## Characteristic Vectors of a Matrix

Let  {jatex options:inline}A= \left( \begin{array}{ccc} 8 & 2 & -2 \\ 3 & 3 & -1 \\ 24 & 8 & -6 \end{array} \right){/jatex}.
The characteristic values - the eigenvalues of  {jatex options:inline}A{/jatex}  are the solutions to  {jatex options:inline}det(A - \lambda I) =0.{/jatex}
{jatex options:inline} \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}{/jatex}
The eigenvalues are  {jatex options:inline}\lambda_1 = 1, \: \lambda_2 =2{/jatex}.
{jatex options:inline}\lambda_1 =1{/jatex}
{jatex options:inline}\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){/jatex}
Multiplying out gives
{jatex options:inline}7x+3y+24z=0{/jatex}
{jatex options:inline}2x+2y+8z=0{/jatex}
{jatex options:inline}-2x-y-7z=0{/jatex}
Solving these gives  {jatex options:inline}(3,1,-1){/jatex}  as a solution with all other solutions being a multiple of this one..
{jatex options:inline}\lambda_2 =2{/jatex}
{jatex options:inline}\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){/jatex}
Multiplying out gives
{jatex options:inline}6x+3y+24z=0{/jatex}
{jatex options:inline}2x+y+8z=0{/jatex}
{jatex options:inline}-2x-y-8z=0{/jatex}
Put nbsp;{jatex options:inline}y=0 \: z=1{/jatex}  then a solution is nbsp;{jatex options:inline}(-4,0,1){/jatex}
Put nbsp;{jatex options:inline}y=2 \: z=0{/jatex}  then a solution is nbsp;{jatex options:inline}(-1,2,0){/jatex}
The characteristic vectors are  {jatex options:inline}\begin{pmatrix}3\\1\\-1\end{pmatrix}, \: \begin{pmatrix}-4\\0\\1\end{pmatrix} , \:\begin{pmatrix}-1\\2\\0\end{pmatrix} {/jatex}.