Page 144 - DMTH502_LINEAR_ALGEBRA

Linear Algebra

Notes These polynomial functions are easily shown to be
(x t )(x t )
p (x) = 2 3
1
(t t )(t t )
1 2 1 3
(x t 1 )(x t 3 )
p (x) =
2 (t 2 t 1 )(t 2 t 3 )

(x t )(x t )
p (x) = 1 2
3
(t 3 t 1 )(t 3 t 2 )
The basis {p , p , p } for V is interesting, because according to (7) we have to each p in V.
1 2 3
t
p
t
p = p ( )p 1 p ( )p 2 p ( ) .
t
3
2
1
3
Thus, if c , c and c are any real numbers, there is exactly one polynomial function p over R
1 2 3
which has degree at most 2 and satisfies ( )p t j c j , j 1,2,3. This polynomial function is
p = c p + c p + c p .
1 1 2 2 3 3
Now let us discuss the relationship between linear functionals and subspaces. If f is a non-zero
linear functional, then the rank of f is 1 because the range of f is a non-zero subspace of the scalar
field and must (therefore) be the scalar field. If the underlying space V is finite-dimensional, the
rank plus nullity theorem tells us that the null space N has dimension
f
dim N = dim V – 1.
f
In a vector space of dimension n, a subspace of dimension n – 1 is called a hyperspace. Such
spaces are sometimes called hyperplanes or subspaces of co-dimension 1. Is every hyperspace
the null space of a linear functional? The answer is easily seem to be yes. It is not much more
difficult to show that each d-dimensional subspace of an n-dimensional space is the intersection
of the null spaces of (n – d) linear functionals (Theorem below).
Definition: If V is a vector space over the field F and S is a subset of V, the annihilator of S is the
set S° of linear functionals on V such that f( ) = 0 for every in S.
It should be clear that S° is a subspace of V*, whether S is a subspace of V or not. If S is the set
consisting of the zero vector alone, then S° = V*. If S = V, then S° is the zero subspace of V*. (This
is easy to see when V is finite-dimensional.)

Theorem 3. Let V be a finite-dimensional vector space over the field F, and let W be a subspace of
V. Then
din W + dim W° = dim V.

Proof: Let k be the dimension of W and { , ..., } a basis for W. Choose vector , ..., , in V
1 k k + 1 n
such that { ,..., } is a basis for V. Let {f , ..., f } be the basis for V* which is dual to this basis for
1 n 1 n
V.
f
This claim is that { f k 1 ,... } is a basis for the annihilator W°. Certainly f belongs to W° for i k
n
i
+ 1, because
( f ) =
i i ij
and ij 0 if i k + 1 and j k; from this it follows that, for i k + 1, ( ) 0f i whenever is a
linear combination of 1 ,..., k . The functionals f k 1 ,..., f are independent, so all we must show
n
is that they span W°. Suppose f is in V*.

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