# Linear Algebra/OLD/Eigenvalues and Eigenvectors

Let ${\displaystyle T:V\rightarrow V}$ be a Linear Map where ${\displaystyle V}$ is a vector space over a field ${\displaystyle K}$. Then if some non-zero vector ${\displaystyle v\in V}$ satisfies the equation ${\displaystyle T(v)=\lambda v}$ with ${\displaystyle \lambda \in K}$ then ${\displaystyle v}$ is an eigen vector of ${\displaystyle T}$ and ${\displaystyle \lambda }$ is its corresponding eigen value.

Equivalently in Matrices we have ${\displaystyle Av=\lambda v}$.

For example, the matrix ${\displaystyle {\begin{pmatrix}2&0\\0&3\end{pmatrix}}}$ will double the vector ${\displaystyle {\begin{pmatrix}1\\0\end{pmatrix}}}$ and triple the vector ${\displaystyle {\begin{pmatrix}0\\1\end{pmatrix}}}$.

## How to find eigenvectors and eigenvalues

Finding the eigenvalues and eigenvectors is fairly simple. First we will find the eigenvalues, and then we will see how to find the eigenvectors that correspond to a specific eigenvalue.

If V is an eigenvector of matrix A with eigenvalue ${\displaystyle \lambda }$ then by definition AV=${\displaystyle \lambda }$V. after subtracting ${\displaystyle \lambda }$V we will get AV-${\displaystyle \lambda }$V=0. since ${\displaystyle \lambda }$V equals ${\displaystyle \lambda }$IV where I is the identity matrix we finally get (A-${\displaystyle \lambda }$I)V=0.
so ${\displaystyle \lambda }$ is an eigenvalue if and only if the matrix (A-${\displaystyle \lambda }$I) sends some vector to 0. This is luckily easy to determine, because a matrix sends some vectors to zero if and only if its determinant is equal to zero.
Let's do an example and find the eigenvalues of ${\displaystyle {\begin{pmatrix}-1&-3&-6\\0&2&0\\3&3&8\end{pmatrix}}}$ we will subtract tI from the matrix (t is a variable) and find out when the determinant |A-tI| equals 0 (this is called the characteristic polynom).

A-tI = ${\displaystyle {\begin{pmatrix}-1-t&-3&-6\\0&2-t&0\\3&3&8-t\end{pmatrix}}}$
|A-tI| is (calculating with the first row): (-1-t)*(2-t)*(8-t)+3*6*(2-t) simplefying a bit we get (2-t)2*(5-t) since |A-tI|=0 when t=2,5, the eigenvalues are 2 and 5.
let's find the eigenvectors space of the eigenvalue 2:
by substracting 2 from the diagonal, we get A-2I: ${\displaystyle {\begin{pmatrix}-3&-3&-6\\0&0&0\\3&3&6\end{pmatrix}}}$
the vectors ${\displaystyle {\begin{pmatrix}2\\0\\-1\end{pmatrix}}}$ and ${\displaystyle {\begin{pmatrix}0\\2\\-1\end{pmatrix}}}$ form a kernel space of A-2I, and therefore any linear combination of them is an eigenvector with an eigenvalue of 2.
for example 10*${\displaystyle {\begin{pmatrix}2\\0\\-1\end{pmatrix}}}$ + 3*${\displaystyle {\begin{pmatrix}0\\2\\-1\end{pmatrix}}}$=${\displaystyle {\begin{pmatrix}20\\6\\-13\end{pmatrix}}}$ and ${\displaystyle {\begin{pmatrix}-1&-3&-6\\0&2&0\\3&3&8\end{pmatrix}}}$*${\displaystyle {\begin{pmatrix}20\\6\\-13\end{pmatrix}}}$=${\displaystyle {\begin{pmatrix}40\\12\\-26\end{pmatrix}}}$=2*${\displaystyle {\begin{pmatrix}20\\-6\\-13\end{pmatrix}}}$
you can similliary find the eigenvectors space of the eigenvalue 5 by computing A-5I (substracting 5 from the diagonal) and finding it's kernel space.

• You can also use a mathematical software like octave to find the eigenvalues:

 octave:1> a=[-1,-3,-6;0,2,0;3,3,8] 

a = 

 -1 -3 -6 0 2 0 3 3 8 octave:2> [v,d]=eig(a) v =  -0.89443 0.70711 -0.28280 0.00000 0.00000 0.90699 0.44721 -0.70711 -0.31209 d =  2 0 0 0 5 0 0 0 2