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diagonalization (Definition)

Let $V$ be a finite-dimensional linear space over a field $K$, and $T:V\rightarrow V$ 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

\begin{displaymath}E_\lambda = \{ u\in V: Tu = \lambda u\}.\end{displaymath}

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
\begin{displaymath}\dim V = \sum_\lambda \dim E_\lambda,\end{displaymath}

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 of $T$ relative to some basis $B$. Let
\begin{displaymath}P=[v_1,\ldots,v_n],\quad n=\dim V,\end{displaymath}

be a matrix whose column vectors are eigenvectors expressed relative to $B$. Thus,
\begin{displaymath}M v_i = \lambda_i v_i,\quad i=1,\ldots,n\end{displaymath}

where $\lambda_i$ is the eigenvalue associated to $v_i$. The above $n$ equations are more succinctly as the matrix equation
\begin{displaymath}MP = PD,\end{displaymath}

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
\begin{displaymath} M=PDP^{-1}. \end{displaymath} (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.



"diagonalization" is owned by rmilson.
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See Also: eigenvector, diagonal matrix

Also defines:  diagonalise, diagonalize, diagonalisation, diagonalization

Attachments:
example of non-diagonalizable matrices (Example) by cvalente
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Cross-references: Jordan canonical form, Jordan blocks, identity, multiple, plus, nilpotent transformation, invariant subspace, strictly, kernel, factor, characteristic polynomial, invertible, diagonal matrix, equations, column vectors, term, sum, eigenspace, eigenvalue, subspace, necessary and sufficient, diagonal, matrix, reflects, diagonalizable, transformation, eigenvectors, basis, linear transformation, field, linear space, finite-dimensional
There are 15 references to this entry.

This is version 13 of diagonalization, born on 2002-02-14, modified 2006-07-26.
Object id is 1960, canonical name is DiagonalizationLinearAlgebra.
Accessed 20733 times total.

Classification:
AMS MSC: 15-00 (Linear and multilinear algebra; matrix theory :: General reference works )
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