Eigen Values/Vectors

  • #1

Homework Statement


Find the eigen values/eigen vectors of

[tex]
\left[
\begin{array}{cc}
3&-1&1&1&0&0\\
1&1&-1&-1&0&0\\
0&0&2&0&1&1\\
0&0&0&2&-1&-1\\
0&0&0&0&1&1\\
0&0&0&0&1&1\\
\end{array}
\right]
[/tex]


Homework Equations





The Attempt at a Solution



The lambda equation is:

[tex]
A - \lambda I =
\left[
\begin{array}{cc}
3 - \lambda&-1&1&1&0&0\\
1&1 - \lambda&-1&-1&0&0\\
0&0&2 - \lambda&0&1&1\\
0&0&0&2 - \lambda&-1&-1\\
0&0&0&0&1 - \lambda&1\\
0&0&0&0&1&1 - \lambda\\
\end{array}
\right]

[/tex]

Finding the determinant yields:

[tex]
-32 \lambda + 80\lambda^2 - 80\lambda^3 + 40\lambda^4 - 10\lambda^5 + \lambda^6
[/tex]

Setting equal to zero and factoring gives the eigen values:

[tex]
\lambda =
\left[
\begin{array}{cc}
0\\
2\\
2\\
2\\
2\\
2\\
\end{array}
\right]
[/tex]


Plugging in lambda = 2 into the lambda equation gives:

[tex]
\left[
\begin{array}{cc}
1&-1&1&1&0&0\\
1&-1&-1&-1&0&0\\
0&0&0&0&1&1\\
0&0&0&0&-1&-1\\
0&0&0&0&-1&1\\
0&0&0&0&1&-1\\
\end{array}
\right]
[/tex]
The columns help for the simultaneous equations:

The bottom four rows reveal that X5, X6 = 0

Adding the first two rows together gives:

2X1 - 2X2 = 0

or

X1 = X2, So I choose X1 = X2 = 1, since the eigen vector must be a non-zero vector.

X3 and X4 cancel out so I choose them to = 0.

-------------
Since the eigen values repeat, I have to use the previous eigen vector as the answer to the simultaneous equations:

Summing the top two rows together:

2X1 - 2X2 = 2

or

X1 = 1 + X2. Everything else is still zero, choose X2 = 0, X1 = 1
--------------
Recursively using the previous eigen vector:

2 equations, 4 unknowns, choose X2 = 0, X4 = 0, top two equations become:

X1 + X3 = 1
X1 - X3 = 0

X1 and X3 = 1/2
--------------
Recursively using the previous eigen vector:

This is where the trouble comes into play:

3rd equation : X5 + X6 = 1/2
4th equation: -X5 - X6 = 0

which cannot be solved....

This keeps happening over and over. I used MATLAB to verify that the eigen values and vectors could be found...but I cannot solve this through.

Any suggestions?
 

Answers and Replies

  • #2
312
0
Your problem is in the block form
[itex][\begin{array}{cc}A&B\\0&C\end{array}][\begin{array}{c}x_1\\x_2\end{array}]=\lambda[\begin{array}{c}x_1\\x_2\end{array}][/itex]
which is
[itex]Ax_1+Bx_2=\lambda x_1[/itex]
[itex]Cx_2=\lambda x_2[/itex]
which you can solve easier because they are 3x3
 
  • #3
A few more questions (forgive my lacking linear algebra skills)

1) You say that the problem was in a special form....why does this method present itself due to the form? Does it not apply if the lower-left 3x3 was not equal to zero?

2) are x1 and x2 both going to be 3x3 matrices, and concatenated together into a 3x6 will be the eigen vectors?

3) In terms of solving for x1 and x2, is this just simultaneous matrix equations, so perform the usual simultaneous-equation solving skills (with inverse matrices instead of division)?

4) Was it just very difficult (or impossible) to solve it the way I had described, due to the fact that I get too much freedom to choose many of the values, and end up getting stuck during the process if I'm not very strategic?
 
  • #4
312
0
A few more questions (forgive my lacking linear algebra skills)

1) You say that the problem was in a special form....why does this method present itself due to the form? Does it not apply if the lower-left 3x3 was not equal to zero?

2) are x1 and x2 both going to be 3x3 matrices, and concatenated together into a 3x6 will be the eigen vectors?

3) In terms of solving for x1 and x2, is this just simultaneous matrix equations, so perform the usual simultaneous-equation solving skills (with inverse matrices instead of division)?

4) Was it just very difficult (or impossible) to solve it the way I had described, due to the fact that I get too much freedom to choose many of the values, and end up getting stuck during the process if I'm not very strategic?
1) The lower-left block being 0 makes the second equation that I gave simpler, otherwise the two equations will be coupled, and not easy to solve by hand. It is just a trick that you play to make a homework (not real life) problem simpler.

2) x1 and x2 are both 3x1 vectors, they concatenate to a 6x1 vector that is the eigenvector, there are multiple solutions, but I recommend that you treat them as vectors not matrices, and find out all the vectors that satisfy the equations.

