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Diagonalizable Matrix

  • Thread starter rjw5002
  • Start date
  • #1
rjw5002
1. 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.

2. Homework Equations

3. 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.
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,258
618
There is a third linearly independent eigenvector, you are just overlooking it. Hint: try (0,1,0). Then do exactly what you propose.
 
  • #3
159
0
Then A^n would just be (S^-1)(D^n)(S)
Wouldn't An = SDnS-1?
 
  • #4
Dick
Science Advisor
Homework Helper
26,258
618
Yep. Sorry, I overlooked that.
 

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