## Characteristic Vectors of a Matrix

Let
$A= \left( \begin{array}{ccc} 8 & 2 & -2 \\ 3 & 3 & -1 \\ 24 & 8 & -6 \end{array} \right)$
.
The characteristic values - the eigenvalues of
$A$
are the solutions to
$det(A - \lambda I) =0.$

\begin{aligned} det(A- \lambda I) &= det( \left( \begin{array}{ccc} 8 & 2 & -2 \\ 3 & 3 & -1 \\ 24 & 8 & -6 \end{array} \right)- \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right)) \\ &= (8- \lambda)((3- \lambda )(-6- \lambda )+8)-3(2(-6 - \lambda )+16)+24(-2 + 2(3- \lambda )) \\ &= - (\lambda-2)^2 (\lambda -1) \end{aligned}

The eigenvalues are
$\lambda_1 = 1, \: \lambda_2 =2$
.
$\lambda_1 =1$

$\mathbf{v} A- I =(x,y,z) \left( \begin{array}{ccc} 7 & 2 & -2 \\ 3 & 2 & -1 \\ 24 & 8 & -7 \end{array} \right) =(0,0,0)$

Multiplying out gives
$7x+3y+24z=0$

$2x+2y+8z=0$

$-2x-y-7z=0$

Solving these gives
$(3,1,-1)$
as a solution with all other solutions being a multiple of this one..
$\lambda_2 =2$

$\mathbf{v} A- I =(x,y,z) \left( \begin{array}{ccc} 6 & 2 & -2 \\ 3 & 1 & -1 \\ 24 & 8 & -8 \end{array} \right) =(0,0,0)$

Multiplying out gives
$6x+3y+24z=0$

$2x+y+8z=0$

$-2x-y-8z=0$

Put nbsp;
$y=0 \: z=1$
then a solution is nbsp;
$(-4,0,1)$

Put nbsp;
$y=2 \: z=0$
then a solution is nbsp;
$(-1,2,0)$

The characteristic vectors are
$\begin{pmatrix}3\\1\\-1\end{pmatrix}, \: \begin{pmatrix}-4\\0\\1\end{pmatrix} , \:\begin{pmatrix}-1\\2\\0\end{pmatrix}$
.