Determine the general solution to the system 𝐱^'=A 𝐱 for the given matrix A. A=[ 4 0

Determine the general solution to the system 𝐱^'=A 𝐱 for the...

Key Concepts

Generalized Eigenvectors and Jordan Canonical Form

If a matrix has repeated eigenvalues and is not diagonalizable—often described as a defective matrix—generalized eigenvectors are used to construct a Jordan chain. The Jordan canonical form organizes the matrix into blocks, each corresponding to an eigenvalue, that include both eigenvectors and generalized eigenvectors, thereby enabling the explicit construction of the general solution.

Systems of Linear Differential Equations

This concept involves studying equations of the form x' = Ax, where x is a vector and A is a constant matrix. The analysis of such systems allows us to understand the dynamics of multiple interacting components over time and is foundational in fields ranging from engineering to physics and economics.

Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors play a crucial role in solving systems of differential equations as they provide a means to decompose the behavior of the system into simpler, independent modes. When the coefficient matrix A is diagonalizable, the solution can be expressed as a linear combination of exponential functions, each scaled by the corresponding eigenvalue and modulated by its eigenvector.

Matrix Exponential

The matrix exponential is a fundamental tool for solving linear differential systems. It provides a way to express the solution of x' = Ax in the form x(t) = e^(At)x(0), capturing both the exponential growth or decay dictated by the eigenvalues and the contributions from the structure of A, including any Jordan block formations in the case of non-diagonalizable matrices.

Transcript

00:01 Hey everyone.

00:01 So today we're looking at a system of differential equations given by this right here.

00:08 And we have a matrix so you can see that these are mixed differential equations.

00:11 And we're asked to solve this using jordan canonical forms and the way we're going to do that.

00:16 And the way you always do that is by starting with a change in variable where x equals s, which is some invertible matrix times a column vector y.

00:24 We'll see that this yields a transformed equation.

00:31 Right.

00:31 And if we choose our s is correctly, which we always, can, then this will equal the jordan canonical form of a, because a is similar to string canonical form.

00:42 So we have the new equation.

00:45 Y equals j, y, oh, sorry, y prime.

00:53 And now if we solve this equation for y and we plug this in and solve for x, we can solve this differential equation for x.

01:02 And yeah, so you'll see that we need two things to start.

01:08 We need s and we we need j.

01:10 Those are the two things we need to find.

01:12 Ok, we'll call these equations 1 and equation 2.

01:19 Now, so let's go about finding j first, which we know how to do.

01:22 So it's diagonal is going to be the column vectors of a.

01:27 So sorry, not the column vectors.

01:29 What am i talking about? the eigenvalues.

01:31 So we have our eigenvalue equation, the characteristic polynomial minus lambda i, that's going to be 0.

01:39 And well, what is a minus lambda i? that's 4 minus lambda, 4 minus lambda, 4 minus lambda on the diagonal, and then everything else is the same.

01:57 Right, perfect.

02:00 So our 3 by 3 determinant is going to be, if you recall, it's going to be the sum of this times the determinant of this matrix, plus the sum of this times the determinant of this matrix, but that's just 0 because 0 times 0 is 0.

02:16 And the sum of this times the determinant of this matrix, which is also zero.

02:21 So it's just the matrix in red times 4 minus lambda.

02:25 So we get that that is going to be, oh, stylus isn't really working here, 4 minus lambda cubed.

02:36 Set that equal to 0.

02:38 And so our eigenvalues, as we can see just very simply here, this is a really nice problem.

02:43 Our lambda, we're just going to have 1, it's going to equal 4.

02:47 And the multiplicity is going to be 3.

02:51 It's cubic.

02:52 Ok, so we have the diagonals for the joint canonical form.

02:57 And now we need to get the how many join blocks there are, which is going to be equal to the number of linearly independent eigenvectors we have.

03:05 So we're going to need to find those.

03:07 And it's going to be eigenvector equation a minus lambda i times v.

03:12 Well, what's lambda it's four? so we can say a minus four i times v equals zero, where v is our eigenvector.

03:20 And so we could solve this.

