00:01
Everyone, so the problem we're looking at today gives us a system of differential equations and it gives us it in the form of two column vectors, one, the derivative of the other with respect to time and a matrix of constants a.
00:15
And the way we're going to solve this with jordan canonical forms is all at the root is the first equation.
00:20
You'll see we'll get a second that directly follows, but one is that x equals some invertible matrix s times y, and the reason we're doing that is that this gives us the new differential equations of the form this in terms of this new variable y.
00:40
And if you recognize this, it's because if we choose the correct s, which we always can give in an a, s inverse a, s will become the jordan canonical form of a.
00:50
So that gives us our second equation two.
00:53
So our big guiding equations.
00:55
This is what you do every time.
00:56
Everything else you can sort of think through based on what you already know.
01:02
It might be hard, it might be messy, but this is the core.
01:05
And we get that y -prime equals j -y.
01:09
We're going to see that for general a, putting it almost diagonally is going to make it much, much easier to solve differential equations that it describes.
01:19
So that's the impetus for this.
01:24
Closed out on my tab for a second.
01:26
Okay, so going forward, what do we need to find? well, we need to find j and solve for y.
01:31
And then once we have y in terms of these differential equations we're going to solve, we need to have s in order to transfer it back to x coordinates, which is going to be our final goal.
01:42
So, well, let's start with j.
01:44
So to determine j, we need to know the eigenvalues, right, because those are going to be on the diagonal.
01:50
So we have our eigenvalue formula, the characteristic polynomial of a, set that to zero.
01:57
So debt of this matrix, 3 minus lambda, 2, minus lambda, negative 5 minus lambda on the diagonals.
02:13
And we have negative 4 -4 -0 -0 -0.
02:19
Well, what's the determinant of a 3x3 -matrix? it's a pretty easy formula.
02:25
It gets pretty messy after 3.
02:26
But for 3 by 3, it's this times the determinant of this little matrix embedded in there.
02:33
And then plus this times the determinant of this next permuted matrix, which is going to be 0.
02:38
So let's not even bother because zero times zero is zero and plus this times the determinant of this matrix which is going to be non -zero so those are two components let's write them out in their colors so we have three minus lambda times two minus lambda times negative five minus lambda we're going to add that to our blue which is four minus minus four times two minus lambda so four times four times four times four 4 times 2 minus lambda.
03:13
All right.
03:14
We're going to sum those together.
03:16
What do we get? well, i'm going to skip through all the algebra for you.
03:20
I trust you can do that and rewrite this.
03:24
Yeah, this is not the conceptually difficult part.
03:29
I will give you the answer so you can just check.
03:32
But yeah, without further ado.
03:37
Ok, so we get this equation all factored out and whatnot.
03:41
And so that gives us two nice eigenvalues.
03:43
We'll call lambda 1 equals negative 1 and that has a multiplicity of 2.
03:47
It because it's squared and lambda 2 equals 2 multiplicity of okay so these are eigenvalues.
03:56
It will be the diagonal of jordan matrix but that doesn't tell us necessarily what the jordan matrix is because we need to know the amount of jordan blocks we're working with and we have a theorem that says that that is going to be equal to the number of linearly independent eigenvectors or in other words, the dimension of the eigenspace for each eigenvalue, which we know is going to be one for anything with multiplicity of one, because they have at least one and max one, right? but for lambda 1 could be up to 2.
04:22
So let's go ahead and calculate the eigenvalues, or sorry, eigenvectors of lambda 1.
04:28
So we use the eigenvector equation, a minus lambda i times our eigenvector.
04:36
Let's call it v1, because we are going to calculate v2 later, cool zero.
04:41
So we can set up the augmented matrix.
04:45
I always forget the word for that.
04:47
To solve for the components of v1.
04:49
So it's going to be 3 plus 1, 2 plus 1, negative 5 plus 1.
04:55
4, 3, negative 4, 0 ,000, 0, 0.
04:59
There's a 4 here.
05:01
I forget the negatives, though.
05:05
Ok, times sum vector equals 0.
05:14
Excellent.
05:16
So we're reducing this.
05:17
Well, just add the first row to the last row and divide everything else.
05:23
We are actually done.
05:27
1 -1 -1.
05:28
Okay, so we get that there is one eigenvector, and we don't actually need to solve it.
05:34
We just see that there is a free variable.
05:38
There's a row of all zeros, so an unconstrained variable.
05:41
And there are two other actual defined constraints.
05:45
So we're going to get something along the lines of v1 equals alpha times some a, b, b c, right? it's just going to be, there's only one free variable, so alpha can change, but a, b, and c need to stay the same.
06:00
It's fixed.
06:01
So one linearly independent eigenvector.
06:04
So that means since there's one here and there's one here, that's going to be a total number of linearly independent eigenvectors 2.
06:12
So we're going to have two jordan blocks, and we can just write out the jordan canonical form right here and now.
06:18
That's minus 1, minus 1, 2.
06:21
Right? that's what we said.
06:23
Yeah.
06:25
Perfect.
06:26
Zero is all below.
06:27
It's every jordan form.
06:29
And well, we need there to be two blocks right now.
06:32
There's three.
06:32
So we need to combine these with a one because you can't put a one there, right? and everything else is going to be zero.
06:38
That's going to be our jordan canonical form.
06:41
That's unique for a.
06:43
But what we need to find now is we need to find our s.
06:46
And so we know how to solve for an s here.
06:49
It's just going to be the same way.
06:50
And then we're going to actually apply it.
06:53
Sometimes we haven't in the past.
06:54
So first what we need.
06:58
So it's going to take the form of, for jordan blocks, the s is going to be the columns that correspond to the jordan blocks are going to be equal to cycles of generalized eigenvectors of the same length.
07:13
So you see this as the size of the block, this jordan block is size 2.
07:17
So we need a cycle of size 2.
07:19
So a minus lambda i, so a plus i times, we'll say v prime 1 because it's the generalized eigenvector of 1.
07:29
Actually, we don't, because we're only doing one generalized eigenvector, comma, v prime, where these are both non -zero and the next term over.
07:38
In order to end the cycle is 0, so we have, let me write that down.
07:45
We have the a plus i squared times v prime is 0, that these are non -zero.
07:51
So those are going to be our constraints here.
07:53
We can solve equations to find that.
07:55
And that's going to be our first two columns, one, two, and it has to be in that order.
07:59
And then as we look left to our matrix, well, we have one jordan block, which we've written the columns for an s, and we have another jordan block that we need to write the columns for.
08:11
And that's just going to be the same as the columns for a, if they were diagonal, right, you guys would know how to do this.
08:20
We've done this before in previous chapters.
08:21
So, and if you want to think about the jordan block of two as being a cycle of length 1, well, that's just v prime, and then this equals 0, and you'll see that just means that v prime has to be an eigenvector.
08:38
All right.
08:39
So comma v2, which is the eigenvector, the eigenvalue of 2.
08:45
Okay.
08:46
So we can calculate this pretty easy.
08:47
We'll do it in a sec, but let's focus on the cycle for now.
08:50
We need this to be true.
08:51
So what's a plus i squared? that is, what was it? and multiply that by itself.
09:10
Okay.
09:11
And we get something that looks like, well, that's going to be something that looks along the lines of this.
09:28
You can check...