\[V\]
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.