# find the annihilator of a subspace

Let $W$ be the subspace of $\mathbb{R}^4$ spanned by $w_1 = (1,0,-1,2)$ and $w_2 = (2,3,1,1)$. Which linear functions $l(x) = c_1x_1 + c_2x_2 + c_3x_3 + c_4x_4$ are in the annihilator of $W$? (specify $W^\perp$)

So far I have that if $x \in W^\perp$, then $l(x) = 0$ for all $x \in W^\perp$

So, since any vector $x \in W$ can be written as $k_1w_1 + k_2w_2$, $k_i$ in field $K$, then

$x_1 = k_1 + 2k_2$
$x_2 = 3k_2$
$x_3 = k_2 - k_1$
$x_4 = 2k_1 + k_2$, and

$l(x) = c_1(k_1 + 2k_2) + c_2(3k_2) + c_3(k_2 - k_1) + c_4(2k_1 + k_2) = 0$

I'm still confused about how to approach this, not sure how to use the span of W to specify the annihilator....

Any help greatly appreciated...

$$\begin{bmatrix} 1 & 0 & -1 & 2 \\ 2& 3 & 1 & 1 \\ \end{bmatrix}\begin{bmatrix} c_1 \\ c_2 \\ c_3 \\ c_4 \\ \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \\ \end{bmatrix}$$
That means $(1, 0, -1, 2)\cdot (c_1,c_2,c_3,c_4)=(0,0,0,0)$ and $(2, 3, 1, 1)\cdot (c_1,c_2,c_3,c_4)=(0,0,0,0)$ and therefore for every linear combination is equal to $(0,0,0,0)$. Of course there is more than $(c_1,c_2,c_3,c_4)$, and given one $(c_1,c_2,c_3,c_4)$ the function is $$l(x,y,z,w)=(x,y,z,w)\cdot (c_1,c_2,c_3,c_4)$$