## A Basis for the Annihilator of a Subspace

Let
$W$
be a subspace of
$\mathbb{R}^4$
spanned by
$\left\{ \begin{pmatrix}-1\\2\\-3\\4\end{pmatrix} , \begin{pmatrix}0\\1\\4\\-1\end{pmatrix} \right\}$
.
The annihilator of
$W$
is the set of linear functions that sends all elements of
$W$
to zero.
$\phi ) \mathbf{w} = \mathbf{0}$
for all
$\mathbf{w} \in W$
. Let
$\phi \begin{pmatrix}w_1\\w_2\\w_3\\w_4\end{pmatrix}=a_1w_1+a_2w_2+a_3w+3+a_4w_40$
then
$\phi \begin{pmatrix}-1\\2\\-3\\4\end{pmatrix}=-a_1+2a_2-3a_3+4a_4=0$
(1)
$\phi \begin{pmatrix}0\\1\\4\\-1\end{pmatrix}=a_2+4a_3-a_4=0$
(2)
(1)-2(2) gives
$\phi \begin{pmatrix}-1\\2\\-3\\4\end{pmatrix}=-a_1-11a_3+6a_4=0$
(3)
$\phi \begin{pmatrix}0\\1\\4\\-1\end{pmatrix}=a_2+4a_3-a_4=0$
(4)
There are 2 equations we 4 unknowns, so we can take 4-2=2 unknowns as arbitrary.
Let
$a_3=0, \; a_4=1$
then
$a_1=6, \; a_2=1, \; a_4=1$
and we can take
$\phi_1(x_1, \; x_2, \; x_3, \; x_4)=6x_1+x_2+x_4$
.
Let
$a_3=1, \; a_4=0$
then
$a_1=-11, \; a_2=-4, \; a_3=1$
and we can take
$\phi_2(x_1, \; x_2, \; x_3, \; x_4)=-11x_1-4x_2+x_3$
.