# Eigenvalues and eigenvectors

1. May 3, 2007

### Neoon

Hi

I came across a problem of eigenvalues and eigenvectors. It was easy and I solved it but one thing made me unsure about the answer. All the three eigenvectors were zero vectors. Here is the question and my answer:

The matrix A=
( -1 0 0 1
0 -2 0 0
0 1 -2 0
0 0 0 1)

I began with finding the eigenvalues. The result is the following fourth order equation (x=lamda=eigenvalue):
x^4+4x^3+x^2-6x-4=0
When i solved this equation using Texas Instruments calculator, it found three solutions:
x1=-(sqrt(5)+1)
x2=(sqrt(5)-1)
x3= -1
x4= -1

So, I had three eignevectors because x3=x4=-1 (multiplicity 2)

When I used these eigenvalues to find the eigenvectors, all of the eigenvectors turned out to be=
(0
0
0
0)

My question: is this reasonable solution to have all of the eigenvectors= 0 or I have made a mistake somewhere.

2. May 3, 2007

### ansrivas

By definition eigenvectors are non-zero. For sure you have made a calculation mistake.

If $$det \left(A-\lambda I \right) = 0$$

then there has to be an non-zero vector that this matrix takes to zero.

3. May 3, 2007

### Office_Shredder

Staff Emeritus
You've made this incredibly complicated in terms of calculating the zeroes.

Expand by cofactors along the first column (expanding A-I*lamda), and you should see a lower triangular 3x3 matrix as the only determinant you need to take. The eigenvalues should fall into your lap at that point

4. May 3, 2007

### HallsofIvy

Staff Emeritus
Since the second row has only one non-zero entry it is simple to expand by the second row. You get $\lambda+ 2$ time a determinant in which, again, has only a single entry. Expanding that determinant by the second row, you get $(\lambda+ 2)^2$ times a 2 by 2 determinant that has a 0 in the lower left corner- it is "upper triangular" so that it is just the product of the diagonal entries: your eigenvalue equation is $(\lambda+ 2)^2(\lambda- 1)^2= 0$ so the only eigenvalues are -2 and 1.
The definition of "eigenvalue" is that $\lambda$ is an eigenvalue of A if and only if the equation $Ax= \lambda x$ is true for non-zero x. The fact that you found that "All the three eigenvectors were zero vectors" should have told you that you did not have the right eigenvalues!

Last edited: May 5, 2007
5. May 3, 2007

### Neoon

Thanks all for your help

I went back and solved it using the hint of the expanding using 2nd row and I have four eigenvalues: -2, -2, 1, -1. You might missed the last eigenvalue.

Then, I found the eigenvectors. It turned out to have only 2 eigenvectors for -2 & 1 since the eigenvector of -1 is a zero-eigenvector.

So, for eigenvalue=(-2) I had the following eigenvector:
(0
0
1
0)

My question now is how to find generalized eigenvectors?

I found the defficincy = d = k - m = 2- 1 = 1
Then, I solved (A - (-2) I)^(d+1) * v = 0
(1 0 0 1 (x1 (0
0 0 0 0 x2 = 0
0 1 0 0 x3 0
0 0 0 3) x4) 0)

of course with raising the quantity (A - (-2) I) to power 2 since d+1=2

Then, I picked a vector such that if I multiply it with the quantity (A - (-2) I)^1 it will give me my eigenvector:
(0
0
1
0)
is my approach for this particular point correct (When I tried more than one vector to have my eigenvector after multiplication)?

I found the generalized eigenvector to be:
(0
1
x
0)

Then I chose x = 2.

Also, does the value of d (difficiency) changes the generalized eigenvectors?

6. May 4, 2007

### Neoon

I appreciate any ubdates on the topic.Neoon

7. May 5, 2007

### Neoon

Just a short question:

If I have d= defficiency = 0
Do I have a generalized eigenvector?

8. May 5, 2007

### HallsofIvy

Staff Emeritus
No, I did not "miss" -1 as an eigenvalue- it is NOT an eigenvalue.

"the eigenvector of -1 is a zero-eigenvector."

You have been told twice here, and I am certain by your teacher and textbook, that a number, $\lambda$ is an eigenvalue of A if and only the equation $Ax= \lambda x$ has non-zero solutions. Please stop talking about "zero-eigenvectors".

The eigenvalue 1 has any multiple of [1, 0, 0, -2] as eigenvector and the eigenvalue -2 has any multiple of the vector you found, [0, 0, 0, 1, 0] as eigenvector.

you haven't actually said what problem it is you are trying to solve but if, for example, you wanted to find a matrix that would change A to its Jordan Normal Form, you will have to find one more "generalized" eigenvector for each eigenvalue.

9. May 6, 2007

### Neoon

Ok, I appreciate your clarification about eigenvectors.

Now, for the generalized eigenvectors, what if we have k, the multiplicity, equal to m, the number of eigenvectors so d, the difficiency, equal to zero? Do we have a gineralized eigenvector in this case?

10. May 6, 2007

### HallsofIvy

Staff Emeritus
If the defficiency is 0- that is, if the number of independent eigenvectors for a given eigenvalue is equal to the multiplicity of that eigenvalue, then you have enough eigenvectors. You don't need "generalized" eigenvectors.

11. May 6, 2007

### Neoon

Thank you for your valuable help

I really appreciate