MatLab: Not Enough Eigenvectors for (Repeated) Eigenvalues
2
$begingroup$
I am trying to make a code that matches the eigenvalues to eigenvectors for one of my projects and I am new to MatLab. I get a 3x3 matrix output when it comes to eigenvalues and a 3x2 when it comes to eigenvectors. I know the issues lies within the fact that one of my eigenvalues is a repeat, but I cannot find any resources telling me how to fix this. My V should definitely be a 3x3 matrix unless I have completely done this wrong when I did it by hand.
When n=2, my answer is how I want and my code works.
% Case n=2.
clear
syms k1 c1 c2 lambda t
A=[-k1 0; k1 0];
[V,D]=eig(A) %V is the eigenvectors, D is the eigenvalues.
p0=[1 0];
eqn11 = c1*V(1,1)+c2*V(1,2) == p0(1,1);
eqn12 = c1*V(2,1)+c2*V(2,2) == p0(1,2);
solcon = solve([eqn11,eqn12], [c1, c2]);
c1Sol=solcon.c1;
c2Sol=solcon.c2;
exp(D(1,1)*t)*c1Sol*V(:,1) + exp(D(2,2)*t)*c2Sol*V(:,2)
However, when n=3, I get an error.
% Case n=3.
clear
syms k1 c1 c2 c3 lambda t
B=[-k1 0 0; k1 -k1 0; 0 k1 0]
[V,D]=eig(B) %V is the eigenvectors, D is the eigenvalues.
p0=[1 0 0];
eqn21 = c1*V(1,1)+c2*V(1,2)+c3*V(1,3) == p0(1,1);
eqn22 = c1*V(2,1)+c2*V(2,2)+c3*V(2,3) == p0(1,2);
eqn23 = c1*V(3,1)+c2*V(3,2)+c3*V(3,3) == p0(1,3);
solcon = solve([eqn21, eqn22, eqn23], [c1, c2,c3]);
c1Sol=solcon.c1;
c2Sol=solcon.c2;
c3Sol=solcon.c3;
exp(D(1,1)*t)*c1Sol*V(:,1) + exp(D(2,2)*t)*c2Sol*V(:,2)+exp(D(3,3)*t)*c3Sol*V(:,3)
V should be a 3x3, right? Please help. Perhaps my code for n=2 is wrong as well and I just lucked out.
linear-algebra matrices matlab matrix-equations
asked Dec 27 '18 at 19:59
hgasuhgasu
111
$endgroup$
|
show 5 more comments
2
$begingroup$
I am trying to make a code that matches the eigenvalues to eigenvectors for one of my projects and I am new to MatLab. I get a 3x3 matrix output when it comes to eigenvalues and a 3x2 when it comes to eigenvectors. I know the issues lies within the fact that one of my eigenvalues is a repeat, but I cannot find any resources telling me how to fix this. My V should definitely be a 3x3 matrix unless I have completely done this wrong when I did it by hand.
When n=2, my answer is how I want and my code works.
% Case n=2.
clear
syms k1 c1 c2 lambda t
A=[-k1 0; k1 0];
[V,D]=eig(A) %V is the eigenvectors, D is the eigenvalues.
p0=[1 0];
eqn11 = c1*V(1,1)+c2*V(1,2) == p0(1,1);
eqn12 = c1*V(2,1)+c2*V(2,2) == p0(1,2);
solcon = solve([eqn11,eqn12], [c1, c2]);
c1Sol=solcon.c1;
c2Sol=solcon.c2;
exp(D(1,1)*t)*c1Sol*V(:,1) + exp(D(2,2)*t)*c2Sol*V(:,2)
However, when n=3, I get an error.
% Case n=3.
clear
syms k1 c1 c2 c3 lambda t
B=[-k1 0 0; k1 -k1 0; 0 k1 0]
[V,D]=eig(B) %V is the eigenvectors, D is the eigenvalues.
p0=[1 0 0];
eqn21 = c1*V(1,1)+c2*V(1,2)+c3*V(1,3) == p0(1,1);
eqn22 = c1*V(2,1)+c2*V(2,2)+c3*V(2,3) == p0(1,2);
eqn23 = c1*V(3,1)+c2*V(3,2)+c3*V(3,3) == p0(1,3);
solcon = solve([eqn21, eqn22, eqn23], [c1, c2,c3]);
c1Sol=solcon.c1;
c2Sol=solcon.c2;
c3Sol=solcon.c3;
exp(D(1,1)*t)*c1Sol*V(:,1) + exp(D(2,2)*t)*c2Sol*V(:,2)+exp(D(3,3)*t)*c3Sol*V(:,3)
V should be a 3x3, right? Please help. Perhaps my code for n=2 is wrong as well and I just lucked out.
linear-algebra matrices matlab matrix-equations
asked Dec 27 '18 at 19:59
hgasuhgasu
111
$endgroup$
1
$begingroup$
A matrix can simply not have enough eigenvectors for a given eigenvalue. Then it isn't diagonalizable. Depending on your application, there are ways around it, but none of them will give you that second eigenvector, because it just doesn't exist.
