# Homework Help: Eigenvalues / Eigenvectors - Is this a legit question to ask?

1. Oct 15, 2012

### eurekameh

-2x + 3y + z = 0
3x + 4y -5z = 0
x -2y + z = -4

Find the characteristic equation, eigenvalues / eigenvectors of the system.
I'm given to understand the eigenvalue problem is Ax = (lamba)x, but lamba doesn't exist in the system above. How can I solve for the eigenvalues when there are none?

2. Oct 15, 2012

### Steve Collins

In the form you have written Ax=λx:
A is your matrix
x is an eigenvector of A
λ is the corresponding eigenvalue

You need to find the characteristic equation to find λ.

Last edited: Oct 15, 2012
3. Oct 15, 2012

### eurekameh

I do not see any eigenvalues in the system:
-2x + 3y + z = 0
3x + 4y -5z = 0
x -2y + z = -4

The A matrix is A = [-2 3 1; 3 4 -5; 1 -2 1], with ";" denoting a new row. My eigenvector x is x = [x;y;z].
Then, we have Ax = [0;0;-4]. Where are the eigenvalues?
Note that I did not come up with the system of equations above. It was the original question, which is why I'm asking if it is a legit question to ask, seeing that there are no eigenvalues to solve for.

4. Oct 16, 2012

### gabbagabbahey

What makes you think that x = [x;y;z] is an eigenvector? You have

$$\begin{pmatrix} -2 & 3 & 1 \\ 3 & 4 & -5 \\ 1 & -2 & 1\end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ -4\end{pmatrix}$$

Does this look like an eigenvalue equation?

The first thing I would suggest you do is to try and get this in the form

$$\begin{pmatrix} -2 & 3 & 1 \\ 3 & 4 & -5 \\ 1 & -2 & 1\end{pmatrix} \begin{pmatrix} x - x_0\\ y -y_0 \\ z - z_0 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0\end{pmatrix}$$

What must be the values of $x_0$, $y_0$ and $z_0$ to get it into this form?

Then, rewrite your eigenvalue equation as $(\mathbf{A}-\lambda\mathbf{I}_{n \times n})\mathbf{x}=\mathbf{0}$ and compare it to the above. How is the characteristic equation related to $(\mathbf{A}-\lambda\mathbf{I}_{n \times n})\mathbf{x}$?

5. Oct 16, 2012

### eurekameh

Why wouldn't x be an eigenvector? Standard form is Ax = (lambda)x. lambda = eigenvalue, x = eigenvector, no?

The thing I don't understand is that there is no lamba in any of the three equations, but it's asking me to solve for it. It's asking me to solve Ax = b, where b = [0;0;-4], which I can do, but it wants me to solve this eigenvalue problem that doesn't look like one at all.

Why are you writing your matrix equation like that? Where did x0,y0,z0 come from? Is x0=0, y0 = 0, z0 = -4?

6. Oct 16, 2012

### gabbagabbahey

Sure, but you don't have an equation of the form Ax = λx.

Nevermind my above suggestion, I don't think it will be useful. After thinking a bit more about the problem, I'm not sure what your instructor is looking for here. What was the exact wording on the original problem statement?

7. Oct 16, 2012

### eurekameh

Consider the following system of equations:
...
...
...
a) Calculate the characteristic equation of the system.
b) Find all the eigenvalue and eigenvectors.

8. Oct 16, 2012

### gabbagabbahey

In that case, I have no idea what the instructor (or whoever posed this problem to you) is looking for. As far as I am aware, in linear algebra, the characteristic polynomial is defined only for a square matrix, not a system of linear equations. It makes no sense to me to ask for the "characteristic equation of the system".

9. Oct 16, 2012

### eurekameh

That's what I thought. In class today, he clarified that he wanted the eigenvalues of the system A. So, Ax = (lambda)x. The b vector above in Ax = b is irrelevant to the problem. So it really is a misguided question. Thanks for your help.