Representing a subspace as intersection of kernels Let $V$ be a linear space of dimension $n$, and let $W$ be a subspace of $V$ of dimension $k$.
Prove that $W$ can be represented as intersection of the kernels of $(n-k)$ linear functionals.
I tried to use the nullity of dual space, but got stuck.
Any ideas?
 A: Hint: in $\mathbb{R}^3$, if $\mathcal{P}$ is a plane through the origin (i.e. two dimensional subspace in a 3-dimensional space), elementary calculus yields:
\begin{align*}
(x_1, x_2, x_3) \in \mathcal{P} \iff ax_1 + bx_2 + cx_3 = 0
\end{align*}
for some $a,b,c$.
If $\mathcal{D}$ is a line through the origin (i.e 1-dimensional subspace in a 3-dimensional space), then there exists $a,b,c,d,e,f$ such that:
\begin{align*}
(x_1, x_2, x_3) \in \mathcal{D} \iff (ax_1 + bx_2 + cx_3 = 0) \, \wedge \,  (dx_1 + ex_2 + fx_3 = 0)
\end{align*}

In the first case, by choosing $\{e_1, e_2\}$a basis of $\mathcal{P}$, we can complete (non-trivial statement) it into a basis of $\mathbb{R}^3$. In fact, we can find $n$ orthonormal to our plane, such that $x$ is in $\mathcal{P}$ exactly when $x$ has no component on $n$ i.e. $\langle x ,n \rangle = 0$. Now notice $x \mapsto \langle x, n \rangle$ is a functional...
A: Let $W$ have the basis $\{x_1,x_2,...,x_k\}$,then extend it to a basis $\{x_1,...,x_n\}$ of $V$,consider the dual basis $\{f_1,...,f_n\}$.$W=span\{x_1,...,x_k\}=\{x\in V: f_i(x)=0 ,\forall i\geq k+1\}=\bigcap\limits_{i\geq k+1}N_{f_i}.$Each of $N_{f_i}$ is a non zero linear functionsl hence the kernel is a hyperspace.
