Question

Defective coefficient matrix:
x'=Ax for A =\(\left[\begin{matrix}15&-32&12\\8&-17&6\\0&0&-1\end{matrix}\right]\)I have eigenvalue =-1triple roots and eigenvectors: \(\left(\begin{matrix}3\\0\\4\end{matrix}\right)\) and \(\left(\begin{matrix}2\\1\\0\end{matrix}\right)\)
How to find the third one by \((A-\lambda I)^2V_1=0 ~~and\(A-\lambda)V_1=V_0\)?
Please, help

Answer 1

I have to use this method to find out the solution for the problem. But I confused everything.

Answer 2

the solution is \(X_1(t) = e^{-t}V_0\\X_2(t) = e^{-t}(V_1+tV_)\\X_3(t)=e^{-t}(V_2+tV_1+\dfrac{t^2}{2!}V_0)\)

Answer 3

@primeralph

Answer 4

Hello, Quebec friend, help me, please. XD either Math or English, I need your help.

Answer 5

hello, of course ^_^

i'm trying to solve this DE (been a while i did matrix DE)

I can help with both :)

Answer 6

I actually never saw this like this. I don't think I can help with this one sorry >.<

Answer 7

there is no t^2 e^-t term in the solution;
v1,v2 are the eigenvectors.
v1={3,0,4}
v2={2,1,0}
x1=v1*e^-t
x2=v2*e^-t
to find x3, start by guessing x3=t*v1 e^-t+v3*e^-t
then (A-(-1)I)v3=v1, where I is the identity matrix
so v3 has no solution; our guess doesn't work.
now let's guess x3=t*v2 e^-t+v3*e^-t
then (A-(-1)I)v3=v2, where I is the identity matrix
so one solution for v3 is {1/8,0,0}
thus x3=v2*t*e^-t+v3*e^-t

Answer 8

Thanks for your instruction. My question is
1/we can construct v3 from either v2 or v1, right?
2/ after calculate (A +I)v3= v2, I get x -2y +3/4z = 1/8. I know from this , you got \(\left(\begin{matrix}1/8\\0\\0\end{matrix}\right)\). My question is: Can we let both y, z which are free, = 0 like this? What if we let either of them =0 ? I am familiar with let y=0, z =1 to get 1 eigenvector, and then let y =1 z =0 to get other one. By that way, I don't know which one I can choose for x3 ,
Again, thanks a lot

Answer 9

1) we can. however, if we try to construct v3 from v1, we will find that there is no v3 satisfying the equation (A-(-1)I)v3=v1.
2) you can chose any vector that satisfies the equation for v3. so for example, v3={-5/8,0,1} will also work to produce a valid solution x3=v2*t*e^-t+v3*e^-t

Answer 10

I have another question, can I have your hand?

Answer 11

sure; what's the question?

Answer 12

Problem 7, I have the correct answer. My question is when choosing V2, I can choose any of them which satisfy the condition. What if I choose other like [1,1,1](transpose) at that time, I have another pair of V1, V0. My prof said that no matter what the eigenvector I choose, the solution must be the same, but I don't see it. How can they are the same?

Answer 13

I sent my question to my prof. hihihi... I must know why he said so. Thanks for reply.