Proof $dim(W)=dim(V)=n \implies W=V$ I want to prove the fact that
If $W\subseteq V$ is a vector subspace of $V$ than 
$dim(W)=dim(V)=n \implies W=V$ 
Without using the fact that if $dim(V)=n$ any set of $n$ indipendent vectors in $V$ is necessarily a basis of $V$, neither the fact that a basis of $\mathscr{L}=(\vec{v_1},...,\vec{v_n})$ is necessarily the set $\big\{ \vec{v_1},...,\vec{v_n} \big\}$. (Since these two facts are consequences of the proof).
By hypotesis I have that $W\subseteq V$ which means that all the vectors in $W$ are in $V$.
I must prove the opposite inclusion $V\subseteq W$: all the vectors in $V$ are a linear combination of $n$ vector of a basis, but I don't know how to prove that all the vectors of that basis are contained in $W$.
Can I say that they must belong to $W$ since $W$ has $n$ vectors in its basis too? (But I think in this way I'm assuming the facts listed above).
Can you help me to finish the proof? 
Thanks for your help
 A: To prove that $V \subset W$, use the fact that $\dim(W)=n$ to choose a set of $n$ independent vectors in $W$, say $\{\vec w_1,…,\vec w_n\}$. That is also a set of $n$ independent vectors in $V$, since $W \subset V$. Therefore, since $\dim(V)=n$, every vector in $V$ is a linear combination of $\{\vec w_1,…,\vec w_n\}$, and is therefore a vector in $W$.
A: You should use this fundamental lemma:

Let $V$ a vector space. Suppose that $v_1 , \dots , v_n \in V$ are linearly independent and span $V$, and let $w_1 , \dots , w_m \in V$ be vectors. If $w_1 , \dots , w_m$ are linearly independent, then $m \le n$ .

This is quite tricky and is proved by induction on $n$. I hope that you have seen it, because this is the lemma allowing you to define the dimension of a vector space.
Now, you can argue by contradiction. Suppose that $V \setminus W$ is not empty. Then there exists some $w_0 \in V \setminus W$. Pick a basis of $W$ with $n$ elements: $w_1 , \dots , w_n$. Then by the lemma the $n+1$ vectors $w_0 , w_1 , \dots , w_n$ are linearly dependent. Let
$$a_0 w_0 + \sum_{i=1}^n a_iw_i = 0$$
be a linear combination with $a_0 , \dots , a_n \in K$ not all zero.
If $a_0 = 0$, then $w_1 , \dots , w_n$ are not linearly independent: a contradiction.
If $a_0 \neq 0$, then $w_0$ is spanned by $w_1 , \dots , w_n$: this contradicts our assumption $w_0 \notin W$.
In any case you get a contradiction, so $V \setminus W$ must be empty.
