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
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)
(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
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.