$endgroup$
– Matt Samuel
Dec 27 '18 at 20:19
$begingroup$
It has been quite some time since I took linear algebra. Does this mean that there are only two eigenvalues that apply (as in, my by-hand method was incorrect), or does this mean that MatLab can simply not do what I want?
$endgroup$
– hgasu
Dec 27 '18 at 21:20
$begingroup$
Try your $n=2$ code with[0 k1;0 0]instead. You won’t have enough eigenvectors there, either. As @MattSamuel points out, not every matrix is diagonalizable. You’ll need to do something else to compute exponentials for those matrices. See “Jordan normal form” for a general method.
$endgroup$
– amd
Dec 27 '18 at 21:20
$begingroup$
MATLAB can do what you need, but you’re going to have to use the Jordan decomposition of the matrix, or something similar. I’m sure that there’s a built-in function for computing the exponential of a matrix, for that matter.
$endgroup$
– amd
Dec 27 '18 at 21:21
$begingroup$
@hgasu It's true that matlab can't find two eigenvectors for that eigenvalue. But that's not matlab's fault. Only one exists. Nothing can change that. It's like asking for a positive real number $x$ such that $2x=0$.
$endgroup$
– Matt Samuel
Dec 27 '18 at 21:22
|
show 5 more comments
2
2
2
$begingroup$
I am trying to make a code that matches the eigenvalues to eigenvectors for one of my projects and I am new to MatLab. I get a 3x3 matrix output when it comes to eigenvalues and a 3x2 when it comes to eigenvectors. I know the issues lies within the fact that one of my eigenvalues is a repeat, but I cannot find any resources telling me how to fix this. My V should definitely be a 3x3 matrix unless I have completely done this wrong when I did it by hand.
When n=2, my answer is how I want and my code works.
% Case n=2.
clear
syms k1 c1 c2 lambda t
A=[-k1 0; k1 0];
[V,D]=eig(A) %V is the eigenvectors, D is the eigenvalues.
p0=[1 0];
eqn11 = c1*V(1,1)+c2*V(1,2) == p0(1,1);
eqn12 = c1*V(2,1)+c2*V(2,2) == p0(1,2);
solcon = solve([eqn11,eqn12], [c1, c2]);
c1Sol=solcon.c1;
c2Sol=solcon.c2;
exp(D(1,1)*t)*c1Sol*V(:,1) + exp(D(2,2)*t)*c2Sol*V(:,2)
However, when n=3, I get an error.
% Case n=3.
clear
syms k1 c1 c2 c3 lambda t
B=[-k1 0 0; k1 -k1 0; 0 k1 0]
[V,D]=eig(B) %V is the eigenvectors, D is the eigenvalues.
p0=[1 0 0];
eqn21 = c1*V(1,1)+c2*V(1,2)+c3*V(1,3) == p0(1,1);
eqn22 = c1*V(2,1)+c2*V(2,2)+c3*V(2,3) == p0(1,2);
eqn23 = c1*V(3,1)+c2*V(3,2)+c3*V(3,3) == p0(1,3);
solcon = solve([eqn21, eqn22, eqn23], [c1, c2,c3]);
c1Sol=solcon.c1;
c2Sol=solcon.c2;
c3Sol=solcon.c3;
exp(D(1,1)*t)*c1Sol*V(:,1) + exp(D(2,2)*t)*c2Sol*V(:,2)+exp(D(3,3)*t)*c3Sol*V(:,3)
V should be a 3x3, right? Please help. Perhaps my code for n=2 is wrong as well and I just lucked out.
linear-algebra matrices matlab matrix-equations
asked Dec 27 '18 at 19:59
hgasuhgasu
111
$endgroup$
I am trying to make a code that matches the eigenvalues to eigenvectors for one of my projects and I am new to MatLab. I get a 3x3 matrix output when it comes to eigenvalues and a 3x2 when it comes to eigenvectors. I know the issues lies within the fact that one of my eigenvalues is a repeat, but I cannot find any resources telling me how to fix this. My V should definitely be a 3x3 matrix unless I have completely done this wrong when I did it by hand.
