Linear Algebra, Introduction to Similarity

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.

{R e p}_{B , D} ( h ) = ( \begin{matrix}\text{𝐼𝑑𝑒𝑛𝑡𝑖𝑡𝑦} & \text{𝑍𝑒𝑟𝑜} \\ \text{𝑍𝑒𝑟𝑜} & \text{𝑍𝑒𝑟𝑜}\end{matrix} )

This representation describes
the map as sending
{c}_{1} {\overset{→}{β}}_{1} + … + {c}_{n} {\overset{→}{β}}_{n} to
{c}_{1} {\overset{→}{δ}}_{1} + … + {c}_{k} {\overset{→}{δ}}_{k} + \overset{→}{0} + … + \overset{→}{0},
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 {R e p}_{B , B} ( t ) 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.

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