Linear Algebra/Characteristic Equation

The matrix definition of an eigen value is very useful since it allows us to find eigen values for a given matrix using the following theorem:

$\lambda$ is an eigen value of $A$ iff $det(A-\lambda I_{n}v)=0.$

Proof:

If $Av=\lambda v$ then

$\Rightarrow Av=\lambda I_{n}v$

$\Rightarrow Av-\lambda I_{n}v=0$

$\Rightarrow (A-\lambda I_n)v=0$

but since $v$ is non-zero we know that $(A-\lambda I_n)$ is singular, ie it's determinant is zero so an eigen value of A will satisfy the equation

$det(A-\lambda I_{n}v)=0.$

which is known as the characteristic equation. (haven't proved the converse, but this is not required when calculating eigenvalues).

In the case $A$ is a $2x2$ matrix, this equation leads to the characteristic polynomial :

$det(\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right]-\lambda \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right])=0$

$\Rightarrow det(\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right]-\left[\begin{array}{cc}\lambda & 0\\ 0& \lambda \end{array}\right])=0$

$\Rightarrow det\left[\begin{array}{cc}{a}_{11}-\lambda & a_{12}\\ a_{21}& a_{22}-\lambda \end{array}\right]=0$

$\Rightarrow (a_{11}-\lambda )(a_{22}-\lambda )-a_{21}{a}_{12}=0$

$\Rightarrow {\lambda }^2-(a_{11}+a_{22})\lambda +a_{11}{a}_{22}-a_{12}{a}_{21}=0$

This is simply a quadratic equation and the roots of this are the eigen values of $A$

In order to find the corresponding eigen vectors, we simply solve the equation $Av=\lambda v$ which will be two simultaneous equations. There will in fact be infinitely many solutions to this equation since any scalar multiple of an eigen vector is also an eigen vector.