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We have a matrix $A = \left( \begin{array}{ccc} 0 & 1 \\ -1 & 0 \\ \end{array} \right)$

I have found the eigenvalues to be simple and equal to $$\lambda_1 = i$$ $$\lambda_2 = -i$$

Computing the eigenvectors we have:$\left( \begin{array}{ccc} -i & 1 \\ -1 & -i \\ \end{array} \right)s_1 = 0$ and :$\left( \begin{array}{ccc} i & 1 \\ -1 & i \\ \end{array} \right)s_2 = 0$

What are the eigenvectors?

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Eigenvectors are solutions to the equation

$A v= \lambda v$.

Since you know both $A$ and $\lambda$, you can solve for $v$:

$$\begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \begin{bmatrix} v_0\\v_1 \end{bmatrix}= \begin{bmatrix} i v_0\\i v_1 \end{bmatrix}$$

In other words, you need to solve the linear equations: $v_1=iv_0$ and $-v_0=iv_1$

The other eigenvector is just as easy.

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  • $\begingroup$ I would say you need to solve either equation, not both. $\endgroup$ – Miguel Jan 8 '16 at 11:08
  • $\begingroup$ Sure, in this case the linear equations are redundant, but in general you need to solve the complete system $\endgroup$ – co9olguy Jan 8 '16 at 12:53
  • $\begingroup$ It is a good practice to work with the complete system, but it is also important to know that it will always be redundant, since any combination of eigenvectors is also an eigenvector. In other words, the eigenvectors of an eigenvalue form a linear space. $\endgroup$ – Miguel Jan 8 '16 at 13:27
  • $\begingroup$ I'm not sure if we're talking about the same thing here. I simply meant that the linear equations to solve for a particular eigenvector will not always be redundant like they are in the above example $\endgroup$ – co9olguy Jan 8 '16 at 13:33
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Hint matrix A is rotation matrix where $\theta=90$ you can write $i=cis(90)$ now do simple caculation which is usually done that is $(A-(cis(90)I_2)x=0$ you will get the eigenvectors.

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