(a) Show that the matrix 𝐀=[ 2 1 6 0 2 5 0 0 2 ] has the repeated eigen

(a) Show that the matrix 𝐀=[ 2 1 6 0 2 5 ...

(a) Show that the matrix $\mathbf{A}=\left[ \begin{array}{lll}{2} & {1} & {6} \\ {0} & {2} & {5} \\ {0} & {0} & {2}\end{array}\right]$ has the repeated eigenvalue $r=2$ with multiplicity 3 and that all the eigenvectors of $\mathbf{A}$ are of the form $\mathbf{u}=s \operatorname{col}(1,0,0)$ (b) Use the result of part (a) to obtain a solution to the system $\mathbf{x}^{\prime}=\mathbf{A} \mathbf{x}$ of the form $\mathbf{x}_{1}(t)=e^{2 t} \mathbf{u}_{1} .$ (c) To obtain a second linearly independent solution to $\mathbf{x}^{\prime}=\mathbf{A} \mathbf{x},$ try $\mathbf{x}_{2}(t)=t e^{2 t} \mathbf{u}_{1}+e^{2 t} \mathbf{u}_{2} .[$Hint$:$ Show that $\mathbf{u}_{1}$ and $\mathbf{u}_{2}$ must satisfy $(\mathbf{A}-2 \mathbf{I}) \mathbf{u}_{1}=\mathbf{0}, \quad(\mathbf{A}-2 \mathbf{I}) \mathbf{u}_{2}=\mathbf{u}_{1} \cdot ]$ (d) To obtain a third linearly independent solution to $\mathbf{x}^{\prime}=\mathbf{A} \mathbf{x},$ try $\quad \mathbf{x}_{3}(t)=\frac{t^{2}}{2} e^{2 t} \mathbf{u}_{1}+t e^{2 t} \mathbf{u}_{2}+e^{2 t} \mathbf{u}_{3}$ [Hint: Show that $\mathbf{u}_{1}, \mathbf{u}_{2},$ and $\mathbf{u}_{3}$ must satisfy $(\mathbf{A}-2 \mathbf{I}) \mathbf{u}_{1}=\mathbf{0}, \quad(\mathbf{A}-2 \mathbf{I}) \mathbf{u}_{2}=\mathbf{u}_{1}$ $(\mathbf{A}-2 \mathbf{I}) \mathbf{u}_{3}=\mathbf{u}_{2} . ]$ (e) Show that $(\mathbf{A}-2 \mathbf{I})^{2} \mathbf{u}_{2}=(\mathbf{A}-2 \mathbf{I})^{3} \mathbf{u}_{3}=\mathbf{0}$

(a) Show that the matrix
$\mathbf{A}=\left[ \begin{array}{lll}{2} & {1} & {6} \\ {0} & {2} & {5} \\ {0} & {0} & {2}\end{array}\right]$
has the repeated eigenvalue $r=2$ with multiplicity 3 and that all the eigenvectors of $\mathbf{A}$ are of the form $\mathbf{u}=s \operatorname{col}(1,0,0)$
(b) Use the result of part (a) to obtain a solution to the
system $\mathbf{x}^{\prime}=\mathbf{A} \mathbf{x}$ of the form $\mathbf{x}_{1}(t)=e^{2 t} \mathbf{u}_{1} .$
(c) To obtain a second linearly independent solution to $\mathbf{x}^{\prime}=\mathbf{A} \mathbf{x},$ try $\mathbf{x}_{2}(t)=t e^{2 t} \mathbf{u}_{1}+e^{2 t} \mathbf{u}_{2} .[$Hint$:$ Show that $\mathbf{u}_{1}$ and $\mathbf{u}_{2}$ must satisfy
$(\mathbf{A}-2 \mathbf{I}) \mathbf{u}_{1}=\mathbf{0}, \quad(\mathbf{A}-2 \mathbf{I}) \mathbf{u}_{2}=\mathbf{u}_{1} \cdot ]$
(d) To obtain a third linearly independent solution to $\mathbf{x}^{\prime}=\mathbf{A} \mathbf{x},$ try
$\quad \mathbf{x}_{3}(t)=\frac{t^{2}}{2} e^{2 t} \mathbf{u}_{1}+t e^{2 t} \mathbf{u}_{2}+e^{2 t} \mathbf{u}_{3}$
[Hint: Show that $\mathbf{u}_{1}, \mathbf{u}_{2},$ and $\mathbf{u}_{3}$ must satisfy
$(\mathbf{A}-2 \mathbf{I}) \mathbf{u}_{1}=\mathbf{0}, \quad(\mathbf{A}-2 \mathbf{I}) \mathbf{u}_{2}=\mathbf{u}_{1}$
$(\mathbf{A}-2 \mathbf{I}) \mathbf{u}_{3}=\mathbf{u}_{2} . ]$
(e) Show that $(\mathbf{A}-2 \mathbf{I})^{2} \mathbf{u}_{2}=(\mathbf{A}-2 \mathbf{I})^{3} \mathbf{u}_{3}=\mathbf{0}$

Key Concepts

Eigenvalues and Eigenvectors

These are fundamental components in linear algebra where an eigenvalue is a scalar for which a square matrix A minus that scalar times the identity matrix becomes singular, meaning there is a non?zero vector (the eigenvector) that satisfies the equation (A - rI)u = 0. In the context of differential equations, eigenvalues and eigenvectors help determine the behavior of solutions, particularly in finding modes that belong to the homogeneous solution for systems like x' = Ax.

