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Example: Calculation of Eigenvalues and Eigenvectors

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In order to find the eigenvalues of matrix

$\displaystyle A=\left(\begin{array}{rrr}0& -2 &-1\\ 2 & 2 & 1 \\ 0 & 2& 2\end{array}\right) $

compute the characteristic polynomial by means of the determinant

$\displaystyle \operatorname{det}(A-\lambda E)=\left\vert\begin{array}{rrr}-\lam... ...t = -\lambda(2-\lambda)^2-4+8-2\lambda = -\lambda^3+4\lambda^2-6\lambda+4\,. $

and find the zeros of this polynomial. By guessing (divisor of the absolute term) we obtain $ \lambda_1=2$ as first zero.

Polynomial division leads to

$\displaystyle (-\lambda^3+4\lambda^2-6\lambda+4)\div(\lambda-2)=-\lambda^2+2\lambda-2 $

and using the quadratic formula yields

$\displaystyle \lambda_{2,3}=\frac{-2\pm \sqrt{4-8}}{-2}=1\pm \mathrm{i}\,. $

Solving the homogeneous LSE

$\displaystyle (A-2 E)v=\left(\begin{array}{rrr}-2& -2 &-1\\ 2 & 0 & 1 \\ 0 & 2 & 0\end{array}\right)v =0 $

we obtain an eigenvector for eigenvalue $ 2$.

It can easily be seen that $ v=(1,0,-2)^{\operatorname t}$ is a solution.

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  automatisch erstellt am 25.  6. 2018