Eigenvalues / Eigenvectors - Is this a legit question to ask?

[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?

 

[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.

 

[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.

 

[gabbagabbahey]

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.

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}?

 

[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?

 

[gabbagabbahey]

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

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

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

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?

 

[eurekameh]

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

 

[gabbagabbahey]

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

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".

 

[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.