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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 $\left \{\lambda \begin{pmatrix}\cos \left (\frac{\alpha}{2}\right )\\ \sin \left (\frac{\alpha}{2}\right )\end{pmatrix}\right \}$

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  • $\begingroup$ I did Second Row= Second Row - First Row * $\frac{\sin (\alpha)}{\cos (\alpha)-1}$. Is that wrong?Why do we have to calculate $\det (M-\lambda I)$ ? I got stuck right now. Do we need the eigenvalues? @Invisible $\endgroup$ – Mary Star May 17 at 4:55
  • $\begingroup$ Sorry. Haven't read the question carefully. You can verify this by multiplying $M$ by your vector. You can't divide by $\cos(\alpha)-1$ if you don't know what $\alpha$ is. $\endgroup$ – Invisible May 17 at 4:56
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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. $(a,b)=k(\cos\alpha-1,\sin\alpha)=K\left(-\sin\frac\alpha2,\cos\frac\alpha2\right)$.

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Another (more transparent, perhaps) approach to finding the kernel.

Since $\det M=0$ the row vectors are linearly dependent so it suffices to solve $$(\cos \alpha -1)x + (\sin\alpha)y = 0. $$ If $\alpha = 2\pi k$, then any $(x,y)\in\mathbb R^2$ satisfies the equation. Otherwise, $\cos\alpha -1 = \sin\alpha = 0$ is impossible. Assume $\alpha \neq 2\pi k$

  1. If $\sin \alpha = 0$, then the kernel is $\{0\}\times\mathbb R$.
  2. Similarly, if $\cos\alpha -1 = 0.$
  3. If both are non-zero, then let $x$ be free, we have $$ \left ( \begin{array}{c} x \\ y \end{array} \right ) = \left ( \begin{array}{c} x \\ \frac{(\cos\alpha-1)}{\sin\alpha}x \end{array} \right ) = x\color{red}{\left ( \begin{array}{c} 1 \\ \frac{\cos \alpha -1}{\sin \alpha} \end{array} \right )},\quad x\in\mathbb R. $$ The kernel is generated by the vector in red.

The image is the linear span of the column vectors of $M$. But we know them to be linearly dependent so the image is $\{\lambda (\cos\alpha -1,\sin\alpha) \mid \lambda\in\mathbb R\}$.


One can use the identities $\sin 2\alpha = 2\sin\alpha\cos\alpha$ and $\sin ^2\alpha = \frac{1}{2}(1-\cos2\alpha)$ to obtain the result provided.

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