How do I find the characteristic polynomial

Click For Summary

Homework Help Overview

The discussion revolves around finding the characteristic polynomial of a linear transformation T defined on the vector space V = C^3. The transformation is given by specific equations relating the standard basis vectors, and participants are exploring methods to determine the eigenvalues of T without directly solving the associated equations.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss various methods to find the characteristic polynomial, including using determinants, row reduction to triangular form, and exploring linear dependence relations among powers of T. Questions are raised about the necessity of finding the characteristic polynomial solely for the purpose of determining eigenvalues.

Discussion Status

The discussion is active, with multiple approaches being considered. Some participants suggest that finding the characteristic polynomial may not be necessary if the minimal polynomial can provide the eigenvalues. There is a recognition of differing methods and some questioning of the validity of certain approaches.

Contextual Notes

Some participants express uncertainty about the methods they were taught for computing eigenvalues and characteristic polynomials, indicating a potential gap in foundational knowledge. There is also mention of a "trivial" root in the characteristic polynomial, suggesting complexity in the problem's structure.

Treadstone 71
Messages
275
Reaction score
0
"Let T be a the transformation on V = C^3 given by the equation

T(x)=-y-2z
T(y)=3x+5y+7z
T(z)=-2x-3y-4z

where (x,y,z) denotes the standard basis. Find the eigenvalues of T and the corresponding eigenspaces."

Is there a way to find the eigenvalues without solving the 3 equations? How do I find the characteristic polynomial of T without resorting to anything related to determinants?
 
Physics news on Phys.org
The determinant of the matrix is one of the coefficients of the characteristic polynomial, and is also the product of the eigenvalues.

So however you do this problem, you have to use a method that is capable of computing the determinant of the matrix.
 
Last edited:
You can find T^2 and T^3, then look for a linear dependence relation amongst I, T, T^2 and T^3 (this won't be difficult here). All the eigenvalues will appear as roots of this polynomial, as the minimal polynomial will divide whatever relation you find (you may in fact find the minimal polynomial this way).
 
Just row reduce the matrix to triangular form. You can find the eigenvalues in, what 3 operations.
 
I was never actually taught HOW to compute the eigenvalue of a matrix other than solving the system of equations. How does one find the characteristic polynomial of a given matrix/transformation?
 
Last edited:
Treadstone 71 said:
I was never actually taught HOW to compute the eigenvalue of a matrix other than solving the system of equations. How does one find the characteristic polynomial of a given matrix/transformation?

What's wrong with that method? It certainly works here. (Yes, the characteristic polynomial is cubic but one of the roots is "trivial".)
 
matt grime said:
Just row reduce the matrix to triangular form. You can find the eigenvalues in, what 3 operations.

Matt, could you explain that more? In general, the eigenvalues of a reduce matrix are not the same as the eigenvalues of the original matrix.
 
Treadstone 71 said:
I was never actually taught HOW to compute the eigenvalue of a matrix other than solving the system of equations. How does one find the characteristic polynomial of a given matrix/transformation?

Do you want to find the characteristic polynomial just to find the eigenvalues? This isn't necessary, each eigenvalue will be a root of the minimal polynomial, so finding this is enough to give all the eigenvalues. Even a polynomial with p(T)=0 will help, you know the minimal polynomial then divides p, so all eigenvalues will be roots of p.
 
HallsofIvy said:
What's wrong with that method? It certainly works here. (Yes, the characteristic polynomial is cubic but one of the roots is "trivial".)

How did you find the characteristic polynomial so fast?
 
  • #10
Sorry, you can find the determinant, is what I meant to say, so there is no reason not to resort to determinental arguments.
 

Similar threads

  • · Replies 17 ·
Replies
17
Views
3K
  • · Replies 19 ·
Replies
19
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
7
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
Replies
1
Views
1K