# Prove that a subspace of dimension $n$ of a vector space of dimension $n$ is the whole space.

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.

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.

• 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. Feb 26, 2015 at 13:01
• I agree, and this means that $v_i$ are a basis in $\mathbb{R}^n$, but why this implies that $\mathbb{R}^n/S = \emptyset$ ? Feb 26, 2015 at 14:44
• 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 $B$. Feb 26, 2015 at 18:01

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 $n.$
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?
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.