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Eigen value eigen vectors in matlab

6 Ansichten (letzte 30 Tage)
Aykut Albayrak
Aykut Albayrak am 21 Jan. 2018
Beantwortet: mercy charles am 19 Feb. 2022
I want to solve this question.I'm going to atteched my code but I don't know this code is true.
if true
clc;
clear all;
A=[4 -5;-2 1];
Eig=eig(A)
F=poly(A)
[V,D]=eig(A)
V1=V(:,1)
V11=V1/V(1,1)
end
  1 Kommentar
Walter Roberson
Walter Roberson am 21 Jan. 2018
What are you doing with the Eig and F values you calculate?
I do not see any initial value problem in the question presented: it gives the initial value. Initial value problems require that information about the original condition is missing and needs to be calculated, which is not the case here.

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Akzeptierte Antwort

Birdman
Birdman am 21 Jan. 2018
Your approach is nice, but not enough of course. First, you need to construct a differential equation for your second question. Third question comes with the second one. The answer for your first question is included in this line:
[V,D]=eig(A)
Full solution may look like as follows:
%%1st
A=[4 -5;-2 1];
[V,D]=eig(A)
%%2nd
syms x1(t) x2(t)
x0=[2.9;2.6];
eq1=diff(x1,t)==A(1,:)*[x1(t);x2(t)];
eq2=diff(x2,t)==A(2,:)*[x1(t);x2(t)];
solx=dsolve([eq1 eq2],[x1(0)==2.9 x2(0)==2.6]);
x1=solx.x1
x2=solx.x2
%%3rd
t=subs(t,0:0.1:1);
x1=subs(x1,t);
x2=subs(x2,t);
plot(t,x1,'o',t,x2,'-*')
  2 Kommentare
Aykut Albayrak
Aykut Albayrak am 21 Jan. 2018
Thanks for helping.I really appreciate for helping
Birdman
Birdman am 21 Jan. 2018
You are welcome. You can officially thank me by accepting my answer.

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Weitere Antworten (2)

johnson wul
johnson wul am 23 Jul. 2019
Nice that. is this formular for eigenvector:
plot(t,x1,'o',t,x2,..,t,x10, '-*')
have been used to for square matrix 10*10 for example if some one want to plot differents eigenvectors?
And how to plot in 3 dimensions the previous eigenvectors? just a step

mercy charles
mercy charles am 19 Feb. 2022
%%1st
A=[4 -5;-2 1];
[V,D]=eig(A)
%%2nd
syms x1(t) x2(t)
x0=[2.9;2.6];
eq1=diff(x1,t)==A(1,:)*[x1(t);x2(t)];
eq2=diff(x2,t)==A(2,:)*[x1(t);x2(t)];
solx=dsolve([eq1 eq2],[x1(0)==2.9 x2(0)==2.6]);
x1=solx.x1
x2=solx.x2
%%3rd
t=subs(t,0:0.1:1);
x1=subs(x1,t);
x2=subs(x2,t);
plot(t,x1,'o',t,x2,'-*')

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