# 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

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

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