User:Bart vanderbeke/sandbox

Eigencircle of a $$ 2\times 2$$ matrix
Eigencircles support a geometric interpretation of linear transformations.

This article only shows the basic properties of an eigencircle.

The referred articles derive more properties

An eigenvalue of a square $$ 2\times 2$$ matrix $$ A$$ is a number $$ \lambda $$ such that $$ AX=\lambda X$$ for an $$ X\neq 0$$. To determine the eigenvalues $$ \lambda $$ of a matrix $$ A$$ a solution for $$ \det(A-\lambda I)=0$$ is to be found.

The collection of eigenvalues $$ EV$$ can be written as:
 * $$ EV=\left\{\lambda \ |\ \exists {\vec {x}}\left[{\begin{matrix}x\\y\\\end{matrix}}\right]and\ \left[{\begin{matrix}\lambda &0\\0&\lambda \\\end{matrix}}\right]\left[{\begin{matrix}x\\y\\\end{matrix}}\right]=A\left[{\begin{matrix}x\\y\\\end{matrix}}\right]\right\}$$

We now extend the concept of eigenvalues by looking for the elements of the set  $$ EC$$ below:


 * $$ EC=\left\{\left(\lambda ,\mu \right)\ |\ \exists \left[{\begin{matrix}x\\y\\\end{matrix}}\right]and\ \left[{\begin{matrix}\lambda &+\mu \\-\mu &\lambda \\\end{matrix}}\right]\left[{\begin{matrix}x\\y\\\end{matrix}}\right]=A\left[{\begin{matrix}x\\y\\\end{matrix}}\right]\right\}$$

Using this broader concept, for every $$ \left[{\begin{matrix}x\\y\\\end{matrix}}\right]$$ a $$ (\lambda ,\mu )$$ can be found such that:
 * $$ L_{\lambda \mu }\left[{\begin{matrix}x\\y\\\end{matrix}}\right]=\left[{\begin{matrix}\lambda &+\mu \\-\mu &\lambda \\\end{matrix}}\right]\left[{\begin{matrix}x\\y\\\end{matrix}}\right]=A\left[{\begin{matrix}x\\y\\\end{matrix}}\right]$$


 * $$ \left(\lambda ,\mu \right)$$ is called a $$ \left(\lambda ,\mu \right)-eigenpair$$ and $$ \left[{\begin{matrix}x\\y\\\end{matrix}}\right]$$ is the corresponding $$ \left(\lambda ,\mu \right)-eigenvector$$.

The condition for $$ \left(\lambda ,\mu \right)-eigenpairs$$ to exist can be written as below:


 * $$ {EC}_{=}\left\{\left(\lambda ,\mu \right)\ |\ \exists {\vec {x}}\left[{\begin{matrix}x\\y\\\end{matrix}}\right]and\ \left[{\begin{matrix}\lambda &+\mu \\-\mu &\lambda \\\end{matrix}}\right]\left[{\begin{matrix}x\\y\\\end{matrix}}\right]=A\left[{\begin{matrix}x\\y\\\end{matrix}}\right]\right\}\neq \left\{\right\}$$


 * $$ det\left(A-L_{\lambda \mu }\right)=\left|{\begin{matrix}a-\lambda &b+\mu \\c-\mu &d-\lambda \\\end{matrix}}\right|=0$$


 * $$ \lambda ^{2}-\left(a+d\right)\lambda +\det {\left(A\right)}-\left(b-c\right)\mu +\mu ^{2}=0$$

Using the following identities, the equation can be simplified:


 * $$ f={\frac {\left(a+d\right)}{2}}$$


 * $$ g={\frac {\left(b-c\right)}{2}}$$


 * $$ r^{2}=f^{2}+g^{2}$$


 * $$ \det {\left(A\right)}=r^{2}-\rho ^{2}$$


 * $$ \rho ^{2}=\left({\frac {a-d}{2}}\right)^{2}+\left({\frac {b+c}{2}}\right)^{2}$$


 * $$ \left(\lambda -f\right)^{2}+\left(\mu -g\right)^{2}-\rho ^{2}=0$$

The set $$EC$$ containing all $$ \left(\lambda ,\mu \right)-eigenvalues$$ is a circle on the $$ \left(\lambda ,\mu \right)-plane$$ with center $$ C(f,g)$$ and radius $$ \rho $$.

This circle is called the eigencircle of $$ A$$.

This is illustrated in Figure 1.


 * $$ EC=\left\{\left(\lambda ,\mu \right)\ |\ \exists {\vec {x}}\left[{\begin{matrix}x\\y\\\end{matrix}}\right]and\ {\mathcal {t}}\left({\vec {x}}\right)=\left[{\begin{matrix}\lambda &+\mu \\-\mu &\lambda \\\end{matrix}}\right]\left[{\begin{matrix}x\\y\\\end{matrix}}\right]=A\left[{\begin{matrix}x\\y\\\end{matrix}}\right]\right\}$$


 * $$ EC=\left\{\left(\lambda ,\mu \right)\ |\ \left(\lambda -f\right)^{2}+\left(\mu -g\right)^{2}-\rho ^{2}=0\right\}$$

Every eigencircle contains four characteristic $$\left(\lambda ,\mu \right)-eigenpairs$$, this is illustrated in Figure 2.

The construction in Figure 3 shows how the eigenvectors of the matrix $$A$$ can be read from the eigencircle.

Figure 4 illustrates how a the existence of a $$\left(\lambda ,\mu \right)-eigenpair$$ $$\left(\lambda_1 ,\mu_1 \right)$$ can be interpreted:

If a $$\left(\lambda_1,\mu_1\right)_{cartesian}=\left(s_1,-\theta_1\right)_{polar}$$ exists on the eigencircle, a vector $$x_1$$, with $$\|x1\|=1$$, must exist such that $$\angle\left(x_1,Ax_1\right)=\theta_1$$ and $$\|A x1\|=s1$$.