TITLE: Two vector spaces with same dimension and same basis, are identical?
QUESTION [7 upvotes]: Let $V$ subspace of $W$ and both have same dimension and same basis. Then can we safely say that $V= W$ ? 
I believe yes. For example there may be an element $x \in V$ written as a linear combination of the basis elements. This linear combination is unique. 
Now let's take another vector $y \in W$. The $y$ is written as a linear combination of the basis vectors. 
We equate the two linear combinations, and since the basis elements are linearly independent, we get that $x=y$. 
Thus $V=W$. 
Do you agree?

REPLY [2 votes]: Here is another proof. We know that $V$ is a subspace of $W$ and $\dim(V) = \dim(W)$. Now suppose that V is a proper subset of W. We know that V must have a complement that is a subspace of $W$. Let that complement be $U$. We have that:
$$U + V = W\\U\cap V = \varnothing$$ 
By Greenman's theorem, we have: 
$$\dim(U+V) = \dim(U) + \dim(V) - \dim(U\cap V)$$ which, using the relations above simplifies to:
$$\dim(W) = \dim(U) + \dim(V) $$
Now, it is obvious that $\dim(U) \ge 1$: Suppose $\dim(V) = 0$, then $V = \varnothing$. But this is impossible since $U + \varnothing = U $, which contradicts the fact that $U$ is a complement of $V$ in $W$. 
It follows immediately that $\dim(V) < \dim(W)$. This is a contradiction. Therefore $V$ is not a proper subset of $W$, in other words:
$$V = W$$