Algebraic and Geometric Multiplicities

Algebraic multiplicity refers to the number of times an eigenvalue appears as a root of the characteristic polynomial, while geometric multiplicity denotes the dimension of the corresponding eigenspace. Problems with repeated eigenvalues, where the algebraic multiplicity exceeds the geometric multiplicity, require special treatment since there are not enough linearly independent eigenvectors to diagonalize the matrix, leading to the necessity of generalized eigenvectors.

Generalized Eigenvectors and Jordan Chains

When a matrix does not have a complete set of eigenvectors due to repeated eigenvalues, generalized eigenvectors are introduced to complete a basis for the state space. These vectors satisfy equations of the form (A - rI)^k u = 0 for some k greater than 1. The chain of vectors (the Jordan chain) is used to construct the Jordan canonical form of the matrix and to build solutions to differential equations by introducing polynomial factors (like t, t², etc.) alongside the exponential term.

Jordan Canonical Form

The Jordan canonical form is a method of simplifying matrices that cannot be diagonalized due to repeated eigenvalues with deficient eigenspaces. By transforming the matrix into a block-diagonal form consisting of Jordan blocks, each corresponding to an eigenvalue, one can systematically handle the contributions of both eigenvectors and generalized eigenvectors. This approach is essential for constructing solutions to systems of differential equations where the matrix is not diagonalizable.

Matrix Exponential Solutions for Differential Equations

In linear systems of differential equations such as x' = Ax, one typically seeks solutions in the form of e^(At). When the matrix A has repeated eigenvalues and lacks a full set of linearly independent eigenvectors, solutions are constructed using the eigenvectors and generalized eigenvectors, leading to terms that involve not only e^(rt) but also polynomial factors (e.g., t, t²) multiplied by e^(rt). This approach accommodates the Jordan block structure of the coefficient matrix.

Nilpotence in the Context of Generalized Eigenvectors

The nilpotence property of the operator (A - rI) is crucial in the theory of generalized eigenvectors. For a given generalized eigenvector of level k, repeated application of (A - rI) eventually yields the zero vector; that is, (A - rI)^k u = 0 for some positive integer k. This property guarantees that the chain of generalized eigenvectors closes and plays a vital role in verifying the validity of the Jordan chain and constructing solutions for the system.

Transcript

00:01 In this video, we're going to go through the answer to question number 37 from chapter 9 .5.

00:05 So there's a few parts of this question, so we'll go through them one at a time.

00:09 We're first given the matrix a that's written on the whiteboard here.

00:14 So we've got to show that it has repeated eigenvalue 2, and to find that all the eigenvectors are of a given form.

00:26 Okay, so part a.

00:31 So to find the eigenvalues, we need to look at the determinant of the matrix a minus r -i, where i is the identity matrix.

00:47 Okay, so what's this? this is going to be 2 minus r -1 -6 -0 -2 -1 -r, 5 -0 -0 -2 -2 -r.

01:07 So then this is very easy to work out because we have a lower diagonal, sorry, the lower triangle is all zero.

01:19 So this is just going to be two minus r cubed.

01:25 Okay, so then the eigenvalues are the r values for which this is equal to zero, which tells us that r is equal to two.

01:34 And because of the cubed here, it's a triple root.

01:45 Let's find that i can vector.

01:51 We need to look at the matrix a minus r, which is 2 times the identity matrix i, and multiply that matrix by a vector u.

02:05 When that's equal to zero, that vector u is r.

02:09 I can vector.

02:10 Okay, so what is this matrix a minus 2i? it's going to be the matrix 0 -1 -6 -0 -0 -0 -5 -0 -0 -0 -0.

02:27 Times by a vector u equals 0.

02:32 Cool.

02:32 Okay, so the second row, what does that tell us, that tells us that the third component is zero.

02:40 If the third component is zero, then the first row tells us that the second component is zero.

02:44 Then the bottom row tells us nothing.

02:47 So we have that the first component is arbitrary.

02:51 So let's just make it a unit vector so that the first component is equal to one.

02:55 So therefore, all the ag vectors of this matrix a are of the form u is equal to 1 .0.

03:08 Part b.

03:18 So now from that we can tell that a solution to the, let's just write it down here, to the system a, prime equals, sorry, x prime equals a, x.

03:35 We know that a solution is going to be of the form e to the 2t, because 2 is the eigen, the eigenvec, that multiplied by the corresponding eigenvector, which is 1 ,00.

03:53 So that has to be a solution.

04:00 Part c, so we have to try and find a second linearly independent solution.

04:08 We can't just write it down because we had a multiple item value.

04:16 So we have to try a different form for a solution.

04:20 So let's try what is suggested, which is x2 is equal to t, e to the 2t times u1 plus e to the 2t u2.

04:42 Okay.

04:43 So let's first.

04:44 So let's first find out what the derivative of this is.

04:50 So differentiating by product rule is going to be, let's see, that's 1 plus 2t, times by u1 plus 2 u2, and then we can just times the whole thing by e to the 2t.

05:15 But we know that this, if it's a solution of the differential equation, then this is going to be equal to a times x2, which is, we can write it as t times a times u1.

05:39 This is just by definition of what x2 is plus a times u2 times e to the 2t, taking that out as a common factor...

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