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:

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

Proof:

If ${\displaystyle Av=\lambda v}$ then

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

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

${\displaystyle \Rightarrow (A-\lambda I_{n})v=0}$

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

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

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

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

${\displaystyle det({\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix}}-\lambda {\begin{bmatrix}1&0\\0&1\end{bmatrix}})=0}$
${\displaystyle \Rightarrow det({\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix}}-{\begin{bmatrix}\lambda &0\\0&\lambda \end{bmatrix}})=0}$
${\displaystyle \Rightarrow det{\begin{bmatrix}a_{11}-\lambda &a_{12}\\a_{21}&a_{22}-\lambda \end{bmatrix}}=0}$
${\displaystyle \Rightarrow (a_{11}-\lambda )(a_{22}-\lambda )-a_{21}a_{12}=0}$
${\displaystyle \Rightarrow \lambda ^{2}-(a_{11}+a_{22})\lambda +a_{11}a_{22}-a_{12}a_{21}=0}$

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

In order to find the corresponding eigen vectors, we simply solve the equation ${\displaystyle Av=\lambda v}$ 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.