We have the matrix \begin{equation*}M=\begin{pmatrix} \cos (\alpha )-1& \sin (\alpha ) \\ \sin (\alpha) & -\cos(\alpha)-1\end{pmatrix}\end{equation*} I want to calculate the kernel and the image of the matrix.
For the kernel we have to solve the system $(s_{\alpha}-u_2)x=0_{\mathbb{R}^2}$.
Using Gauss elimination algorithmwe get
\begin{equation*}\begin{pmatrix} \cos (\alpha )-1& \sin (\alpha ) \\ \sin (\alpha) & -\cos(\alpha)-1\end{pmatrix} \rightarrow \begin{pmatrix} \cos (\alpha )-1& \sin (\alpha ) \\ 0 & 0\end{pmatrix}\end{equation*} or not?
Is the kernel $\left \{\lambda \begin{pmatrix}\cos (\alpha)-1\\ \sin (\alpha)\end{pmatrix}\right \}$? Can we write this vector in respect of $\frac{\alpha}{2}$ instad of $\alpha$ ?
The solution must be $\left \{\lambda \begin{pmatrix}\cos \left (\frac{\alpha}{2}\right )\\ \sin \left (\frac{\alpha}{2}\right )\end{pmatrix}\right \}$
Best Answer
For the kernel you get $y\sin\alpha=x(1-\cos\alpha)$ i.e. $(x,y)=k(\sin\alpha,1-\cos\alpha)=k\left(2\sin\frac\alpha2\cos\frac\alpha2,2\sin^2\frac\alpha2\right)=K\left(\cos\frac\alpha2,\sin\frac\alpha2\right)$.
For the image let $M[x,y]^T=[a,b]^T$. We get$$\begin{bmatrix} \cos (\alpha )-1& \sin (\alpha ) &|&a \\ 0 & 0&|&b-\frac{a\sin\alpha}{\cos\alpha-1}\end{bmatrix}$$This system is solvable iff $b-\frac{a\sin\alpha}{\cos\alpha-1}=0$, i.e. $(a,b)=k(\cos\alpha-1,\sin\alpha)=K\left(-\sin\frac\alpha2,\cos\frac\alpha2\right)$.