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. Thetransformation is called diagonalizable if such a basis exists.The choice of terminology reflects the fact thatthe matrix of a linear transformation relative to a given basis is diagonalif and only if that basis consists ofeigenvectors.

Next, we give necessary and sufficient conditions for $T$ to bediagonalizable. 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 isnon-trivial if and only if $\lambda$ is an eigenvalue of $T$. In that case, $E_{\lambda }$ is called the eigenspaceassociated 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 relativeto $B$. 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 $i$-thposition. Now the eigenvectors in question form a basis, if and onlyif $P$ is invertible. In that case, we may write

$$M=PD{P}^{-1}.$$(1)

Thus in the matrix-based approach, to “diagonalize” a matrix $M$ isto find an invertible matrix $P$ and a diagonal matrix $D$ such thatequation (1) is satisfied.

Subtleties.

There are two fundamental reasons why a transformation $T$ can fail tobe diagonalizable.

1. The characteristic polynomial of $T$ does not factor into linearfactors 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 theidentity. 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