What Are Eigenvalues and Eigenspaces in Linear Transformations?

  • Thread starter Thread starter derryck1234
  • Start date Start date
  • Tags Tags
    Eigenvectors
Click For Summary

Homework Help Overview

The discussion revolves around the concepts of eigenvalues and eigenspaces in the context of linear transformations, specifically focusing on a transformation defined on 2x2 matrices. Participants are exploring how to derive the eigenvalues and eigenspaces from a given transformation matrix.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the initial steps to form the transformation matrix and question the correct terminology for certain matrices. There are attempts to derive the characteristic equation and inquiries about methods for solving it.

Discussion Status

Some participants have made progress in determining the transformation matrix and its characteristic equation. There is a mix of confirmations and corrections regarding the matrix representations and calculations, indicating an ongoing exploration of the topic.

Contextual Notes

There are indications of confusion regarding matrix terminology and the formation of the transformation matrix, as well as expressions of frustration about the learning process. Participants are actively engaging with the material and seeking clarification on their understanding.

derryck1234
Messages
55
Reaction score
0

Homework Statement



Let T: M22 - M22 be defined by

T(a b = 2c a + c
c d) b-2c d

(a) Find the eigenvalues T.
(b) Find bases for the eigenspaces of T.

Homework Equations



det(lambdaI - A) = 0

The Attempt at a Solution



I just wanted to know how to start here. I am used to using vector transformations...

Would I use identity matrices 1 0 0 1 0 0 0 0
0 0 0 0 1 0 0 1

And then form a matrix somehow? I just don't know how big or how this matrix would be formed...?
 
Physics news on Phys.org
One small thing: you shouldn't call the matrices

\left(\begin{array}{cc} 1 & 0\\ 0 & 0\end{array}\right), \left(\begin{array}{cc} 0 & 1\\ 0 & 0\end{array}\right), \left(\begin{array}{cc} 0 & 0\\ 1 & 0\end{array}\right), \left(\begin{array}{cc} 0 & 0\\ 0 & 1\end{array}\right)

identity matrices. We reserve the term identity matrix for something else.
You can call them elementary matrices however.

Anyway, your approach is good: try to find the images of the elementary matrices to find the matrix form of T. T will have the form of a 4x4-matrix.
 
Ok. So I find the matrix for T to be:

0 1 0 0
0 0 1 0
2 1 0 0
-2 0 0 1

The characteristic equation of which I find to be:

(lambda)4 - (lambda)3 - (lambda)2 - (lambda) + 2

Just wondered if there was an easy way to solve this?
 
derryck1234 said:
Ok. So I find the matrix for T to be:

0 1 0 0
0 0 1 0
2 1 0 0
-2 0 0 1

The characteristic equation of which I find to be:

(lambda)4 - (lambda)3 - (lambda)2 - (lambda) + 2

Just wondered if there was an easy way to solve this?

Hmm, that's not what I get:

T\left(\begin{array}{cc} 1 & 0\\ 0 & 0\end{array}\right)=\left(\begin{array}{cc} 0 & 1\\ 0 & 0\end{array}\right)

So the first column needs to be (0,1,0,0), no?
 
O ok. So I must turn the resulting matrix into a column vector? I just put them together as matrices...my bad...ok I shall try this agen...shall get back to you to confirm my answer...thanks a lot...

Ciao...I feel mentally ill that I am such a dumb ***!...but then again...have to learn somehow...
 
Ok I think I have solved it:

The standard matrix for T is:

0 0 2 0
1 0 1 0
0 1 -2 0
0 0 0 1

Whose characteristic equation is:

(lambda - 1)2(lambda + 1)(lambda + 2) = 0

This right?

Thanks

Derryck
 
derryck1234 said:
Ok I think I have solved it:

The standard matrix for T is:

0 0 2 0
1 0 1 0
0 1 -2 0
0 0 0 1

Whose characteristic equation is:

(lambda - 1)2(lambda + 1)(lambda + 2) = 0

This right?

Thanks

Derryck

Yes, that's also what I've got!
 
Thanks a lot micromass

Tell you what if it wasn't for physicsforums I don't know how I would have passed my correspondence linear algebra course...

Cheers

Derryck
 

Similar threads

Replies
2
Views
2K
Replies
5
Views
2K
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 19 ·
Replies
19
Views
4K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 24 ·
Replies
24
Views
4K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
4
Views
2K