Eigenvector for Complex Eigenvalue help

Click For Summary

Discussion Overview

The discussion revolves around the process of finding eigenvectors corresponding to complex eigenvalues, specifically focusing on the eigenvalue \( c = 1 + i \). Participants explore the derivation of a basis for the eigenvector associated with this eigenvalue, including matrix construction and row reduction techniques.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions the validity of constructing a matrix solely from an eigenvalue, suggesting that the original matrix is necessary for understanding the context.
  • Another participant clarifies that they meant "basis for the subspace of all eigenvectors corresponding to the given eigenvalue," acknowledging the confusion in terminology.
  • A participant provides the original matrix \( A = \begin{pmatrix} 4 & 2 \\ -5 & -2 \end{pmatrix} \) and describes the process of finding the characteristic polynomial, which has roots \( 1+i \) and \( 1-i \).
  • It is noted that for real operators, eigenvalues and eigenvectors come in complex conjugate pairs, linking the eigenvalue \( 1+i \) with its corresponding eigenvector \( (3+i, -5) \) and the eigenvalue \( 1-i \) with the eigenvector \( (3-i, -5) \).
  • A participant explains that row reducing a singular \( 2 \times 2 \) matrix leads to a specific form, indicating that non-zero solutions can be derived from this structure.

Areas of Agreement / Disagreement

Participants express differing views on the initial steps of the eigenvector derivation process, particularly regarding the construction of the matrix from the eigenvalue. While some clarify their terminology, there is no consensus on the initial matrix's role or the method of deriving the eigenvector basis.

Contextual Notes

Participants highlight the importance of the original matrix and the characteristic polynomial in understanding the eigenvalue-eigenvector relationship. There is an acknowledgment of the complexity involved in row reduction and the implications of singular matrices.

SeannyBoi71
Messages
84
Reaction score
0
In my lecture notes my prof used the eigenvalue c= 1 + i and ended up with the matrix with (5 3+i) as row 1, and the second row is zeroes. After that, he simply wrote that the basis for this eigenvalue c is (3+i,-5) (in column form) without explaining. How did he get that basis? I tried working it out and not sure what he did.
 
Physics news on Phys.org
I really don't understand what you are trying to say. You cannot start with an eigenvalue and just make a matrix out of it. Possibly, you are saying that 1+i is an eigenvalue of a given matrix and that the new matrix was constructed of corresponding eigenvectors. Also, it makes no sense to say "a basis for an eigenvalue" since only vector spaces have bases and an eigenvalue is a number. Perhaps you mean "basis for the subspace of all eigenvectors corresponding to the given eigenvalue".


What was the original matrix for which 1+ i was an eigenvalue? That is essential to understanding this.

(It is certainly true that the vectors (5, 3+i) and (3+ i, -5) are orthogonal. That may be relevant.)
 
HallsofIvy said:
Perhaps you mean "basis for the subspace of all eigenvectors corresponding to the given eigenvalue".)

Yes, this is what I meant. Sorry for any confusion.

In the example, A=(4 2) (row 1) and (-5 -2) (row 2) (my apologies, I don't know how to put actually make that into a matrix here). We found the characteristic polynomial, and that the roots were 1+i and 1-i. He took the eigenvalue 1+i, and proceeded to get the eigenvector for that eigenvalue. When he plugged in [tex]((i+1)I_2 - A))[/tex] he then row reduced and got the matrix I said in my first post. He then wrote that the basis for this eigenvalue was (3+i, -5). I hope that clarifies. I was just confused on the arithmetic he did to get that basis.
 
It is because for a real operator eigenvalues and eigenvectors come in complex conjugate pairs.
so if we have the eigen pair
1 + i;(3+i,-5)
we also have the eigen pair
1 - i;(3-i,-5)
 
If you row reduce a 2x2 singular matrix, you will always get something of the form
[tex]\left(\begin{array}{c}a & b \\ 0 & 0\end{array}\right)[/tex]
And if you want a non-zero solution to
[tex]\left(\begin{array}{c}a & b \\ 0 & 0\end{array}\right) <br /> \left(\begin{array}{c}x \\ y \end{array}\right)= <br /> \left(\begin{array}{c}0 \\ 0 \end{array}\right)[/tex]
If sould be obvious it will be a multiple of
[tex]\left(\begin{array}{c}-b \\ a\end{array}\right)[/tex]
 
Ok this all makes sense. Thank you
 

Similar threads

  • · Replies 33 ·
2
Replies
33
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
11K