3) The second equation is a 3x3 eigenvalue problem, solve as you did for the 6x6 problem, only much simpler (by hand); then sub into the first equation and solve as a linear system (also 3x3, try to find all the solutions, whatever way you prefer)

4) Honestly I didn't read your description, but eigenvalue problems are hard problems, usually handled by computers, any general form beyond 6x6 is probably too much for a human :)(unless it has special structures, such as convolution matrix, etc. )
 
  • #5
Alright, well here's how far I got:

[tex]
|C - \lambda I| = (2-\lambda)(-2\lambda + \lambda^2)
[/tex]

So the eigen values for the C Matrix are 2, 2, and 0.

I wont show the exact steps, but my eigen vectors (x2) for C are:
[tex]
\left[
\begin{array}{cc}
0\\
1\\
-1\\
\end{array}
\right]
[/tex]

[tex]
\left[
\begin{array}{cc}
1\\
0\\
0\\
\end{array}
\right]
[/tex]

[tex]
\left[
\begin{array}{cc}
0\\
-\frac{1}{2}\\
-\frac{1}{2}\\
\end{array}
\right]
[/tex]

Solving the first equation for x1:

[tex]
x1 = (A-\lambda I)^{-1} (-B x2)
[/tex]

but now the problem is that when I plug in 2 for lambda, I end up with a matrix that I cannot take the inverse of.

Any suggestions on moving forward??
 
  • #6
Office_Shredder
Staff Emeritus
Science Advisor
Gold Member
3,837
149
but now the problem is that when I plug in 2 for lambda, I end up with a matrix that I cannot take the inverse of.
I think you're a bit confused. The whole point of eigenvalues is that those are the values which, when you plug in for lambda, you get a non-invertible matrix. Let's look at a smaller example to see how to find the corresponding eigenvector. Given the matrix

[tex]\left( \begin{array}{cc}1&1\\0&0\end{array} \right)[/tex]

You can check that it has eigenvalues of 0 and 1. Let's find the eigenvector corresponding to 1:

[tex]\left( \begin{array}{cc}1&1\\0&0\end{array} \right) \left( \begin{array}{cc} x\\ y \end{array} \right) = 1*\left( \begin{array}{cc} x\\ y \end{array} \right)[/tex]

This gives us two equations: x+y=x and 0+0=0. So the "solution" to this system of equations is y=0 and x can be anything. In general you won't get a unique solution, because if v is an eigenvector, you can scale v to get another eigenvector. So you expect to get a one dimensional set of solutions
 
  • #7
Okay. I see. You're right I was confused about the eigen-vectors creating a non-invertible matrix.

The example you described was the original method I was trying to do, but got stuck with impossible to solve system of equations (on the 4th repeated eigen-value "plug-in", see first post)

Following sunjin's suggestion of partitioning the matrix and creating linear system to solve, I got stuck on not being able to take the inverse (because of what you just told me about eigen-vectors).

Sooooo...is there a different way to solve that linear system..is there a different way to do the original approach of simultaneous equations...or is there another method altogether??
 
Last edited:
  • #8
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
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If you row-reduce your matrix a bit more, you get
\begin{bmatrix}
1&-1&0&0&0&0\\
0&0&1&1&0&0\\
0&0&0&0&1&0\\
0&0&0&0&0&1\\
0&0&0&0&0&0\\
0&0&0&0&0&0\\
\end{bmatrix} which corresponds to the four equations
\begin{align*}
x_1 - x_2 &= 0 \\
x_3+x_4 &= 0 \\
x_5 = 0 \\
x_6 = 0
\end{align*} You analyzed the first, third, and fourth equations already and found one eigenvector from this system, but there's a second one. From the first eigenvector, you had (1,1,0,0,0,0), (1,0,0,0,0,0), and (1/2,0,1/2,0,0,0). The fact you couldn't find a solution using the last vector in the chain means you've reached the end. Now you have to find a new chain starting with the second eigenvector.
 
Last edited:
  • #9
Okay, I continued forward by doing the reduced-row echelon form, and now I am verifying my answers.

When I multiply the original A matrix by the eigenvector (1,0,0,0,0,0), I get an answer of (3,1,0,0,0,0)....Shouldn't the answer be a scalar multiple of the eigenvector? Does this mean I've made a mistake?
 
  • #10
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
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The matrix A doesn't have five linearly independent eigenvectors corresponding to the eigenvalue 2; it has only two, which you can find by solving the original system (A-λI)x=0.

The vector (1,0,0,0,0,0) is a generalized eigenvector of A. You found it by solving (A-λI)x = (1,1,0,0,0,0)T so you have Ax = (1,1,0,0,0,0)T + λx = (3,1,0,0,0,0)T.
 
  • #11
Great! So that means everything is gravy now!

Thanks all, this has REALLY given me a lot of insight on some of these concepts. :)
 

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