When n=2, my answer is how I want and my code works.
% Case n=2.
clear
syms k1 c1 c2 lambda t
A=[-k1 0; k1 0];
[V,D]=eig(A) %V is the eigenvectors, D is the eigenvalues.
p0=[1 0];
eqn11 = c1*V(1,1)+c2*V(1,2) == p0(1,1);
eqn12 = c1*V(2,1)+c2*V(2,2) == p0(1,2);
solcon = solve([eqn11,eqn12], [c1, c2]);
c1Sol=solcon.c1;
c2Sol=solcon.c2;
exp(D(1,1)*t)*c1Sol*V(:,1) + exp(D(2,2)*t)*c2Sol*V(:,2)
However, when n=3, I get an error.
% Case n=3.
clear
syms k1 c1 c2 c3 lambda t
B=[-k1 0 0; k1 -k1 0; 0 k1 0]
[V,D]=eig(B) %V is the eigenvectors, D is the eigenvalues.
p0=[1 0 0];
eqn21 = c1*V(1,1)+c2*V(1,2)+c3*V(1,3) == p0(1,1);
eqn22 = c1*V(2,1)+c2*V(2,2)+c3*V(2,3) == p0(1,2);
eqn23 = c1*V(3,1)+c2*V(3,2)+c3*V(3,3) == p0(1,3);
solcon = solve([eqn21, eqn22, eqn23], [c1, c2,c3]);
c1Sol=solcon.c1;
c2Sol=solcon.c2;
c3Sol=solcon.c3;
exp(D(1,1)*t)*c1Sol*V(:,1) + exp(D(2,2)*t)*c2Sol*V(:,2)+exp(D(3,3)*t)*c3Sol*V(:,3)
V should be a 3x3, right? Please help. Perhaps my code for n=2 is wrong as well and I just lucked out.
linear-algebra matrices matlab matrix-equations
linear-algebra matrices matlab matrix-equations
asked Dec 27 '18 at 19:59
hgasuhgasu
111
asked Dec 27 '18 at 19:59
hgasuhgasu
111
asked Dec 27 '18 at 19:59
hgasuhgasu
111
asked Dec 27 '18 at 19:59
hgasuhgasu
111
asked Dec 27 '18 at 19:59
hgasuhgasu
111
111
1
$begingroup$
A matrix can simply not have enough eigenvectors for a given eigenvalue. Then it isn't diagonalizable. Depending on your application, there are ways around it, but none of them will give you that second eigenvector, because it just doesn't exist.
$endgroup$
– Matt Samuel
Dec 27 '18 at 20:19
$begingroup$
It has been quite some time since I took linear algebra. Does this mean that there are only two eigenvalues that apply (as in, my by-hand method was incorrect), or does this mean that MatLab can simply not do what I want?
$endgroup$
– hgasu
Dec 27 '18 at 21:20
$begingroup$
Try your $n=2$ code with[0 k1;0 0]instead. You won’t have enough eigenvectors there, either. As @MattSamuel points out, not every matrix is diagonalizable. You’ll need to do something else to compute exponentials for those matrices. See “Jordan normal form” for a general method.
$endgroup$
– amd
Dec 27 '18 at 21:20
$begingroup$
MATLAB can do what you need, but you’re going to have to use the Jordan decomposition of the matrix, or something similar. I’m sure that there’s a built-in function for computing the exponential of a matrix, for that matter.
$endgroup$
– amd
Dec 27 '18 at 21:21
$begingroup$
@hgasu It's true that matlab can't find two eigenvectors for that eigenvalue. But that's not matlab's fault. Only one exists. Nothing can change that. It's like asking for a positive real number $x$ such that $2x=0$.
$endgroup$
– Matt Samuel
Dec 27 '18 at 21:22
|
show 5 more comments
1
$begingroup$
A matrix can simply not have enough eigenvectors for a given eigenvalue. Then it isn't diagonalizable. Depending on your application, there are ways around it, but none of them will give you that second eigenvector, because it just doesn't exist.
$endgroup$
– Matt Samuel
Dec 27 '18 at 20:19
$begingroup$
It has been quite some time since I took linear algebra. Does this mean that there are only two eigenvalues that apply (as in, my by-hand method was incorrect), or does this mean that MatLab can simply not do what I want?
