# Problem with null Eigen vectors.

$$\begin{pmatrix} 3 & 0 & 1\\ 1 & 3 & 0\\ 0 & 1 & 3 \end{pmatrix}$$

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

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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
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
Did you compute the characteristic polynomial of the matrix? –  Yury Dec 21 '12 at 6:00
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
@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

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 $x^2-5x+7=0$.)
Yes, solve $x^2-5x+7=0$. –  copper.hat Dec 21 '12 at 6:18