Question
Find eigenvectors of matrix:
$$ A = \left[ \begin{matrix}0&3&0&0&0\\3&0&0&0&0\\0&0&4&1&1\\0&0&1&4&1\\0&0&1&1&4\end{matrix} \right] $$Answer
The eigenvectors and eigenvalues of matrix A are:
| Eigenvalue | Eigenvector |
| $ 6 $ | $ \begin{bmatrix} 0 \cr 0 \cr 1 \cr 1 \cr 1 \end{bmatrix} $ |
| $ 3 $ | $ \begin{bmatrix} 1 \cr 1 \cr 0 \cr 0 \cr 0 \end{bmatrix} $ |
| $ 3 $ | $ \begin{bmatrix} 0 \cr 0 \cr 1 \cr 0 \cr -1 \end{bmatrix} $ |
| $ 3 $ | $ \begin{bmatrix} 0 \cr 0 \cr 0 \cr 1 \cr -1 \end{bmatrix} $ |
| $ -3 $ | $ \begin{bmatrix} 1 \cr -1 \cr 0 \cr 0 \cr 0 \end{bmatrix} $ |
Explanation
Step 1 : Find characteristic polynomial $ p(\lambda) $:
$$ p(\lambda) = -\lambda^5+12\,\lambda^4-36\,\lambda^3-54\,\lambda^2+405\,\lambda-486 $$Step 2 : Find the eigenvalues by solving the characteristic equation $ -\lambda^5+12\,\lambda^4-36\,\lambda^3-54\,\lambda^2+405\,\lambda-486 = 0. $
The eigenvalues are
$$ \lambda_1 = 6 ~ , ~ \lambda_2 = 3 ~ , ~ \lambda_3 = 3 ~ , ~ \lambda_4 = 3 ~ , ~ \lambda_5 = -3 $$( 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 = 6 $ we have $ A x = 6 I $.
For $ \lambda_ 1 = 3 $ we have $ A x = 3 I $.
For $ \lambda_ 4 = -3 $ we have $ A x = -3 I $.
This page was created using
Eigenvectors calculator