$endgroup$
– hgasu
Dec 27 '18 at 21:20
$begingroup$
Try your $n=2$ code with[0 k1;0 0]instead. You won’t have enough eigenvectors there, either. As @MattSamuel points out, not every matrix is diagonalizable. You’ll need to do something else to compute exponentials for those matrices. See “Jordan normal form” for a general method.
$endgroup$
– amd
Dec 27 '18 at 21:20
$begingroup$
MATLAB can do what you need, but you’re going to have to use the Jordan decomposition of the matrix, or something similar. I’m sure that there’s a built-in function for computing the exponential of a matrix, for that matter.
$endgroup$
– amd
Dec 27 '18 at 21:21
$begingroup$
@hgasu It's true that matlab can't find two eigenvectors for that eigenvalue. But that's not matlab's fault. Only one exists. Nothing can change that. It's like asking for a positive real number $x$ such that $2x=0$.
$endgroup$
– Matt Samuel
Dec 27 '18 at 21:22
1
1
$begingroup$
A matrix can simply not have enough eigenvectors for a given eigenvalue. Then it isn't diagonalizable. Depending on your application, there are ways around it, but none of them will give you that second eigenvector, because it just doesn't exist.
$endgroup$
– Matt Samuel
Dec 27 '18 at 20:19
$begingroup$
A matrix can simply not have enough eigenvectors for a given eigenvalue. Then it isn't diagonalizable. Depending on your application, there are ways around it, but none of them will give you that second eigenvector, because it just doesn't exist.
$endgroup$
– Matt Samuel
Dec 27 '18 at 20:19
$begingroup$
It has been quite some time since I took linear algebra. Does this mean that there are only two eigenvalues that apply (as in, my by-hand method was incorrect), or does this mean that MatLab can simply not do what I want?
$endgroup$
– hgasu
Dec 27 '18 at 21:20
$begingroup$
It has been quite some time since I took linear algebra. Does this mean that there are only two eigenvalues that apply (as in, my by-hand method was incorrect), or does this mean that MatLab can simply not do what I want?
$endgroup$
– hgasu
Dec 27 '18 at 21:20
$begingroup$
Try your $n=2$ code with
[0 k1;0 0] instead. You won’t have enough eigenvectors there, either. As @MattSamuel points out, not every matrix is diagonalizable. You’ll need to do something else to compute exponentials for those matrices. See “Jordan normal form” for a general method.$endgroup$
– amd
Dec 27 '18 at 21:20
$begingroup$
Try your $n=2$ code with
[0 k1;0 0] instead. You won’t have enough eigenvectors there, either. As @MattSamuel points out, not every matrix is diagonalizable. You’ll need to do something else to compute exponentials for those matrices. See “Jordan normal form” for a general method.$endgroup$
– amd
Dec 27 '18 at 21:20
$begingroup$
MATLAB can do what you need, but you’re going to have to use the Jordan decomposition of the matrix, or something similar. I’m sure that there’s a built-in function for computing the exponential of a matrix, for that matter.
$endgroup$
– amd
Dec 27 '18 at 21:21
$begingroup$
MATLAB can do what you need, but you’re going to have to use the Jordan decomposition of the matrix, or something similar. I’m sure that there’s a built-in function for computing the exponential of a matrix, for that matter.
$endgroup$
– amd
Dec 27 '18 at 21:21
$begingroup$
@hgasu It's true that matlab can't find two eigenvectors for that eigenvalue. But that's not matlab's fault. Only one exists. Nothing can change that. It's like asking for a positive real number $x$ such that $2x=0$.
$endgroup$
– Matt Samuel
Dec 27 '18 at 21:22
$begingroup$
@hgasu It's true that matlab can't find two eigenvectors for that eigenvalue. But that's not matlab's fault. Only one exists. Nothing can change that. It's like asking for a positive real number $x$ such that $2x=0$.
$endgroup$
– Matt Samuel
Dec 27 '18 at 21:22
|
show 5 more comments
1 Answer
1
active
oldest
votes
0
$begingroup$
The "secret" is to use the kernel of $B-lambda_k I$ for the different eigenvalues $lambda_k$.
