^(options)

Similarity (pdf)

While studying matrix equivalence, we have shown that for any homomorphism there are bases $B$ and $D$ such that the representation matrix has a block partial-identity form.

$\begin{equation*} \rep{h}{B,D} = \begin{pmat}{c|c} \text{\textit{Identity}} &\text{\textit{Zero}} \\ \hline \text{\textit{Zero}} &\text{\textit{Zero}} \end{pmat} \end{equation*}$

This representation describes the map as sending $c_1\vec{\beta}_1+\dots+c_n\vec{\beta}_n$ to $c_1\vec{\delta}_1+\dots+c_k\vec{\delta}_k+\zero+\dots+\zero$ , where

$n$ is the dimension of the domain and $k$ is the dimension of the

range. So, under this representation the action of the map is easy to understand because most of the matrix entries are zero.

This chapter considers the special case where the domain and the codomain are equal, that is, where the homomorphism is a transformation. In this case we naturally ask to find a single basis $B$ so that $\rep{t}{B,B}$ is as simple as possible (we will take ‘simple’ to mean that it has many zeroes). A matrix having the above block partial-identity form is not always possible here. But we will develop a form that comes close, a representation that is nearly diagonal.

« | Table of Contents | »

CHAPTER 5

Similarity (pdf)