TITLE: Geometric multiplicity of repeated Eigenvalues
QUESTION [9 upvotes]: I am still finding it difficult to determine the geometric multiplicity for repeated eigenvalues and the resultant eigenspace. For example, I am not quite sure what to do with the following matrix, where repeated Eigenvalues $\lambda_1 = \lambda_2 = 5$:
$$A=\begin{bmatrix}
5 & -4 & 0\\ 
1 & 0 & 2\\
0 & 2 & 5
\end{bmatrix}, [A-\lambda I] = \begin{bmatrix}
0 & -4 & 0\\ 
1 & -5 & 2\\
0 & 2 & 0
\end{bmatrix}$$
It is not obvious how to determine the Eigenvectors from this, as there are no free variables, and moving $e_2$ for example (for the first row) such that $-4e_2 = 0 \rightarrow e_2 = 0$ shows that all other values result to zero as well (which is not a valid eigenvector). How does one go about determining the geometric multiplicity and the Eigenspace with such a matrix?

REPLY [2 votes]: The usual procedure to find the eigenspace and GM(geometric multiplicity) for a particular eigenvector of a matrix A  is as follows:
1).Solve A's characteristic polynomial |A-lamda*I|, which yields an n-th polynomial in regard to lamda.
2)If you have obtained a group real solutions, say, lamda 1,lamda 2,lamda 3...lamda r, from the previous step, for each of the lamda s, put it back into the homogeneous equation (A-lamda*I)x= 0 in regard to x. If you find the equation has only trivial solution, then this lamda is not an eigenvalue, so you can just quit and go to the next lamda. Otherwise, this lamda is indeed an eigenvalue. Solve the equation to obtain the kernel of (A- lamdaI), whose basic vectors are just the eigenvectors corresponding to A and whose dimension is just the GM for this lamda.
And note that for any eigenvalues GM<=AM(algebraic multiplicity) always holds.