Uniqueness of basis in a Vector Space I know that any vector in a vector space can be uniquely represented as a linear combination of its basis i.e. components of a vector are unique. My question is that if my components of a vector are unique then how to prove that my basis would be unique?
My attempt to the question assuming finite n-dimensional space :
As we know that in a finite-dimensional space there can be many basis but the cardinality of all these basis has to be same ( in this case n) i.e dimensionality = n. 
Let {$ v_1, v_2, .... ,v_n $} and {$v'_1, v'_2, .... ,v'_n$} be the basis of the V.S.
v= $a_1 v_1 + a_2 v_2 +...... + a_n v_n$
v= $a_1 v'_1 + a_2 v'_2 +...... + a_n v'_n$
Now, writing {$v'_1, v'_2, .... ,v'_n$} in terms of {$ v_1, v_2, .... ,v_n $} considering latter the basis of V.S.
I got the whole bunch of equations from which I am not able to see how to prove $v_i =v'_i$ $ \forall$ i.
Thanks in advance.
 A: From the comments, let me rephrase the question:

Let $V$ be a vector space, and let $\beta=\{v_i\}_{i\in I}$ and $\beta’=\{v’_i\}_{i\in I}$ be two ordered bases (same index bset because they have the same cardinality), with the property that for every vector $v\in V$, if we express $v$ as a linear combination of vectors in $\beta$,
  $$v = \sum_{i\in I}\alpha_iv_i,\qquad \alpha_i=0\text{ for almost all }i\in I,$$
  and we do the same in terms of $\beta’$,
  $$v = \sum_{i\in I}\alpha’_iv’_i,\qquad \alpha’_i=0\text{ for almost all }i\in I$$
  then $\alpha_i=\alpha’_i$ for every $i\in I$. Then $\beta=\beta’$.

This is true. To verify this, fix $i_0\in I$, and consider how to express $v_{i_0}$ as a linear combination of $\beta$: 
$$v_{i_0} = 1v_{i_0} + \sum_{i\in I-\{i_0\}}0v_i.$$
By assumption, the way to express $v_{i_0}$ in terms of $\beta’$ is
$$v_{i_0} = 1v’_{i_0} + \sum_{i\in I-\{i_0\}}0v’_i.$$
But the latter expression is just $v’_{i_0}$. Hence, $v_{i_0}=v’_{i_0}$.
As this holds for an arbitrary $i_0\in I$, it follows that $\beta=\beta’$. 

By the way, the other interpretation, that the assertion holds for a single (nonzero) vector $v$, is false if $\dim(V)\gt 1$. To see this, take $\beta$, and take two vectors in $\beta$, $v_1,v_2$. Let $\beta’$ be the basis obtained by replacing $v_2$ with $v_1+v_2$. Then every vector in $\mathrm{span}(\beta-\{v_2\})$ has the same expansion relative to $\beta$ as relative to $\beta’$, but $\beta\neq \beta’$. 
