# Kernel and image of matrix

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

• 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 – Mary Star May 17 at 4:55
• 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. – Invisible May 17 at 4:56

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. $$(a,b)=k(\cos\alpha-1,\sin\alpha)=K\left(-\sin\frac\alpha2,\cos\frac\alpha2\right)$$.

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.