diagonalization
Let $V$ be a finitedimensional linear space^{} over a field $K$, and $T:V\to V$ a linear transformation. To diagonalize $T$ is to find a basis of $V$ that consists of eigenvectors^{}. The transformation is called diagonalizable if such a basis exists. The choice of terminology reflects the fact that the matrix of a linear transformation relative to a given basis is diagonal if and only if that basis consists of eigenvectors.
Next, we give necessary and sufficient conditions for $T$ to be diagonalizable. For $\lambda \in K$ set
$${E}_{\lambda}=\{u\in V:Tu=\lambda u\}.$$ 
It isn’t hard to show that ${E}_{\lambda}$ is a subspace^{} of $V$, and that this subspace is nontrivial if and only if $\lambda $ is an eigenvalue^{} of $T$. In that case, ${E}_{\lambda}$ is called the eigenspace^{} associated to $\lambda $.
Proposition 1
A transformation is diagonalizable if and only if
$$dimV=\sum _{\lambda}dim{E}_{\lambda},$$ 
where the sum is taken over all eigenvalues of the transformation.
The Matrix Approach.
As was already mentioned, the term “diagonalize” comes from a matrixbased perspective. Let $M$ be a matrix representation^{} (http://planetmath.org/matrix) of $T$ relative to some basis $B$. Let
$$P=[{v}_{1},\mathrm{\dots},{v}_{n}],n=dimV,$$ 
be a matrix whose column vectors^{} are eigenvectors expressed relative to $B$. Thus,
$$M{v}_{i}={\lambda}_{i}{v}_{i},i=1,\mathrm{\dots},n$$ 
where ${\lambda}_{i}$ is the eigenvalue associated to ${v}_{i}$. The above $n$ equations are more succinctly as the matrix equation
$$MP=PD,$$ 
where $D$ is the diagonal matrix^{} with ${\lambda}_{i}$ in the $i$th position. Now the eigenvectors in question form a basis, if and only if $P$ is invertible^{}. In that case, we may write
$$M=PD{P}^{1}.$$  (1) 
Thus in the matrixbased approach, to “diagonalize” a matrix $M$ is to find an invertible matrix $P$ and a diagonal matrix $D$ such that equation (1) is satisfied.
Subtleties.
There are two fundamental reasons why a transformation $T$ can fail to be diagonalizable.

1.
The characteristic polynomial^{} of $T$ does not factor into linear factors over $K$.

2.
There exists an eigenvalue $\lambda $, such that the kernel of ${(T\lambda I)}^{2}$ is strictly greater than the kernel of $(T\lambda I)$. Equivalently, there exists an invariant subspace where $T$ acts as a nilpotent transformation plus some multiple^{} of the identity^{}. Such subspaces manifest as nontrivial Jordan blocks^{} in the Jordan canonical form of the transformation.
Title  diagonalization 
Canonical name  Diagonalization 
Date of creation  20130322 12:19:49 
Last modified on  20130322 12:19:49 
Owner  rmilson (146) 
Last modified by  rmilson (146) 
Numerical id  16 
Author  rmilson (146) 
Entry type  Definition 
Classification  msc 1500 
Related topic  Eigenvector 
Related topic  DiagonalMatrix 
Defines  diagonalise 
Defines  diagonalize 
Defines  diagonalisation 
Defines  diagonalization 