Matrix diagonalization & eigenvectors

viko · #1 $Old$ October 6th, 2009, 10:17 PM

Hi

Given a matrix $A \in \mathbb{R}^{3x3}$ . it's eigenvalues are $\lambda_{1}=1 \ \ \lambda_{2}=1 \ \ \lambda_{3}=2$

I know that if i found two eigenvectors with $\lambda_{1} = \lambda_{2}$ and one eigenvector with $\lambda_{3} \Rightarrow A$ is diagonlizable.

But what happend when i get just one eigenvector with $\lambda_{1} = \lambda_{2}$ but two with $\lambda_{3}$ and the three are linearly independent?

is A diagonalizable?

thanks very much

tonio · #2 $Old$ October 7th, 2009, 05:55 AM

Quote:

Originally Posted by viko $View Post$

Hi

Given a matrix $A \in \mathbb{R}^{3x3}$ . it's eigenvalues are $\lambda_{1}=1 \ \ \lambda_{2}=1 \ \ \lambda_{3}=2$

I know that if i found two eigenvectors with $\lambda_{1}$ and one eigenvector with $\lambda_{2} \Rightarrow A$ is diagonlizable.

== I suppose you meant that if you have two lin. indep. eigenvectors corresponding to the eigenvalue 1 then A is diagonalizable. Yes, this is true because then you've a basis of your space with all its components eigenvectors.

But what happend when i get just one eigenvector with $\lambda_{1}$ but two with $\lambda_{2}$ and the three are linearly independent?

is A diagonalizable?

== You realize that Lamba_1 = Lambda_2 and thus eigenvectors corresponding to the first one correspong also to the second one and the other way around??

Anyway: A is diagonalizable iff there are three lin. indep. eignevectors of A.

Tonio

thanks very much

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viko · #3 $Old$ October 7th, 2009, 08:22 AM

hi tonio

i made a mistake in my first post.

what i wonder is: can i make the base with only one vector from the 'double' eigenvalue and two vectors from the 'single' one?

(i'm editing my first post too)

tonio · #4 $Old$ October 7th, 2009, 11:38 AM

Quote:

Originally Posted by viko $View Post$

hi tonio

i made a mistake in my first post.

what i wonder is: can i make the base with only one vector from the 'double' eigenvalue and two vectors from the 'single' one?

(i'm editing my first post too)

Nop. According to the given info your matrix's characteristic polynomial is
(x-2)(x-1)^2 , and since the algebraic multiplicity of an eigenvalue (= the power to which the corresponding linear factor is raised in the char. pol.) is greater than or equal to its geometric multiplicity (= the dimension of the eigenspace of all eigenvalues , together with the vector zero, corresponding to that eignevalue), there can't be more than 1 lin. indep. eigenvalue corresponding to 2

Tonio

HallsofIvy · #5 $Old$ October 7th, 2009, 01:58 PM

If your characteristic equation gives $\lambda_1$ as a "double" eigenvalue and $\lambda_2$ as a "single" eigenvalue, you can't have two independent eigenvalues corresponding to $\lambda_2$ . The number of independent eigenvectors corresponding to an eigenvalue (the "geometric multiplicity") cannot be larger than the multiplicity of the eigenvalue (the "algebraic multiplicity").

If $\lambda_1$ is a double eigenvalue ("algebraic multiplicity" two) with only one independent eigenvector ("geometric multiplicty one), then the matrix is not diagonalizable.

viko · #6 $Old$ October 12th, 2009, 07:58 AM

Quote:

Originally Posted by HallsofIvy $View Post$

If your characteristic equation gives $\lambda_1$ as a "double" eigenvalue and $\lambda_2$ as a "single" eigenvalue, you can't have two independent eigenvalues corresponding to $\lambda_2$ . The number of independent eigenvectors corresponding to an eigenvalue (the "geometric multiplicity") cannot be larger than the multiplicity of the eigenvalue (the "algebraic multiplicity").

If $\lambda_1$ is a double eigenvalue ("algebraic multiplicity" two) with only one independent eigenvector ("geometric multiplicty one), then the matrix is not diagonalizable.

Thanks hallsofivy, i can't find it now, but i think i had a matrix where the the eigenvectors corresponding to a simple eigenvalue were 2, but that isn't posible, so for sure those vectors were linearly dependent.