General Question(s) about Eigenvalue Problem

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Homework Help Overview

The discussion revolves around the eigenvalue problem related to a specific matrix. Participants are exploring the process of finding eigenvalues and eigenvectors, particularly focusing on the implications of the equations derived from the eigenvalue condition.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to understand how to derive eigenvectors from the eigenvalue equation, expressing confusion about the implications of the resulting system of equations.
  • Some participants question the choice of variables in the eigenvector solution, specifically why one variable is treated as arbitrary while others are set to zero.
  • There is a discussion about the interpretation of the nullspace and its relation to the eigenvalue problem.

Discussion Status

The conversation is ongoing, with participants providing insights and clarifications regarding the eigenvalue problem. Some guidance has been offered on interpreting the equations, but there is still uncertainty and further questions about the concepts involved.

Contextual Notes

Participants are working within the context of a Mathematical Methods for Engineers course, which may influence their familiarity with certain terms and concepts related to linear transformations and eigenvalues.

Saladsamurai
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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. :smile:

In one problem, I am given some matrix

<br /> \begin{bmatrix}<br /> 1 &amp; 0 &amp; 1 \\<br /> 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; -2<br /> \end{bmatrix}<br />

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 =<br /> \begin{bmatrix}<br /> 0 &amp; 0 &amp; 1 \\<br /> 0 &amp; -1 &amp; 0 \\<br /> 0 &amp; 0 &amp; -3<br /> \end{bmatrix}<br /> \begin{bmatrix}<br /> x_1 \\<br /> x_2 \\<br /> x_3<br /> \end{bmatrix}<br /> =<br /> \begin{bmatrix}<br /> 0 \\<br /> 0 \\<br /> 0<br /> \end{bmatrix}<br /> \qquad(2)<br />

I am not really sure what to do with (2). At first glance, it suggests to me that all of the xi's are zero. But the whole point of the eigenvalue problem is that all of the xi'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

<br /> e_1 = <br /> \begin{bmatrix}<br /> 1 \\<br /> 0 \\<br /> 0<br /> \end{bmatrix}<br />

How do they get this from (2)?
 
Last edited:
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How do you find (a basis for) the nullspace (aka kernel) of a linear transformation?
 
Hi Tedjin :smile: 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.
 
Saladsamurai said:
For \lambda_1 = 1:

A =<br /> \begin{bmatrix}<br /> 0 &amp; 0 &amp; 1 \\<br /> 0 &amp; -1 &amp; 0 \\<br /> 0 &amp; 0 &amp; -3<br /> \end{bmatrix}<br /> \begin{bmatrix}<br /> x_1 \\<br /> x_2 \\<br /> x_3<br /> \end{bmatrix}<br /> =<br /> \begin{bmatrix}<br /> 0 \\<br /> 0 \\<br /> 0<br /> \end{bmatrix}<br /> \qquad(2)<br />

I am not really sure what to do with (2). At first glance, it suggests to me that all of the xi's are zero. But the whole point of the eigenvalue problem is that all of the xi'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

<br /> e_1 = <br /> \begin{bmatrix}<br /> 1 \\<br /> 0 \\<br /> 0<br /> \end{bmatrix}<br />

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.
 
fzero said:
(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 :smile: 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 :smile: 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.
 

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