Hi
Given a matrix . it's eigenvalues are
I know that if i found two eigenvectors with and one eigenvector with is diagonlizable.
But what happend when i get just one eigenvector with but two with and the three are linearly independent?
is A diagonalizable?
thanks very much
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
If your characteristic equation gives as a "double" eigenvalue and as a "single" eigenvalue, you can't have two independent eigenvalues corresponding to . 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 is a double eigenvalue ("algebraic multiplicity" two) with only one independent eigenvector ("geometric multiplicty one), then the matrix is not diagonalizable.