Let $T_1$ be the linear transformation corresponding to a counterclockwise rotation of $120$ degrees and let $T_2$ be the linear transformation corresponding to a clockwise rotation of $45$ degrees.
Let $u = \begin{bmatrix} 4 \\ 0 \end{bmatrix}$. Evaluate $T_2(T_1(u))$?
My work
I used $\begin{pmatrix} cos\theta & -sin\theta \\ sin\theta & cos\theta \end{pmatrix}$
T1 = $\begin{pmatrix} cos(\frac { 2\Pi }{ 3 } ) & -sin(\frac { 2\Pi }{ 3 } ) \\ sin(\frac { 2\Pi }{ 3 } ) & cos(\frac { 2\Pi }{ 3 } ) \end{pmatrix}$
T1=$\begin{pmatrix} \frac { -1 }{ 2 } & \frac { -\sqrt { 3 } }{ 2 } \\ \frac { \sqrt { 3 } }{ 2 } & \frac { -1 }{ 2 } \end{pmatrix}$
T1(u)=$\begin{pmatrix} -2 & \\ 2\sqrt { 3 } & \end{pmatrix}$
Similarly, T2 = $\begin{pmatrix} \frac { 1 }{ \sqrt { 2 } } & \frac { 1 }{ \sqrt { 2 } } \\ \frac { -1 }{ \sqrt { 2 } } & \frac { 1 }{ \sqrt { 2 } } \end{pmatrix}$, Here I took $\theta$=$\frac { 7\pi }{ 4 }$
Now, T2(T1(u)) = $\begin{pmatrix} \frac { 1 }{ \sqrt { 2 } } & \frac { 1 }{ \sqrt { 2 } } \\ \frac { -1 }{ \sqrt { 2 } } & \frac { 1 }{ \sqrt { 2 } } \end{pmatrix}$ $\begin{pmatrix} -2 & \\ 2\sqrt { 3 } & \end{pmatrix}$
Finally, I got T2(T1(u)) = $\begin{pmatrix} \frac { -2 }{ \sqrt { 2 } } +\frac { 2\sqrt { 3 } }{ 2 } & \frac { }{ } \\ \frac { 2 }{ \sqrt { 2 } } +\frac { 2\sqrt { 3 } }{ 2 } & \end{pmatrix}$
Can anyone please verify my work. Thanks in advance
You made a mistake as you compute $T_1(u)$:$$T_1(u)=\begin{pmatrix} -2 \\ 2\sqrt { \color{red}3 } \end{pmatrix}$$
$$\begin{pmatrix} \frac { 1 }{ \sqrt { 2 } } & \frac { 1 }{ \sqrt { 2 } } \\ \frac { -1 }{ \sqrt { 2 } } & \frac { 1 }{ \sqrt { 2 } } \end{pmatrix}\begin{pmatrix} -2 \\ 2\sqrt { 3 } \end{pmatrix}=\begin{pmatrix} -\sqrt2+\sqrt6 \\ \sqrt2+\sqrt6 \end{pmatrix}$$