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Maybe this is a stupid question.

I was brought to this from the observation that an infinite dimensional vector space can have proper subspace that have the same dimension of the whole space.

But, for a finite dimensional vector space, as $\mathbb{R}^n$ it seems obvious that, if $S$ is a subspace of dimension $n$, than $S=\mathbb{R}^n$.

I've tempted to give a proof this statement, thinking that was a simple exercise, but I am no able to find a good way.
I suppose that I'm lost in a teacup so I need an help.


Added after the first answers.

The answer of @Mark Joshi was my first idea, so I llustrate better where is my trouble.

Suppose $S \in \mathbb{R}^n$ be a subspace of dimension $n$. This means, by definition, that:

1) there exists a set of linearly independent vectors $\{v_1, v_2, \cdots, v_n\}$ with $v_i \in S$

and

2) $\forall v \in S$ we have $v \ne v_i \; \forall i \Rightarrow \{v, v_1,\cdots,v_n\} $ are linearly dependent.

Note that 2) is valid for $v\in S$ and all $v_i \in S$.

Since $\mathbb{R}^n$ is a vector space, we know that there exists a set $\{u_1,u_2,\cdots,u_n\}$ of vectors $u_i \in \mathbb{R}^n $ such that:

3) $\{u_1,u_2,\cdots,u_n\}$ are linearly independent

and

4) $\forall u \in \mathbb{R}^n$ we have $u \ne u_i \; \forall i \Rightarrow \{u, u_1,\cdots,u_n\} $ are linearly dependent.

Now, if $u, u_i \in \mathbb{R}^n/S$ I can not conclude from 1) and 2) that $u$ is a linear combination of $\{v_i\}$ and this means that I cannot decide if it is in the span of $ \{v, v_1,\cdots,v_n\} $. Surely it is a linear combination of $\{u_1,u_2,\cdots,u_n\}$ , but some (or all) of these $u_i$ can well not be elements of $S$.

So it seams that we can have a vector $u$ that is a linear combination of $n$ linearly independent $\{u_i\}$ but we connot proof that it is also a linear combination of $\{v_i\}$. And this is my trouble.

It seems to me that the way suggested by @Awllower has the some problem when we search to show that $f :S \rightarrow \mathbb{R}^n$ defined by the correspondence $v_i \rightarrow u_i$ is bijective, because we have to proof that $\forall u \in \mathbb{R}^n$ there exists $v \in S$ such that $v=\sum_i a_i v_i \Rightarrow u=\sum_i a_i u_i$ and this come to the same trouble.

At last, I am sure that there is something of logically wrong in my reasoning, but it is so subtle or so stupid that i cannot see it.

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    $\begingroup$ The important question underlying what you are asking is, "Is the dimension of a finite dimensional vector space a well defined number?" That is, are different collections of independent spanning sets the same size? Any linear algebra book should answer this when they introduce dimension. $\endgroup$ – Aaron Feb 26 '15 at 13:01
  • $\begingroup$ I agree, and this means that $v_i$ are a basis in $\mathbb{R}^n$, but why this implies that $\mathbb{R}^n/S = \emptyset$ ? $\endgroup$ – Emilio Novati Feb 26 '15 at 14:44
  • $\begingroup$ You may find it helpful to establish a lemma: If $B$ is a linearly independent subset of a vector space $V$, and if $v \in V\setminus B$, then: $B \cup \{v\}$ is linearly independent if and only if $v$ is not in the span of $B$. $\endgroup$ – Andrew D. Hwang Feb 26 '15 at 18:01
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suppose we have a subspace of dimension then it has a basis of $n$ linearly independent vectors $v_1, v_2, \dots, v_n.$ If we have a vector $v \in R^n$ either it is in their span or it is not. if it is not then we have $n+1$ linearly independent vectors which is impossible since the dimension of $R^n$ is $n.$

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Hint:
Consider the linear map $L$ that embedds the subspace $S$ in $V.$ What is the rank of the matrix representation of $L?$ And what does this mean?

Another approach:
Suppose the dimension of the vector-space has dimension $n.$ Then there are $n$ linearly independent vectors $\{v_1,\cdots, v_n\}$ in $V.$
Since $\text{dim}S=\text{dim}V=n,$ there are $n$ linearly independent vectors $\{w_1,\cdots, w_n\}$ in $S.$
Then define a linear map $f: S\rightarrow V$ by sending $w_i$ to $v_i, \forall i=1, \cdots, n.$ All that is left to do is verify that $f$ is an isomorphism.
Hope this helps.

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