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

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In summary, to calculate the matrix A^n with entries depending on n, we first need to show that A is diagonalizable and find a matrix S such that D = (S^-1)(A)(S) is diagonal. To find S, we can use the eigenvalues and eigenvectors of A, and if there are n linearly independent eigenvectors, then A^n would just be (S^-1)(D^n)(S). It is important to not overlook any potential eigenvectors.
  • #1
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.
 
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  • #2
There is a third linearly independent eigenvector, you are just overlooking it. Hint: try (0,1,0). Then do exactly what you propose.
 
  • #3
Then A^n would just be (S^-1)(D^n)(S)

Wouldn't An = SDnS-1?
 
  • #4
Yep. Sorry, I overlooked that.
 

1. What is a diagonalizable matrix?

A diagonalizable matrix is a square matrix that can be transformed into a diagonal matrix through a similarity transformation. This means that the matrix can be written as A = PDP^-1, where P is an invertible matrix and D is a diagonal matrix.

2. How do you determine if a matrix is diagonalizable?

A matrix is diagonalizable if and only if it has n distinct eigenvalues, where n is the size of the matrix. This means that the matrix must have n linearly independent eigenvectors.

3. What is the significance of diagonalizable matrices?

Diagonalizable matrices have many applications in mathematics and science. They can be used to simplify calculations, solve systems of linear equations, and model real-world phenomena. They also have important properties that make them easier to work with compared to non-diagonalizable matrices.

4. How do you compute A^n with n as an integer?

To compute A^n, first diagonalize the matrix A by finding its eigenvalues and eigenvectors. Then, use the formula A^n = P D^n P^-1, where D^n is the diagonal matrix with the eigenvalues raised to the nth power. This formula can also be written as A^n = c1^n v1 + c2^n v2 + ... + cn^n vn, where c1, c2, ..., cn are the eigenvalues and v1, v2, ..., vn are the corresponding eigenvectors.

5. Can a matrix be diagonalizable if it has repeated eigenvalues?

Yes, a matrix can still be diagonalizable even if it has repeated eigenvalues. However, the matrix must have n linearly independent eigenvectors, where n is the size of the matrix. This means that the matrix must have distinct eigenvectors for each repeated eigenvalue.

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