Change of basis to find coordinates 
This was my previous post: https://math.stackexchange.com/posts/1243265 I changed my question quite drastically and I didn't feel I was asking the question correctly.
I attempted with change of basis method, I'm unsure if my values for the vectors are correct as I've just assumed that the length of the vector is 1 as it isn't explicitly stated.
S = {i, j} T = {u1, u2}
some vector w coordinates in respect to basis T:  [w]T = C * [w]S and the coordinates [w]S were the coordinates listed (a), (b) etc.
where C is the change of basis vector. 
I assume that the vector length is 1 since it's not explicitly stated??
so u1 = (1*cos(30), 1*sin(30)) u2 is orthogonal so u2 = (1 * cos(90), 1*sin(90)) = (0, 1)
i and j are orthogonal so they both = (0, 1) ?
then putting i, j, u1, u2 vectors into columns of the below matrix and using guassian elimination method to get the inverse of S to get the change of basis vector:
$$ [u1 \;  u2 \; | \; i  \; j] = \begin{bmatrix}
\cos(30) & 0 & \mid & 0 & 0 \\ 
\sin(30) & 1 & \mid & 1 & 1 \\ 
 \end{bmatrix} $$
^all one matrix (sorry about format)
I wanted to find the inverse of [i, j] to get the change of basis vector but it seems that [i j] was not invertible so my coordinates for the vectors i and j are wrong. Where have I gone wrong?
 A: At first, let me clarify some thing for you, whenever you have basis, the vector in this basis will never be equal. Accordingly, as $S=\{ i,j\}$  is a basis, then of course $i \neq j$. In fact, if you have a quick look at your figure, you directly recognize that $i$ must be $(1,0)$, and $j$ must be $(0,1)$. Indeed, they are also orthogonal, see that $$ i . j = 1\times 0 + 0\times 1= 0 .$$
Coming now to the new coordinate system $T=\{ u_1 , u_2\}$. As I see in the figure, we have $u_2=j$, so $ u_2=(0,1)$. On the other hand, the vector $u_1$, is just rotating the vector $i $ by $30$° anticlockwise. So as you state above $u_2=(\cos(30), \sin(30))$, assuming their length is $1$.
Note that the the vectors $u_1$ and $u_2$ are not orthogonal.
So in $T $in terms  of $S$ can be written ussing the matrix $$ \begin{bmatrix}  \cos(30) & 0 \\ \sin(30) & 1 \\ \end{bmatrix}  $$
Accordingly,$$ \begin{bmatrix}  \cos(30) & 0 \\ \sin(30) & 1 \\ \end{bmatrix}   \begin{bmatrix}  \sqrt3  \\  1 \\ \end{bmatrix} $$
are the  coordinates of $(\sqrt3, 1) $ in system $T$. similarly for the resting parts. 
A: Ok, first we have to find the transition  matrix from the basis $(u_1,u_2)$ $\rightarrow$ $(i,j)$. Its clear as you said that, $u_1\ = \cos(30)i + \sin(30)j$, $u_2\ = j$, hence $i \ = \frac{u_1\ - \sin(30)u_2}{\cos(30)}$, $j=u_2$ the transition matrix is $$ 
                                               \begin{pmatrix}
                            \frac{2}{\sqrt3}& 0  \\
                             -\frac{1}{\sqrt3} & 1  \\
\end{pmatrix}$$
and so You can write that $$ 
                                               \begin{pmatrix}
x' \\
y' \\
\end{pmatrix}=\bigl( \begin{smallmatrix}   \frac{2}{\sqrt3} & 0 \\ -\frac{1}{\sqrt3} & 1 \end{smallmatrix} \bigr)*\bigl( \begin{smallmatrix}  x \\ y\end{smallmatrix} \bigr)$$ In the first case for example: $$    \begin{pmatrix}
                              \frac{2}{\sqrt3}& 0  \\
                             -\frac{1}{\sqrt3} & 1 \\
\end{pmatrix}$$
$$ 
                                               \begin{pmatrix}
x' \\
y' \\
\end{pmatrix}=\bigl( \begin{smallmatrix}   \frac{2}{\sqrt3} & 0 \\ - \frac{1}{\sqrt3} & 1 \end{smallmatrix} \bigr)*\bigl( \begin{smallmatrix}  \sqrt3 \\ 1\end{smallmatrix} \bigr)=\bigl( \begin{smallmatrix} 2 \\ 0 \end{smallmatrix} \bigr)$$ written in $(u_1,u_2)$ basis. 
