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Question

Find the eigenvalues and eigenvectors of the matrix in the following problem

(234320402)\left( \begin{array}{ccc} 2 & -3 & 4 \\ -3 & 2 & 0 \\ 4 & 0 & 2% \end{array}% \right)

Solution

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For the following 3×33\times 3 square matrix AA defined by

A=(234320402)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 λ\lambda which satisfy the characteristic equation of the matrix AA\,, thus the values of λ% \lambda must satisfy

det(AλI)=AλI=0\det \left( A-\lambda I\right) =\left\vert A-\lambda I\right\vert =0% where II is the 3×33\times 3 square identity matrix.

Computing the matrix AλIA-\lambda I implies that

AλI=(234320402)λ(100010001)=(234320402)(λ000λ000λ)=(2λ3432λ0402λ)\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 AλIA-\lambda I is just equal to AA with λ% \lambda subtracted from each element on the main diagonal elements.

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