Basic Linear Algebra - Computer Graphics and Geometric Modeling - page 873

This equation and the definition of the trace function implies (1). To prove (2), simply
substitute 0 for t in (C.6). Part (3) follows immediately from (1) and (2).

Here is a fundamental theorem about the diagonalizability of transformations.

C.4.10. Theorem. Let T : V Æ V be a linear transformation. Assume that
l 1 , l 2 ,..., l m are the distinct eigenvalues of T and let d 1 , d 2 ,..., d m be the dimen-
sions of the corresponding eigenspaces. Let p(t) be the characteristic polynomial of
T. Then T is diagonalizable if and only if

d d d m

() =- ( )

1

( )

2

( )

pt t l t - l ◊◊◊ t - l .

1 2 m

Proof. This follows easily from Theorem C.4.2 and Theorem C.4.7.

Much more can be said about the diagonalizability of transformations. For
example, see [HofK71].

C.5 The Dual Space

Given vector spaces V and W over a field k, let

( ) =

{ }

L VW , T : V Æ W T is a linear transformation .

If S, T ΠL( V , W ) and a Πk, then define

STaS + , : VW Æ

by

( )( ) = () + ()

ST + v S v T v

( () = ( () )

aS v a S v .

C.5.1. Theorem. The maps S + T and aS are linear transformation and this addi-
tion and scalar multiplication make L( V , W ) into a vector space over k.

Proof. Easy.

Definition. Let V be a vector space over a field k. A linear transformation T : V Æ k
is called a linear functional on V . The vector space of linear functionals on V is called
the dual space of V and is denoted by V *.

Let V be a vector space over a field k and let v 1 , v 2 ,..., v n be a basis for V . Define
linear functionals

i * :

vVR Æ

by