General Question(s) about Eigenvalue Problem

[Saladsamurai]

Homework Statement

I am having some trouble with this procedure and I am not exactly sure how to phrase my questions; so I will procede with one particular problem that is giving me trouble and perhaps someone can help to shed light on it.

In one problem, I am given some matrix

$$\begin{bmatrix} 1 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & -2 \end{bmatrix}$$

and I am asked to find the eigenvalues (the lambdas) as well as the eigenvectors corresponding to them.

Homework Equations

Here I will just include a brief description of what "the eigenvalue problem" is. We are essentially looking for values of $\lambda$ for which the equation $(\mathbf{A} - \lambda\mathbf{I})\mathbf{x} = 0\qquad(1)$ admits nontrivial solutions (i.e. $\mathbf{x}\ne0$)

The Attempt at a Solution

Since the matrix A is upper triangular, it's eigenvalues are simply the elements along the main diagonal:

$$\lambda_{1,2,3} = \{1, 0, -2\}$$

For each of these, there is a corresponding eigenvector that we find by plugging each value back into (1).

For $\lambda_1 = 1$:

$$A = \begin{bmatrix} 0 & 0 & 1 \\ 0 & -1 & 0 \\ 0 & 0 & -3 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \qquad(2)$$

I am not really sure what to do with (2). At first glance, it suggests to me that all of the x_i's are zero. But the whole point of the eigenvalue problem is that all of the x_i's are not zero. So, how can I reconcile my system of equations with this?

The solution says that the eigenvector corresponding to $\lambda_1 =1$ is

$$e_1 = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$$

How do they get this from (2)?

 

[Tedjn]

How do you find (a basis for) the nullspace (aka kernel) of a linear transformation?

 

[Saladsamurai]

Hi Tedjin I should have prefaced this with the fact that it is from a Mathematical Methods for Engineers class; so most of those words don't mean anything to me. I know what a basis is, but that's it. I'll definitely look them up now, but I think that the answer should be simpler or more intuitive then what you suggest. It appears that they let x2 = x3 = 0 and let x1 remain arbitrary, but I don't know why they chose x1 to be arbitrary over the others.

 

[fzero]

For $\lambda_1 = 1$:

$$A = \begin{bmatrix} 0 & 0 & 1 \\ 0 & -1 & 0 \\ 0 & 0 & -3 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \qquad(2)$$

I am not really sure what to do with (2). At first glance, it suggests to me that all of the x_i's are zero. But the whole point of the eigenvalue problem is that all of the x_i's are not zero. So, how can I reconcile my system of equations with this?

The solution says that the eigenvector corresponding to $\lambda_1 =1$ is

$$e_1 = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$$

How do they get this from (2)?

(2) represents the set of equations

$$0 x_1 + 1 x_3 =0, ~ -1 x_2 = 0 ,~ 0 x1 - 3 x_3 =0.$$

You know how to solve these. There's no way to solve them with $$x_3$$ arbitrary.

 

[Saladsamurai]

(2) represents the set of equations

$$0 x_1 + 1 x_3 =0, ~ -1 x_2 = 0 ,~ 0 x1 - 3 x_3 =0.$$

You know how to solve these. There's no way to solve them with $$x_3$$ arbitrary.

Hi fzero I am not sure what you are driving at? Yes, (2) is the system:

$x_3 = 0$
$-x_2 = 0$
$-3x_3 = 0$

but I don't know how to solve this. It clearly suggests that $x_2 = x_3 = 0$, but there is no mention of x₁ in this system. Hmmm ... I think the light bulb just went off Since there is no mention of x₁ in this system, it means that it does not matter what value x₁ takes on ... it does not effect the system and hence it is arbitrary.


I know I have more question regarding eigenvalue problems, but I cannot think of what they are at the moment. So I will post back in a while with more. Thanks for your help so far.