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# Find the eigenvalues and eigenvectors of the matrix in the following problem$\left( \begin{array}{ccc} 2 & -3 & 4 \\ -3 & 2 & 0 \\ 4 & 0 & 2% \end{array}% \right)$

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For the following $3\times 3$ square matrix $A$ defined by

$A=\left( \begin{array}{ccc} 2 & -3 & 4 \\ -3 & 2 & 0 \\ 4 & 0 & 2% \end{array}% \right)$

Firstly, to find its eigenvalues, we find the values of $\lambda$ which satisfy the characteristic equation of the matrix $A\,$, thus the values of $% \lambda$ must satisfy

$\det \left( A-\lambda I\right) =\left\vert A-\lambda I\right\vert =0$% where $I$ is the $3\times 3$ square identity matrix.

Computing the matrix $A-\lambda I$ implies that

\begin{aligned} A-\lambda I &=&\left( \begin{array}{ccc} 2 & -3 & 4 \\ -3 & 2 & 0 \\ 4 & 0 & 2% \end{array}% \right) -\lambda \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1% \end{array}% \right) \\ &=&\left( \begin{array}{ccc} 2 & -3 & 4 \\ -3 & 2 & 0 \\ 4 & 0 & 2% \end{array}% \right) -\left( \begin{array}{ccc} \lambda & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda% \end{array}% \right) \\ &=&\left( \begin{array}{ccc} 2-\lambda & -3 & 4 \\ -3 & 2-\lambda & 0 \\ 4 & 0 & 2-\lambda% \end{array}% \right) \end{aligned}

% where we find that the matrix $A-\lambda I$ is just equal to $A$ with $% \lambda$ subtracted from each element on the main diagonal elements.

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