero is faster in this term than it is in this one so intuitively sort of in this in this direction corresponding to this line the component of solutions in that
14:50
new axis skewed access Direction has to go to zero faster than in this direction
14:58
and if you think about it that means the solution's got to be tangent to this one because when you're tangent to this one
15:05
their components relative to this skewed axis are a lot closer to zero
15:11
when you when you're there say right there on this solution curve right about there say
15:18
if you think about the skewed coordinate axes you're a lot closer to this red line than you are to this red line meaning the coordinate of this point in the skewed coordinate system with respect to this line is a lot closer to zero than the coordinate with respect to this red line
15:38
that's a very vague description of what's really a very algebraic idea
15:46
not expecting that you're going to necessarily remember what I just said there so you have to think about it after class the book actually gives a more rigorous explanation of it using limits I don't want to go there
16:00
what I think would be most worthwhile for our last 12 minutes here
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is to relate both try to relate this to flows and coordinate changes as quickly as we can in 12 minutes here let's relate it to coordinate changes first actually
16:19
so in the lectures I don't really go here quite yet I'm previewing
16:25
what starts really in lecture 24b which is right before the exam
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but comes into a full fruition after exam three but it's still related to things that you do need to know for exam three it's related to change accordance matrices and bases
16:44
it's related to all that and what I mentioned last Friday if you create a matrix p
16:54
whose columns are the eigenvectors
17:00
what were the eigenvectors one one and two one
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you have created a matrix that's first of all invertible
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because those columns are basis vectors for R2 they are linearly independent this is invertible I made this determinant the opposite of the example from last Friday the determinant of this one's negative one instead of positive one but that's still okay
17:35
here's the inverse Matrix
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I swap the numbers on the main diagonal well they're both ones so it still stays both ones and I negated the numbers in the off diagonal and I multiplied by 1 over the determinant and this is just a negative one
17:55
multiply the negative one through you get negative 1 positive two positive one negative one this is p inverse if I didn't make a mistake
18:04
yeah you can check it
18:09
I claim if I calculate P inverse times AP
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I get a diagonal matrix let's see if that happens let's do a quick calculation
18:22
a was this Matrix
18:27
P inverse AP
18:32
doing this calculation as quick as I can here uh taking dot products again the negative 1 minus two is negative three negative two minus two is negative four hopefully I don't make a mistake one minus four is negative three two minus four is negative two
18:53
what are they doing
18:57
here I get positive three times negative six
19:03
is negative three I better get a zero up here let's see positive four times negative four is zero I better get a zero down here negative three times plus positive three is zero I better get negative two here negative four plus two is negative two how did I know that would happen because those are the eigenvalues I knew that had to be the case if I didn't make a mistake that's a diagonal matrix call it d
19:34
what did we do last Friday last Friday we we reversed this process I started with a diagonal matrix you did a completely decoupled system
19:46
and I told you what P was and you found an a that had the same eigenvalues that's exactly what we did here in reverse
19:58
this Matrix p is a change of coordinates Matrix that converts our original linear system to a completely decoupled system
20:09
in the new coordinate system which corresponds to the
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foreign s the straight line Solutions being a skewed coordinate system
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the system is completely decoupled and very easy to solve
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in particular The Matrix exponential is very easy to solve for
20:35
e to the t d
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is also a diagonal matrix looking like this
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and it then turns out that e to the TA
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since P inverse AP is D it turns out that D is p a p inverse
21:02
oops no sorry a is p d p inverse you get confused there and it turns out without explaining this this means the Matrix exponential e to the TA turns out to equal p times e to the T D times P inverse this is a super useful thing allowing you to compute The Matrix exponential of a using the diagonal matrix D whose Matrix exponential we can find very easily
21:33
so far with Matrix exponentials we're only using technology here's the key to doing it by hand
21:46
how is this related to flows
21:55
um
21:58
remember the beautiful generalization
22:04
the Y DT equals a times y
22:10
of zero equals y Sub Zero that is a generic initial value problem
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it's got unique solution not General solution unique solution
22:25
foreign y equals I'll go ahead and use Phi Capital Phi sub y zero of T equals I wrote e to the a t last time I'll write e to the T A this time that's actually more how often I more how I do it in the lectures
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what this means is if you can find this Matrix exponential you can solve the system
22:54
it can also relate it to flows
22:59
bad thing you find with Mathematica but we also could do it now this way too and get the same answer
23:10
the flow I'll go to using a curvy fee here
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has the same formula the exact same formula we're just thinking of it differently
23:25
the time one map you just replace t with one and you get this and this then you think of as generating a difference equation not a differential equation a difference equation
23:40
y n equals ultimately e to the a times y sub n minus 1.
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and these two things are equal whoops yeah okay that's that's the difference equation itself whose solution
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is y n equals e to the N times a times y zero
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which matches this when T is n
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so there's a relationship between the differential equation and the difference equation
24:23
got a finished class here before we look at what you're doing for number two let me also mention something that can help you for number one
24:33
on the linear algebra portion
24:37
this I forgot to mention it earlier this kind of relationship between a and d mean A and D are rather called similar matrices when this kind of relationship holds for some invertible Matrix p could be anything any invertible Matrix and you would say A and D are similar
24:55
that's the definition of similar matrices that's relevant for number one on the linear algebra
25:02
number one is what kind of matrices are similar to Lambda times the identity Matrix and the answer is and I even tell you is it's only similar to itself and the hint I'll give you there is that any Matrix commutes with the identity Matrix in general matrix multiplication is not commutative but with the identity Matrix it is then that's a hint and also the definition of similarity to number one number one is actually a very easy problem if you know what you're doing
25:31
to Finish class
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how do you do number number two
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I know I got a lot of complicated ideas in a short amount of time here well where to go
25:56
[Music] so okay we only have time to just outline it here so you're going to find the Matrix exponential
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that's going to be with Matrix X Mathematica and you can make them make it the output look like a matrix with Matrix form [Music] use your answer to find
26:20
a formula for the time one map well that's just exactly what I was showing you there
26:26
it's e to the a times y zero and you just plug in what e to the a is
26:33
it's a linear transformation y 0 is the variable
26:38
the initial condition is the variable I know it's strange but that's what it is
26:43
then I say find the eigenvalues of eigenvectors and determine it of e to the a not a but e to the A it's a Matrix Matrix exponential is a matrix it's got eigenvalues eigenvectors in the determinant you can compute those with Mathematica
27:00
so there's no problem Computing it what's more important is there's a relationship between the eigenvalues eigenvectors and determinant of e to the A and the eigenvalues eigenvectors and determine an a itself there's a relationship that comes out in the lectures
27:16
also in the lecture I talk about
27:20
the image of a set under the time one map and what happens to its area so s is just some subset of the plane maybe you imagine it to be a circle and all the points inside a circle it's got a certain area I'm saying it's ten thousand
27:40
B1 of s represents all possible outputs of the time one map as you plug in different points and F's what is the image of that Circle P1 is a linear transformation
27:54
this is stuff we did talk about when we talked about images of linear transformation
28:01
what's its area it comes on the lecture I tell you what to do okay now again you could just look at the PowerPoint maybe you could figure it out from the PowerPoint but I go into detail in the lecture
28:15
not enough time to really really go in depth as much as I wanted to but it keep thinking about it okay have a good day
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