We could start by finding both of the separate transformation matrices and then multiplying them together which would give us the general transformation matrix T. However since we only need T([1,0]), the transformation of the unit vector in the x-direction, it will suffice to just consider how that single vector is transformed and not deal with matrices at all.
We first rotate the vector [1,0] counter-clockwise by 
to get a new vector 
. After doing so, the new vector forms a right triangle with a vector in the x-direction and a vector in the y-direction. We can get its components by using trigonometry. The new vector will have length 1 (since it is a rotation of [1,0] which also has length 1) so its components are
![{\displaystyle \mathbf {v} _{1}=\left[\cos \left({\frac {\pi }{6}}\right),\sin \left({\frac {\pi }{6}}\right)\right]=\left[{\frac {\sqrt {3}}{2}},{\frac {1}{2}}\right].}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/71b9004f10d84237c7ec0c4e2438789f67d608c9)
We then must project this vector onto u=[1,2]. Recall that a projection of a vector v onto a vector u is
Using our vector and projecting it onto u which has length 
, we have that our new vector, 
, is
![{\displaystyle {\begin{aligned}\mathbf {v_{2}} &={\textrm {proj}}_{\mathbf {u} }\mathbf {v} _{1}={\frac {\left[{\frac {\sqrt {3}}{2}},{\frac {1}{2}}\right]\cdot \left[1,2\right]}{5}}[1,2]\\&={\frac {{\frac {\sqrt {3}}{2}}(1)+({\frac {1}{2}})2}{5}}[1,2]={\frac {{\sqrt {3}}+2}{10}}[1,2].\end{aligned}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/69f3f2088014408f341732e0d4951a5c788ddf25)
Therefore, the transformation on the vector [1,0] is
