diagonalization

Let $V$ be a finite-dimensional 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 non-trivial 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 matrix-based 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 matrix-based 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 non-trivial Jordan blocks in the Jordan canonical form of the transformation.

Title	diagonalization
Canonical name	Diagonalization
Date of creation	2013-03-22 12:19:49
Last modified on	2013-03-22 12:19:49
Owner	rmilson (146)
Last modified by	rmilson (146)
Numerical id	16
Author	rmilson (146)
Entry type	Definition
Classification	msc 15-00
Related topic	Eigenvector
Related topic	DiagonalMatrix
Defines	diagonalise
Defines	diagonalize
Defines	diagonalisation
Defines	diagonalization