1. ## Diagonalization

A is a square matrix and from its characteristic polynomial I am to figure out if is diagonalizable over all real numbers, not diagonalizable over all real numbers or if it's impossible to say.

x = lambda

f(x)= (x-3)^2 (x-5)

I know that the eigenvalues are 3 (with multiplicity 2) and 5 (with multiplicity 1)

I'm not sure what to do from here.

Another example is f(x)= (x^2-1)(x^2-2)
With this one i assume that there are no eigenvalues in the lR of A, so A would not be diagonalizable, but im not sure if it is correct

2. Originally Posted by Arita
A is a square matrix and from its characteristic polynomial I am to figure out if is diagonalizable over all real numbers, not diagonalizable over all real numbers or if it's impossible to say.

x = lambda

f(x)= (x-3)^2 (x-5)

I know that the eigenvalues are 3 (with multiplicity 2) and 5 (with multiplicity 1)

I'm not sure what to do from here.

Another example is f(x)= (x^2-1)(x^2-2)
With this one i assume that there are no eigenvalues in the lR of A, so A would not be diagonalizable, but im not sure if it is correct
Your square matrix is diagonliziable if you have an eigenvector for each eigenvalue. You will need 3 eigenvectors for your first case.

$x^2=1\Rightarrow x=\pm 1, \ \ x^2=2\Rightarrow x=\pm\sqrt{2}$

3. I know that there needs to be a eigenvector for each eigenvalue, but when your only given the characteristic polynomial how are you suppose to know?

4. Originally Posted by Arita
I know that there needs to be a eigenvector for each eigenvalue, but when your only given the characteristic polynomial how are you suppose to know?
You would need to know the geometric multiplicity, i.e. the dimension of the eigenspace.

5. So for the first example the multiplicity would be 2 for the eigenvalue 3 and 1 for the eigenvalue 5. Since there would be 3 eigenvectors, does that mine that the square matrix is diagonalizable?