## Annihilation By a Set of Linear Functionals

Let
$V$
be a vector space over a field
$F$
. The annihilator of a subspace
$W$
of
$V$
is the set
$W^0$
of all linear functionals
$f: V \rightarrow F$
such that
$f( \alpha)=0$
for all
$\alpha \in W$
.
Obviously the zero function is in
$W^0$
.
Also, if
$\alpha , \; \beta \in W^0, \; a, \; b \in F$
then
$f( a \alpha + b \beta ) = a f( \alpha ) + b f( \beta )=0 \rightarrow a \alpha + b \beta \in W^0$

$W^0$
is closed under addition and scalar multiplication, so the set
$W^0$
is a subspace of
$V^*$
, the space dual to
$V$
.
To find the annihilator for
$f_1(x_1, \; x_2, \; x_3, \; x_4)=x_1+2x_2+2x_3+x_4$

$f_2(x_1, \; x_2, \; x_3, \; x_4)=2x_2+x_4$

$f_1(x_1, \; x_2, \; x_3, \; x_4)=-2x_1-4x_3+3x_4$

We solve the simultaneous equations
$x_1+2x_2+2x_3+x_4=0$
(1)
$2x_2+x_4=0$
(2)
$-2x_1-4x_3+3x_4=0$
(3)
(1)-(2) gives the system
$x_1+2x_3+=0$
(4)
$2x_2+x_4=0$
(5)
$-2x_1-4x_3+3x_4=0$
(6)
(6)+2(1) gives the system
$x_1+2x_3+=0$
(7)
$2x_2+x_4=0$
(8)
$3x_4=0$
(9)
Hence
$x_4=0$
then from (5)
$x_2=0$
. In (1) put
$x_3=s$
then
$x_1=-2s$
so that vectors of the form
$\begin{pmatrix}-2s\\0\\s\\0\end{pmatrix}$
are annihilated by the functionals above.