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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?

Kishalay Sarkar
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Itay4
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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 $V$. – user60589 Jan 31 '17 at 09:13
  • That's how I started, but couldn't get further.. – Itay4 Jan 31 '17 at 09:19
  • 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. – user60589 Jan 31 '17 at 09:29

2 Answers2

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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...

Hermès
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  • I understand the intuition, but can't see how to imply it in my problem. – Itay4 Jan 31 '17 at 09:18
  • @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? – Hermès Jan 31 '17 at 09:26
  • I'm afraid not.. – Itay4 Jan 31 '17 at 09:28
  • 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. $\mathcal{P} = \text{ker}(\langle \cdot, n \rangle)$ ? – Hermès Jan 31 '17 at 09:32
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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.

Kishalay Sarkar
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