Let me explain it on a numerical example :
B=[2 1 1
1 2 1
1 1 2];
eig(B),% gives 1 (double eigenvalue) and 4 (which is simple)
I=eye(3);
null(B-1*I), % provides a basis of 2 vectors for the eigenspace associated with $lambda_1=1$
null(B-4*I), % provides a unique vector generating the eigenspace associated with $lambda_1=1$
answered Jan 11 at 19:11
Jean MarieJean Marie
30.7k42154
$endgroup$
add a comment |
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$begingroup$
The "secret" is to use the kernel of $B-lambda_k I$ for the different eigenvalues $lambda_k$.
Let me explain it on a numerical example :
B=[2 1 1
1 2 1
1 1 2];
eig(B),% gives 1 (double eigenvalue) and 4 (which is simple)
I=eye(3);
null(B-1*I), % provides a basis of 2 vectors for the eigenspace associated with $lambda_1=1$
null(B-4*I), % provides a unique vector generating the eigenspace associated with $lambda_1=1$
answered Jan 11 at 19:11
Jean MarieJean Marie
30.7k42154
$endgroup$
add a comment |
0
$begingroup$
The "secret" is to use the kernel of $B-lambda_k I$ for the different eigenvalues $lambda_k$.
Let me explain it on a numerical example :
B=[2 1 1
1 2 1
1 1 2];
eig(B),% gives 1 (double eigenvalue) and 4 (which is simple)
I=eye(3);
null(B-1*I), % provides a basis of 2 vectors for the eigenspace associated with $lambda_1=1$
null(B-4*I), % provides a unique vector generating the eigenspace associated with $lambda_1=1$
answered Jan 11 at 19:11
Jean MarieJean Marie
30.7k42154
$endgroup$
add a comment |
0
0
0
$begingroup$
The "secret" is to use the kernel of $B-lambda_k I$ for the different eigenvalues $lambda_k$.
Let me explain it on a numerical example :
B=[2 1 1
1 2 1
1 1 2];
eig(B),% gives 1 (double eigenvalue) and 4 (which is simple)
I=eye(3);
null(B-1*I), % provides a basis of 2 vectors for the eigenspace associated with $lambda_1=1$
null(B-4*I), % provides a unique vector generating the eigenspace associated with $lambda_1=1$
answered Jan 11 at 19:11
Jean MarieJean Marie
30.7k42154
$endgroup$
The "secret" is to use the kernel of $B-lambda_k I$ for the different eigenvalues $lambda_k$.
Let me explain it on a numerical example :
B=[2 1 1
1 2 1
1 1 2];
eig(B),% gives 1 (double eigenvalue) and 4 (which is simple)
I=eye(3);
null(B-1*I), % provides a basis of 2 vectors for the eigenspace associated with $lambda_1=1$
null(B-4*I), % provides a unique vector generating the eigenspace associated with $lambda_1=1$
answered Jan 11 at 19:11
Jean MarieJean Marie
30.7k42154
edited Jan 11 at 20:41
answered Jan 11 at 19:11
Jean MarieJean Marie
30.7k42154
answered Jan 11 at 19:11
Jean MarieJean Marie
30.7k42154
answered Jan 11 at 19:11
Jean MarieJean Marie
30.7k42154
30.7k42154
add a comment |
add a comment |
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1
$begingroup$
A matrix can simply not have enough eigenvectors for a given eigenvalue. Then it isn't diagonalizable. Depending on your application, there are ways around it, but none of them will give you that second eigenvector, because it just doesn't exist.
$endgroup$
– Matt Samuel
Dec 27 '18 at 20:19
$begingroup$
It has been quite some time since I took linear algebra. Does this mean that there are only two eigenvalues that apply (as in, my by-hand method was incorrect), or does this mean that MatLab can simply not do what I want?
$endgroup$
– hgasu
Dec 27 '18 at 21:20
$begingroup$
Try your $n=2$ code with
[0 k1;0 0]instead. You won’t have enough eigenvectors there, either. As @MattSamuel points out, not every matrix is diagonalizable. You’ll need to do something else to compute exponentials for those matrices. See “Jordan normal form” for a general method.$endgroup$
– amd
Dec 27 '18 at 21:20
$begingroup$
MATLAB can do what you need, but you’re going to have to use the Jordan decomposition of the matrix, or something similar. I’m sure that there’s a built-in function for computing the exponential of a matrix, for that matter.
$endgroup$
– amd
Dec 27 '18 at 21:21
$begingroup$
@hgasu It's true that matlab can't find two eigenvectors for that eigenvalue. But that's not matlab's fault. Only one exists. Nothing can change that. It's like asking for a positive real number $x$ such that $2x=0$.
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– Matt Samuel
Dec 27 '18 at 21:22