03:22 For a, solve this as an augmented matrix for the components of v.

03:35 It's just a system of equations.

03:37 And you'll see this is already in a reduced form if we switch the rows, but that doesn't really matter.

03:41 So we get that v equals, well, the first and the second element have to be zero, zero, and the third can be whatever it wants.

03:50 So we'll say one and multiply by some alpha.

03:53 So that's our eigenvector.

03:55 We can see there is one linearly independent eigenvector.

03:58 So there's one jordan block.

04:00 So that's pretty easy.

04:01 We can just write the jordan canonical form, the eigenvalues down the center, and combining them all went to one jordan block.

04:15 Okay, so that's good.

04:18 So that's the key to equation two.

04:21 Now equation one, we need to find s.

04:23 Well, we also know how to do that, because we know to create an s such that this happens if we have the a and the j.

04:32 So for a non -diagy matrix, right? for diagonal matrix, s, the columns of s would just be the eigenvectors.

04:42 But since j is non -diagonal, we're going to need to use generalized eigenvectors.

04:46 And it's of size 3.

04:47 So we have a theorem that says the the s is going to be, its columns are going to be the cycle elements of the generalized eigenvector, and the size is going to equal the size of the jordan block you're doing it for.

05:03 So we need a size cycle of three.

05:06 So well, it's a cycle, a minus 4i, squared times v prime, which is a generalized eigenvector.

05:21 Perfect.

05:22 So that's going to be our cycle.

05:23 So this is going to be the first column, the second column, and the third column of s, and it's important that we keep it in that order.

05:30 And well, so to find a v prime, which is all we really need, we know that all of these have to be non -zero.

05:39 So this just says it can't be a trivial vector.

05:43 That being non -zero.

05:44 This says it can't be an eigenvector because that's the eigenvector equation.

05:47 We set it to zero, and this is its own constraint that can equal zero.

05:52 And the one after it has to equal zero in order to limit, so it's not a size four cycle, right? so let's just start calculating powers of a minus 4i.

06:03 Well, that's 0 -0 -1 -0 -0 .0.

06:11 That's a minus 4i, and we're going to multiply it by this.

06:18 To get a minus 4i squared and that's going to be equal to what top row is going to be 0 that's going to be 0 and finally we get a 1 right here and the rest is 0 okay so that's a minus 4 i squared now we need to calculate a minus 4 i cubed so that's a minus 4i squared times a minus 4i all right and might be able to see where this is going already so if you can see it that's just going to be well, that's going to be a 3x3 -0 matrix.

07:23 None of these line up.

07:24 They're all in different rows and columns.

07:27 Okay.

07:29 So if every vector satisfies that a minus 3 iq times it is going to be zero, our new conditions are that these are non -zero.

07:40 That's the only conditions we have.

07:43 So just deleting all this, actually deleting this.

07:49 So we don't need that.

07:59 All right so a minus 4 i squared which we have right here we are going to find a vector such that it does not equal zero so we can basically just look at it um why not set up a generalized uh augmented sorry matrix uh times v prime and we want that not to equal zero so let's find the things that do equal zero and make sure it's not part of that.

08:35 So the things that do equal 0, well we can see immediately are any vectors of the form, call it w for bad vectors, anything of the form, we can have anything in the first slot and anything in the second slot and it has to have a zero in the third slot.

08:56 So we can see that if we want a vector that's not like w, we just need a vector with anything in the third slot.

09:02 Okay, and this being non -zero gives us a different approach, right? so we can't have just something in the third slot because then it's an eigenvector.

09:17 So let's write out our random generalized eigenvector.

09:22 Well, that works.

09:22 It doesn't have, it has something in the third slot, but not only in the third slot.

09:27 And so that's actually every condition that we have.

09:30 And v -prime isn't unique.

09:32 There are other generalized eigenvectors.

09:34 But that's all we need.

09:37 So now we can go about calculating these two things.

09:41 Just directly brute force.

10:10 Perfect.

10:32 Right, there we go.

10:33 Okay, so these are going to be our columns, and we have to work backwards, right? so this is our third column.