Matrix $A$ is given by;

$A = \begin{bmatrix} 0 & 1 \\ -4 & 0 \end{bmatrix}$

First I find the eigen values as $\lambda_1$ and $\lambda_2$ as $2i$ and $-2i$. And then to find the eigen vector of the eigen value $2i$, I do;

$\begin{bmatrix}2i & -1\\4&2i\end{bmatrix}\cdot\begin{bmatrix}w_1\\w_2\end{bmatrix} = \begin{bmatrix}0\\0\end{bmatrix}$

from this point equations look like;

$\begin{align*}2iw_1 - w_2 &= 0 \\[0.4em] 4w_1+2iw_2 &= 0\end{align*}$

and I cannot do the elimination at this step because when I try to cancel one variable, the other one is cancelled out too..

  • $\begingroup$ By the very definition of eigenvalue, the system you get must be underdetermined. If it has only one solution it would be the zero vector, so the number couldn't be an eigenvalue. $\endgroup$ – egreg May 9 '14 at 16:41

Basically, the two equations are the same and you get from them, $$w_1=\frac{w_2}{2i}$$ So one eigenvector corresponding to the eigenvalue $2i$ is $[1\quad 2i]^T$ because it satisfies the equations. Remember, there can be infinitely many eigenvectors corresponding to a single eigenvalue.

  • $\begingroup$ I was missing the point that the two equations are the same! thanks you Samrat! $\endgroup$ – Yirmidokuz May 9 '14 at 16:23
  • $\begingroup$ @Yirmidokuz: If two equations simultaneously cancel out with each other, they are the same. $\endgroup$ – user88595 May 9 '14 at 16:27

You are supposed to get an under-determinate system. Eigenvectors are defined up to a constant. Add one more equation, say $w_1 + w_2 = 1$


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