Page 153 - DMTH502_LINEAR_ALGEBRA

Unit 11: The Double Dual

Definition: If V is a vector space, a hyperspace in V is a maximal proper subspace of V. Notes
Theorem 3. If f is a non-zero linear functional on the vector space V, then the null space of f is a
hyperspace in V. Conversely, every hyperspace in V is the null space of a (not unique) non-zero
linear functional on V.
Proof: Let f be a non-zero linear functional on V and N its null space. Let be a vector in V which
f
is not in N , i.e., a vector such that f 0. We shall show that every vector in V is in the
f
subspace spanned by N and . That subspace consists of all vectors
f
c , in N f , in .
c
F
Let be in V. Define
f
c
f
which makes sense because f 0. Then the vector c is in N since
f
f f c

f cf

0. ...(7)

So is in the subspace spanned by N and .
f
Now let N be a hyperspace in V. Fix some vector which is not in N. Since N is a maximal proper
subspace, the subspace spanned by N and is the entire space V. Therefore each vector in V
has the form

N
c
c , in , in .
F
The vector and the scalar c are uniquely determined by . If we have also
N
' c ' , ' in , ' in . ...(8)
F
c
c
then ( ' c ) '
If 'c c 0, then would be in N; hence, 'c c and ' = . Another way to phrase our conclusion is
this: If is in V, there is a unique scalar c such that c is in N. Call that scalar g . It is easy
to see that g is a linear functional on V and that N is the null space of g.
Lemma: If f and g are linear functionals on a vector space V, then g is a scalar multiple of f if and
only if the null space of g contains the null space of f, that is, if and only if f 0 implies g 0.

Proof: If f = 0 then g = 0 as well and g is trivially a scalar multiple of f. Suppose f 0 so that the

null space N is a hyperspace in V. Choose some vector in V with f 0 and let
f
g
c . ...(9)
f

The linear functional h g cf is 0 on N since both f and g are 0 there, and h g cf 0.
f

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