Eigen value eigen vectors in matlab
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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 件のコメント
Walter Roberson
2018 年 1 月 21 日
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.
採用された回答
Birdman
2018 年 1 月 21 日
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 件のコメント
その他の回答 (2 件)
johnson wul
2019 年 7 月 23 日
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
0 件のコメント
mercy charles
2022 年 2 月 19 日
%%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,'-*')
0 件のコメント
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