All basis holds same number of elements

Well, the proof is easy. We will prove it by contradiction.

Given statement: "All bases for a vector space V, the number of elements in each basis is unique."

Note: I take this statement as "Given a finitely generated vector space V, then all the bases of the vector space have the same number of elements". This is because the second part of the statement does not make really some sense.

So, the given statement to prove is, "All bases in a finitely generated vector space V, have the same number of elements".

Let us assume that the statement is false. Therefore, consider two bases of a vector space V, say, $\left\lbrace v_1, v_2, ..., v_n \right\rbrace$ and $\left\lbrace w_1, w_2, ..., w_m \right\rbrace$ where, $m \neq n$.

We will exploit the fact that all the vectors in a basis for a vector space are linearly independent and also span the vector space. Also, we have a result that if for a vector space with n - vectors in the basis, any set of m vectors such that $m > n$ is linearly dependent.

So, we have two sets: $S_1 = \left\lbrace v_1, v_2, ..., v_n \right\rbrace$ and $S_2 = \left\lbrace w_1, w_2, ..., w_m \right\rbrace$ in which we claim that $m \neq n$ and both are bases for the vector space.

First, consider set $S_1$ to be the basis for the vector space and let $S_2$ be a set of m - linearly independent vectors (since we claim that they are basis vectors, they must be linearly independent). Therefore, we know that $m \leq n$.

Similarly, if we take the set $S_2$ to be a set of basis vectors and the set $S_1$ to be a set of n - linearly independent vectors, then we have $n \leq m$.

Thus, from these two statements, we have $n \leq m$ and $m \leq n$. This is true only if $n = m$.

Thus, all the bases of the vector space have the same number of elements.

For finite-dimensional spaces, one can argue as follows: Let $(v_1, \ldots, v_n)$ be a basis of $V$. Then we can write the identity map on $V$ as $\operatorname{Id} = v_1^* \otimes v_1 + \cdots + v_n^* \otimes v_n$, so $$ \operatorname{tr}(\operatorname{Id}) = n, $$ which is the number of elements in the basis. But the trace of the identity map (or any map) is independent of the basis chosen, so all bases have the same number of elements.