Page 142 - DMTH502_LINEAR_ALGEBRA

Linear Algebra

Notes
Let ( 1 , 2 ,... n ) be a basis for V. According to theorem 1 of unit 7, there is (for each i) a
unique linear functional f on V such that
i
f ( ) = ...(4)
i i ij
In this way we obtain from a set of n distinct linear functionals f , f , ... f on V. These functionals
1 2 n
are also linearly independent. For, suppose
n
f = c f ...(5)
i i
i 1
n
Then f( ) = c ( f )
j i i j
i 1

n
= c
i ij
i 1
= c .
j
In particular, if f is the zero functional f( ) = 0 for each j and hence the scalars c are all 0. Now
j j
f 1 ,...f are n linearly independent functionals, and since we know that V* has dimension n, it
n
must be that * { ,...,f 1 f n } is a basis for V*. This basis is called the dual basis of .
Theorem 2: Let V be a finite-dimensional vector space over the field F, and let  { 1 ,..., n } be
a basis for V. Then there is a unique dual basis  * { ,..., f n } for V* such that f i ( j ) ij . For
f
1
each linear functional f on V we have
n
f = ( f i )f i ...(6)
i 1
and for each vector in V we have
n
= f 1 ( ) i . ...(7)
i 1
Proof: We have shown above that there is a unique basis which is ‘dual’ to . If f is a linear
functional on V, then f is some linear combination (5) of the f , and as we observed after (5) the
i
scalars c must be given by c = f( ). Similarly, if
j j j
n
= x i i
i 1

is a vector in V, then
n
f j ( ) = x i i ( f i )
i 1
n
= x i ij
i 1
= x
j

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