Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Please help me calculate the Eigen vectors of this matrix.
$$\begin{pmatrix} 3 & 0 & 1\\ 1 & 3 & 0\\ 0 & 1 & 3 \end{pmatrix}$$

The first vector comes out to be null, no clue how to find out the other two.

share|improve this question
3  
By null do you mean the zero vector? In that case you must've made a mistake somewhere. –  EuYu Dec 21 '12 at 5:56
2  
This matrix has three distinct eigenvectors (two complex). It has a complete set of non-zero eigenvectors. –  copper.hat Dec 21 '12 at 5:59
2  
Did you compute the characteristic polynomial of the matrix? –  Yury Dec 21 '12 at 6:00
2  
The eigenvector that is obvious is $(1,1,1)$, with eigenvalue $4$, since each row has the same sum ($4$). –  mjqxxxx Dec 21 '12 at 6:00
3  
@AdnanZahid Eigenvectors by definition cannot be the zero vector, after all the main interest in eigenvectors is for forming eigenbasis in which zero vectors are a big no. Also, eigenvalues themselves are defined as the values for which $\det(A-\lambda I) = 0$ so there must be a non-trivial solution. –  EuYu Dec 21 '12 at 6:07

1 Answer 1

up vote 1 down vote accepted

Try $v_1=(1,1,1)^T$, $v_2=(1-i\sqrt{3},-2,1+i\sqrt{3})^T$, $v_3 = \overline{v_2}$.

$A v_i = \lambda_i v_i$, where $A$ is the matrix above and $\lambda_i$ can be found by solving $\lambda^3-9\lambda^2+27 \lambda -28 = 0$. (By inspection, $4$ is a solution, and synthetic division results in $x^2-5x+7=0$.)

share|improve this answer
    
Can you suggest how to evaluate its complex Eigen values? –  Adnan Zahid Dec 21 '12 at 6:16
    
Yes, solve $x^2-5x+7=0$. –  copper.hat Dec 21 '12 at 6:18
    
Okay, thank you so much. –  Adnan Zahid Dec 21 '12 at 6:19
    
You are very welcome. –  copper.hat Dec 21 '12 at 6:19
    
Also suggest how to compute eigen vectors for identity matrix. –  Adnan Zahid Dec 21 '12 at 6:20

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.