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Let V be a linear space of dimension n, and let K be a subspace of V of dimension k.

Prove that K 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?

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    Just look at $\Bbb R^n$ and the projections to the $k+1, \dots , n$th component. By the right choice of a base of $V$ you then can extend this to $V$. – 2017-01-31
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    That's how I started, but couldn't get further.. – 2017-01-31
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    Well, if you intersect the kernels of those projections you get the subspace of $\Bbb R^n$ spanned by $\{e_1, \dots, e_k\}$. Choose a base $\{x_1,\dots , x_k \}$ of $K$ and then, by extending the basis, you find an isomorphism $f\colon \Bbb R^n \to V$ with $f(e_i)=x_i$ for $i=1,\dots,k$. Then $p_i\circ f$ for $i= k+1 ,\dots, n$ are one choice of the maps you are looking for. – 2017-01-31

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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...

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    I understand the intuition, but can't see how to imply it in my problem. – 2017-01-31
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    @Itay4 What about now? If you get the idea in $\mathbb{R}^3$, it's basically finished in higher dimensions. Are you aware of Gram-Schmidt? – 2017-01-31
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    I'm afraid not.. – 2017-01-31
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    Gram-Schmidt tells us we can make any basis spawning a subspace $W$ into a orthonormal basis. Did you understand the idea in two dimensions, i.e. $\mathcal{P} = \text{ker}(\langle \cdot, n \rangle)$ ? – 2017-01-31