How do you calculate the geometric multiplicities? I know that these are equal to the dimension of the kernel of $A-\lambda_i I$ for eigenvalues $\lambda_i$ of $A$, but for some reason I'm still struggling to figure this out.
Suppose we have a matrix like $\begin{pmatrix}5&0\\0&5 \end{pmatrix}$ and $\begin{pmatrix}5&1\\0&5 \end{pmatrix}$. Is there any simple way to find the geometric multiplicities of each? An explanation that helps me extend to more complicated matrices will be greatly appreciated.
 A: You know, that for a $n\times n$ matrix $A$ you have
$$\dim (\ker(A-\lambda_i I)) = n - \dim(\operatorname{im}(A-\lambda_i I)) = n - \operatorname{rank}(A-\lambda_i I)$$
Depending on the matrix $A-\lambda_i I$ it might be easy "to see" its rank (for example if $A-\lambda_i I$ is already in a row echelon form).
Example: Take $A=\left(\begin{matrix} 5 & 0 \\ 0 & 5 \end{matrix}\right)$ and $B=\left(\begin{matrix} 5 & 1 \\ 0 & 5 \end{matrix}\right)$. You see easily, that both matrices have the only eigenvalue $\lambda = 5$. While $A-\lambda I$ has rank 0, $B-\lambda I$ has rank 1. So the geometric multiplicity of $A$ for $\lambda$ is $2-0=2$ while it is for $b$ equal to $2-1=1$.
Obviously this "method" is not easy for each matrix and eigenvalue, but it is easy in some cases. I am not aware of a better method for calculating the geometric multiplicities... 
A: Are you allowed to use a calculator? I know that ti-83 has a function called ${\bf rref}$ to row reduce a matrix.  $\bf rref$ the matrix $A - \lambda I$ and count the number of leading $1$'s, called pivots, on each row. Subtract this number of pivots from the size of the matrix $A$ and that would be the geometric multiplicity of eigenvalue $\lambda$
