geometrical explanation for dimension of a subspace Let $S$ be a subspace of a $\mathbb R^n$ which has a basis of $k$ vectors. How can one geometrically justify that every other basis of $S$ has precisely $k$ vectors without using algebraic arguments like system of linear equations or induction.
I am looking for a purely geometric argument like since only $k$ specific directions are sufficient to pinpoint a vector in $S$ if more than $k$ different directions are used there will be some contradiction.
 A: I'm going to interpret "geometrically" as meaning you want to be able to draw a picture, which limits us to $n=0,1,2,3$, and $0\leq k\leq n$. So I'm going to pick $k=2=n$, by considering $S=\mathbb{R}^2$ to be the $xy$-plane. I'm going to rely on you drawing the picture, however. Also, this is a long post since I can't know your level of familiarity with linear algebra.
So suppose I have three linearly independent vectors in $S$. Then I can name them $a,b,c$, and I know they cannot be zero (since the zero vector lies in every direction), and no two of them can lie on the same line. Draw any three vectors you like fitting this description.
Now I'm going to have to use equations for this part, but these are easily "artistically-renditionable". Look at $a$ and $b$, rotating the picture (or your head, but not too far!) so that $a$ points directly to your right. You may notice that $b$ is then hanging a little over/under $a$, i.e. they are not perpendicular. We can fix this by considering the vector $b'=b-a$, which on the picture will be exactly the vector perpendicular to $a$ which goes the same amount directly up or down (remember your tilted picture!) as $b$. It should be clear that $a$ and $b'$ hit all the same points as $a$ and $b$ do. 
Now we reach the problem of what to do with $c$. On your tilted picture, it should be easy to see that $c$ is reachable with some stretched/shrunk vector in the direction of $a$ added to another in the direction of $b'$. Thus, $a,b,c$ cannot be linearly independent, and since these were chosen arbitrarily, it's wont work for any three vectors in $\mathbb{R}^2$.
From here, it's up to you to try to imagine this on any 2D subspace of $\mathbb{R}^3$, and then to see how the picture looks for $\mathbb{R}^3$ with four vectors. 
A: Well, another m-basis generates an isomorphic copy of $\mathbb{R}^m$ for some $m$. Then if $m \not = k$, we have $\mathbb{R}^{k} \not \simeq \mathbb{R}^m$ (as vector-spaces). I guess from a geometric perspective: if $m<k$ then this m-basis generates a hyperplane in $\mathbb{R}^k$ i.e can't span since it generates $\mathbb{R}^m \subset \mathbb{R}^k$. And if $m>k$ then the basis with $k$-vectors only generates a $k$-dimensional hyperplane (copy of $\mathbb{R}^k$) which is just a subset of $\mathbb{R}^m$. 
A: There is a complementary space $S^{\perp}$ to $S$ in $\mathbb{R}^n$, which has dimension $n-k$. Any basis of $S$ can be extended by any basis of  $S^{\perp}$ into a basis of $\mathbb{R}^n$. There are also projection linear transformations $\pi_S,\pi_{S^\perp}$ which fix their subscripted space (and in particular its basis vectors) but map the orthogonal complement space to zero, which is therefore the kernel of each transformation. Perhaps the rank-nullity theorem can then be invoked to give the result.  
