Diagonalizable Matrix Homework: Compute A^n with n as Integer

  • Thread starter Thread starter rjw5002
  • Start date Start date
  • Tags Tags
    Matrix
Click For Summary

Homework Help Overview

The problem involves determining the matrix A raised to the power of n, where A is given as a 3x3 matrix. The context is centered around the concept of diagonalization and the computation of powers of matrices.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to establish whether the matrix A is diagonalizable by finding its eigenvalues and eigenvectors. They express confusion regarding the number of linearly independent eigenvectors needed for diagonalization.

Discussion Status

Participants are actively engaging with the original poster's concerns. One participant suggests a potential third eigenvector, while another clarifies the relationship between the matrices involved in the diagonalization process. There is a recognition of oversight in the original poster's understanding.

Contextual Notes

The original poster is working under the assumption that three linearly independent eigenvectors are necessary for diagonalization, and they are exploring the implications of this requirement on their computations.

rjw5002

Homework Statement


Let A =
[4 0 1;
2 3 2;
1 0 4]
Let n >= 1 be an integer. Compute the matrix A^n with entries depending on n.

Homework Equations



The Attempt at a Solution


First I need to show that A is diagonalizable, and find a matrix S such that D = (S^-1)(A)(S) is diagonal. I am having serious trouble finding S. First I found the eigenvalues of A to be 3,3, and 5. Next I found the eigenvectors to be for [tex]\lambda = 3[/tex]: (1 0 -1) and [tex]\lambda = 5[/tex] : (1 2 1). There's a theorem in my book that says that a matrix is only diagonalizable if and only if there are n linearly independent eigenvectors. I only have 2. I'm not quite sure how to get the third.
Now, assuming I had three eigenvectors, I would combine them to form the matrix S, and if the [tex]det(S)\neq 0[/tex] then the eigenvectors would be linearly independent, and
i could find S^-1. Then A^n would just be (S^-1)(D^n)(S). Does this all sound correct? I guess I'm just a little confused about the eigenvalue/eigenvector situation...
Thanks for any help.
 
Physics news on Phys.org
There is a third linearly independent eigenvector, you are just overlooking it. Hint: try (0,1,0). Then do exactly what you propose.
 
Then A^n would just be (S^-1)(D^n)(S)

Wouldn't An = SDnS-1?
 
Yep. Sorry, I overlooked that.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
9
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
2
Views
2K
Replies
8
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K