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  
\[det(A - \lambda I) =0.\]

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

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  
\[\begin{pmatrix}3\\1\\-1\end{pmatrix}, \: \begin{pmatrix}-4\\0\\1\end{pmatrix} , \:\begin{pmatrix}-1\\2\\0\end{pmatrix} \]
.

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