Basis of a Subspace - Meaning of the Term 'Unique' Following on from a previous question - Definition for Basis of a Subspace; it is often said that a basis of a subspace is a set of vectors that can be used to uniquely represent any vector in the subspace. 
I am having some trouble understanding the meaning of the word 'uniquely' in this context
In this article - http://mathworld.wolfram.com/VectorBasis.html; it is said that 'a vector space has many different vector bases'. However in this item - https://www.quora.com/What-is-a-basis-in-linear-algebra; it is said that 'each vector in the vector space can be written exactly in one way as a linear combination of the basis vectors'.
I find the above language rather confusing as the first quote seems to suggest that there is no 'unique' way to represent any given vector - there are 'many different' bases. And then the second quote states that 'each vector in the vector space can be written in exactly one way'. These quotes seem to contradict each other. 
My confusion is likely routed in a misunderstanding of what is meant by 'unique' in this context, so I seek the advice of Stack Exchange members on this topic.
Perhaps if I work through an example, users will be able to detect the error in my thinking. 

$S$ is a linearly independent subset of $\mathbb  R^{2}$, that spans all of the subspace:
$S = \{\begin{bmatrix}2\\5\end{bmatrix}, \begin{bmatrix}0\\6\end{bmatrix}\}$
We should be able to find any given vector in the subspace. Given the arbitrary vector $\vec x$:
$\vec x = \begin{bmatrix}x_1\\x_2\end{bmatrix}$
We can use a linear combination of the vectors in $S$ to arrive at $\vec x$:
$2(b_1) + 0(b_2) = x_1$
$5(b_1) + 6(b_2) = x_2$
By solving for $b_1$ and $b_2$, we arrive at:
$b_1 = \frac{x_1}2$
$b_2 = \frac{x_1 + x_2 * (-2 \div 5)}{-12 \div 5}$
Suppose that we want to find the vector $\vec e$:
$\vec e = \begin{bmatrix}-4\\5\end{bmatrix}$
We can use the following linear combination to get us there: 
$2(\frac{-4}2) + 0(\frac{-4 + 5 * (-2 \div 5)}{-12 \div 5}) = -4$
$5(\frac{-4}2) + 6(\frac{-4 + 5 * (-2 \div 5)}{-12 \div 5}) = 5$

Hopefully I have provided enough information as for the reader to decipher my misunderstanding. 
Thanks in advance.
 A: 
I find the above language rather confusing as the first quote seems to suggest that there is no 'unique' way to represent any given vector - there are 'many different' bases. And then the second quote states that 'each vector in the vector space can be written in exactly one way'. These quotes seem to contradict each other.

You may be confused because you are mixing two things:

*

*a basis for a subspace is not unique in the sense that a subspace can have many different bases; e.g. $\left\{ \begin{pmatrix}1 \\0 \end{pmatrix} , \begin{pmatrix}0 \\1 \end{pmatrix} \right\}$ and $\left\{ \begin{pmatrix}1 \\1 \end{pmatrix} , \begin{pmatrix}2 \\-1 \end{pmatrix} \right\}$ are two possible bases for $\mathbb{R}^2$;


*once a specific basis is chosen, then every vector can be written in only one way as a linear combination of these basis vectors, which gives rise to its coordinates relative to that basis and we call this a  unique representation with respect to that basis.


Suppose that we want to find the vector $\vec e$:
$\vec e = \begin{bmatrix}-4\\5\end{bmatrix}$
We can use the following linear combination to get us there:

What you find now is the only way to write $\vec x$ as a linear combination of the vectors in the basis $S$. We say that the coordinates of $\vec x$ with respect to the basis $S$ are unique.
However, you could choose another basis $T$ instead of $S$ and also with respect to that new basis, the vector $\vec x$ will have a unique representation.
A: For each basis $B$, there is one and only one way of representing each vector as a linear combination of vectors of $B$. But, of course, if you choose another basis $B^\star$, then the way of representing a vector as a linear combination of elements of $B^\star$ can be (and usually is) different from the way of representing it as a linear combination of elements of $B$.
