Transformation matrix - Are the matrices correct? I want to describe for the following $\mathbb{R}$-vector spaces $V$with basis $B$ the matrix $A_{f, B, B}$ for $f\in \text{End}_K(V)$: 


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*$V=\mathbb{R}^2, B=((1,0)^T, (0,1)^T)$ and $f$ the rotation by 45 degrees clockwise. 

*$V=\mathbb{C}, B=(1,i)$ and $f(z)=a^2\overline{z}$ with $a=\cos \alpha+i\sin\alpha, \alpha \in [0,2\pi)$. 
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The matrix $A_{f, B, B}$ contains the coefficients $a_{ij}$ of $f(b_j)=\sum a_{ij}b_i$, right? 
We have the following: 


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*We have that $f(x)=\begin{pmatrix}\cos\frac{\pi}{4} & -\sin\frac{\pi}{4} \\ \sin\frac{\pi}{4} & \cos\frac{\pi}{4}\end{pmatrix}\cdot x=\frac{1}{\sqrt{2}}\begin{pmatrix}1 & -1 \\ 1 & 1\end{pmatrix}\cdot x$, right? 
Therefore we have the following: 
$$f(b_1)=f\begin{pmatrix}1\\0\end{pmatrix}=\frac{1}{\sqrt{2}}\begin{pmatrix}1 & -1 \\ 1 & 1\end{pmatrix}\cdot\begin{pmatrix}1\\0\end{pmatrix}=\frac{1}{\sqrt{2}}\begin{pmatrix}1 \\ 1 \end{pmatrix}=\frac{1}{\sqrt{2}}\begin{pmatrix}1 \\ 0 \end{pmatrix}+\frac{1}{\sqrt{2}}\begin{pmatrix}0 \\ 1 \end{pmatrix}=\frac{1}{\sqrt{2}}\cdot b_1+\frac{1}{\sqrt{2}}\cdot b_2 \\ f(b_2)=f\begin{pmatrix}1\\0\end{pmatrix}=\frac{1}{\sqrt{2}}\begin{pmatrix}1 & -1 \\ 1 & 1\end{pmatrix}\cdot\begin{pmatrix}0\\1\end{pmatrix}=\frac{1}{\sqrt{2}}\begin{pmatrix}-1 \\ 1 \end{pmatrix}=-\frac{1}{\sqrt{2}}\begin{pmatrix}1 \\ 0 \end{pmatrix}+\frac{1}{\sqrt{2}}\begin{pmatrix}0 \\ 1 \end{pmatrix}=-\frac{1}{\sqrt{2}}\cdot b_1+\frac{1}{\sqrt{2}}\cdot b_2$$ 
So, do we get $$A_{f, B, B}=\begin{pmatrix}\frac{1}{\sqrt{2}}& -\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}\end{pmatrix}$$ ? 

*We have the following: $$f(1)=a^2=(\cos\alpha+i\sin\alpha)^2=\cos^2\alpha-\sin^2\alpha-2i\cos\alpha\sin\alpha= (\cos^2\alpha-\sin^2\alpha)\cdot 1+(-2\cos\alpha\sin\alpha) \cdot i\\  f(i)=-a^2=-(\cos\alpha+i\sin\alpha)^2=-\cos^2\alpha+\sin^2\alpha+2i\cos\alpha\sin\alpha=-(\cos^2\alpha-\sin^2\alpha)\cdot 1+(2\cos\alpha\sin\alpha) \cdot i$$ 
So, do we get $$A_{f, B, B}=\begin{pmatrix}\cos^2\alpha-\sin^2\alpha& -(\cos^2\alpha-\sin^2\alpha)\\ -2\cos\alpha\sin\alpha&2\cos\alpha\sin\alpha\end{pmatrix}$$ ? 
How could we justify that $f$ is a reflection at the line $\{at: t\in \mathbb{R}\}$ ?
 A: For the first, note that rotation matrix by $\theta$ clockwise is:
$$f(x)=\begin{pmatrix}\cos\ \theta & \sin\theta  \\ -\sin\theta  & \cos\theta \end{pmatrix}$$
Thus:
$$f(x)=\begin{pmatrix}\cos\frac{\pi}{4} & \sin\frac{\pi}{4} \\ -\sin\frac{\pi}{4} & \cos\frac{\pi}{4}\end{pmatrix}\cdot x=\frac{1}{\sqrt{2}}\begin{pmatrix}1 & 1 \\ -1 & 1\end{pmatrix}\cdot x$$
For the second we have:
$$a=\cos \alpha+i\sin\alpha\implies a^2= \cos^2 \alpha -\sin^2\alpha+2i\cos \alpha\sin \alpha $$
$$z=1\implies a^2 \bar z= a^2= \cos^2 \alpha -\sin^2\alpha+2i(\cos \alpha\sin \alpha)$$
$$z=i\implies a^2 \bar z= -ia^2= -i(\cos^2 \alpha -\sin^2\alpha)+2(\cos \alpha\sin \alpha) $$
Thus:
$$A_{f, B, B}=\begin{pmatrix}\cos^2\alpha-\sin^2\alpha&2(\cos\alpha\sin\alpha)\\ 2(\cos\alpha\sin\alpha)&-(\cos^2\alpha-\sin^2\alpha)\end{pmatrix}$$
To justify that $f$ is a reflection at the line $\{at: t\in \mathbb{R}\}$ note that:
$z\to f(z)=a^2 \bar z\to f(a^2 \bar z)=a^2 \bar a^2 z=z$
We can also verify this by matrix $A_{f, B, B}$, indeed:
$$A_{f, B, B}\cdot a=
\cos^3\alpha-\sin^2\alpha \cos\alpha+2(\cos\alpha\sin^2\alpha)+2i(\cos^2\alpha\sin\alpha)-i(\cos^2\alpha \sin \alpha-\sin^3\alpha)=
\cos^3\alpha+\sin^2\alpha\cos\alpha+i(\cos^2\alpha\sin\alpha+\sin^3\alpha)=\cos\alpha+i\sin\alpha=a$$
