In this section, we analyse the diagonalization process in detail.
To diagonalize an n by n matrix A, we first find the eigenvalues. As a root of the characteristic polynomial p(λ) = det(A  λI), any eigenvalue λ_{i} has a multiplicity m_{i}^{A} defined by (see here for more details)
p(λ_{i}) = (λ_{i}  λ)^{mi}^{A} s(λ) for integer m_{i}^{A} > 0 and another polynomial s satisfying s(λ_{i}) ≠ 0.
Basically, m_{i}^{A} is the number of times λ_{i} appears as a root of the characteristic polynomial. This multiplicity m_{i}^{A} is called the algebraic multiplicity of the eigenvalue λ_{i}. Moreover, this fact about polynomial and the fact that the top degree term of det(A  λI) is (1)^{n}λ^{n} tell us the following.
Let λ_{1}, λ_{2}, ..., λ_{k} be all the distinct eigenvalues of A, with algebraic multiplicities m_{1}^{A}, m_{2}^{A}, ..., m_{k}^{A}. Then
det(A  λI) = (λ_{1}  λ)^{m1A} (λ_{2}  λ)^{m2A} ... (λ_{k}  λ)^{mk}^{A}.
In particular, m_{1}^{A} + m_{2}^{A} + ... + m_{k}^{A} = n.
In the second step of the diagonalization process, we compute a basis B_{i} = {v_{i1}, v_{i2}, ..., v_{imiG}} for the eigenspace E_{λi} = nul(A  λ_{i}I) of each eigenvalue λ_{i}. The number of vectors in B_{i} is m_{i}^{G} = dimE_{λi}, called the geometrical multiplicity of the eigenvalue λ_{i}.
Example Based on the computations in this example, this example, and this example, we have the following algebraic and geometric multiplicities.
matrix  characteristic polynomial 
eigenvalue  algebraic multiplicity 
eigenvector  geometric multiplicity 


(5  λ)(15  λ)  λ_{1} = 5  m_{1}^{A} = 1  (1,2)  m_{1}^{G} = 1  
λ_{2} = 15  m_{2}^{A} = 1  (2, 1)  m_{2}^{G} = 1  

(4  λ)^{2}( 2  λ)  λ_{1} = 4  m_{1}^{A} = 2  (1, 1, 0), (1, 0, 1)  m_{1}^{G} = 2  
λ_{2} = 2  m_{2}^{A} = 1  (1, 1, 2)  m_{2}^{G} = 1  

(2  λ)^{2}(1  λ)  λ_{1} = 2  m_{1}^{A} = 2  (1, 0, 0)  m_{1}^{G} = 1  
λ_{2} = 1  m_{2}^{A} = 1  (6, 3, 1)  m_{2}^{G} = 1 
We observe that the algebraic multiplicity is always no smaller than the geometric multiplicity: m_{i}^{A} ≥ m_{i}^{G}. Moreover, the first two matrices are diagonalizable, for which we have m_{i}^{A} = m_{i}^{G}. However, the third matrix is not diagonalizable, for which we see the equality does not always hold. The observation will be generalized in the next part.