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Finding eigenvectors of a matrix

  • Thread starter JaysFan31
  • Start date
JaysFan31
1. Homework Statement
I need to find the eigenvectors for the matrix shown below.


2. Homework Equations
Nothing relevant.


3. The Attempt at a Solution
I have the matrix
2 1 -1 0
0 4 -2 0
0 3 -1 0
0 3 -2 1

The eigenvectors are 1,1,2,2.

To get the eigenvectors for 1,1, I use the matrix
-1 1 -1 0
0 -3 -2 0
0 3 2 0
0 3 -2 0

When I bring this to RREF, I have
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 0

So I know that the vector
0
0
0
1
is an eigenvector.

But the book also says that
1
2
3
0
is an eigenvector for the eigenvalue 1.

If the matrix only has one solution, how do I get this other eigenvector?
 

Answers and Replies

218
0
2. Homework Equations
Nothing relevant.
See if this equations help:

[tex]det(A-{\lambda}I) = 0[/tex]

[tex]A{\cdot}x = {\lambda}x[/tex]

where [tex]A[/tex] is the matrix, [tex]x[/tex] is an eigenvector and [tex]\lambda[/tex] is it's corresponding eigenvalue
 
D H
Staff Emeritus
Science Advisor
Insights Author
15,329
681
When an eigenvalue has a multiplicity > 1 it means that there multiple linearly independent eigenvectors corresponding to that eigenvalue. In particular, the eigenvectors span an N-dimensional subspace, where N is the multiplicity of the eigenvalue.

As a trivial example, the n-dimensional identity matrix [itex]I_n[/itex] has eigenvalue 1 with multiplicity n. The eigenvalues span the original space. (In other words, any vector in [itex]\mathbb R^n[/itex] is an eigenvector of [itex]I_n[/itex]).
 
HallsofIvy
Science Advisor
Homework Helper
41,736
894
1. Homework Statement
I need to find the eigenvectors for the matrix shown below.


2. Homework Equations
Nothing relevant.


3. The Attempt at a Solution
I have the matrix
2 1 -1 0
0 4 -2 0
0 3 -1 0
0 3 -2 1

The eigenvectors are 1,1,2,2.

To get the eigenvectors for 1,1, I use the matrix
-1 1 -1 0
0 -3 -2 0
0 3 2 0
0 3 -2 0
This is, I take it, your original matrix with 1 subtracted from the diagonal values? Or, at least it's supposed to be: check the second row: 4-1= 3, not -3.

When I bring this to RREF, I have
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 0

So I know that the vector
0
0
0
1
is an eigenvector.

But the book also says that
1
2
3
0
is an eigenvector for the eigenvalue 1.

If the matrix only has one solution, how do I get this other eigenvector?
1 1 -1 0
0 3 -2 0
0 3 -2 0
0 3 -2 0

row reduces to
1 0 -1/3 0
0 1 -2/2 0
0 0 0 0
0 0 0 0

That gives both (1, 2, 3, 0) as eigen value.
 

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