Show that $M=\{x\in\mathbb{R}^n; f_{1}(x)=...=f_{n-m}(x)=0\}$ 
Let $\mathbb{R}^n$ be an n dimensional space and $M\subset\mathbb{R}^n$ an $m$-dimensional subspace. Prove that there exist linear functionals $f_1,\ldots,f_{n-m}\in(\mathbb{R}^n)^*$ such that $$M=\{x\in\mathbb{R}^n; f_{1}(x)=...=f_{n-m}(x)=0\}$$

Define $F:\mathbb{R}^n\to M$ as $F(x)=(f_{1}(x),...,f_{m}(x))$. Let $\{\pi_{1},...,\pi_n\}$ basis for the dual space of $\mathbb{R}^n$. Now take $f_{i}(x)=\pi_{i}(F(x))$. Also by dimension theorem, $\dim(Im F)\leq\dim(M)=m<n$, so $k\geq \dim(Ker F)\geq 1$ where $k$ is st $m+k=n$, i.e, $k=n-m$. Then $$\ker(F)=\bigcap_{k=1}^{n-m}{f_k}\neq\{0\}$$
Therefore, the exist some $x\neq0$such that $x\in\bigcap_{k=1}^{n-m}{f_k}$, i.e., $f_{1}(x)=...=f_{n-m}(x)=0$
 A: You could also try the following:
First observe that every linear functional $f$ in the dual space $(\mathbb{R}^n)^*$ can be characterized as $f(x) = c^\top x$ for some $c \in \mathbb{R}^n$. Then prove that $M \oplus M^\perp = \mathbb{R}^n$ for
$$ M^\perp := \lbrace x \in \mathbb{R}^n: x^\top y = 0~\forall y \in M \rbrace.$$
$M \cap M^\perp = \lbrace 0 \rbrace$ is almost automatic.
Show $M+M^\perp = \mathbb{R}^n$ (for us it actually suffices to prove $\dim M^\perp = n-m$) by choosing a basis $\lbrace \mu_1, …, \mu_m \rbrace \subseteq M$ and taking a look at $\Gamma: \mathbb{R}^n \rightarrow \mathbb{R}^m$,
$$ \Gamma(x) := (x^\top \mu_1, …, x^\top \mu_m)^\top.$$
Then apply the Rank-Nullity Theorem. 
Afterwards, choose $c_i \in \mathbb{R}^n$ so that the $f_i(x) = c_i^\top x$ serve your purpose.
(This basically means that $\mathrm{span} \lbrace f_i \rbrace_{i = 1}^{n-m}$ and $M^\perp$ are isomorphic.)
A: Your proof does not make sense as written. One proof of the result can be written by beginning with the following:
Hint: Let $\{x_1,\dots,x_m\}$ be a basis for $M$. Extend this to a basis $\mathcal B = \{x_1,\dots,x_n\}$ of $\Bbb R^n$, and let $\{f_{n},f_{n-1},\dots, f_1\}$ denote the dual basis associated with $\mathcal B$.
