Question
Find eigenvectors of matrix:
$$ A = \left[ \begin{matrix}2&-1&0\\-1&2&-1\\0&-1&2\end{matrix} \right] $$Answer
The eigenvectors and eigenvalues of matrix A are:
| Eigenvalue | Eigenvector |
| $ 2-\sqrt{2} $ | $ \begin{bmatrix} 1 \cr \sqrt{2} \cr 1 \end{bmatrix} $ |
| $ \sqrt{2}+2 $ | $ \begin{bmatrix} 1 \cr -\sqrt{2} \cr 1 \end{bmatrix} $ |
| $ 2 $ | $ \begin{bmatrix} 1 \cr 0 \cr - 1 \end{bmatrix} $ |
Explanation
Step 1 : Find characteristic polynomial $ p(\lambda) $:
$$ p(\lambda) = -\lambda^3+6\lambda^2-10\lambda+4 $$Step 2 : Find the eigenvalues by solving the characteristic equation $ -\lambda^3+6\lambda^2-10\lambda+4 = 0. $
The eigenvalues are
$$ \lambda_1 = 2-\sqrt{2} ~ , ~ \lambda_2 = \sqrt{2}+2 ~ , ~ \lambda_3 = 2 $$( click here to view an explanation on how to solve this equation.)
Step 3 : To find the associated eigenvectors ( $ x $ ), we have to solve the equation $ A x = \lambda I $.
For $ \lambda_ 0 = 2-\sqrt{2} $ we have $ A x = 2-\sqrt{2} I $.
For $ \lambda_ 1 = \sqrt{2}+2 $ we have $ A x = \sqrt{2}+2 I $.
For $ \lambda_ 2 = 2 $ we have $ A x = 2 I $.
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