# Eigenvectors of $\left( \begin{array}{ccc} 0 & -b \\ a & 0 \end{array} \right)$

This is similar to my previous question in that I when I form a system of simultaneous equations and solve them all the terms cancel and I don't get any information on the eigenvectors. The matrix in this case is -

$J = \left( \begin{array}{ccc} 0 & -b \\ a & 0 \end{array} \right)$

I get complex eigenvalues, $\lambda_{1,2} = \pm \sqrt{ab}i$

The answer to my previous question is not applicable in this situation. How do I get the eigenvectors here?

## 3 Answers

The eigenvectors will be the vectors which, by definition, verify the relation:

$(J-\lambda·I)·w=0$

As we have $2$ eigenvalues $(\lambda_1,\lambda_2)$ we are going to have $2$ eigenvectors $(u,v)$:

$(J-\lambda_1·I)·u=\left( \begin{array}{ccc} -\sqrt{ab}i & -b \\ a & -\sqrt{ab}i \end{array} \right)\left( \begin{array}{ccc} u_1 \\ u_2 \end{array} \right)=0$

and

$(J-\lambda_2·I)·v=\left( \begin{array}{ccc} \sqrt{ab}i & -b \\ a & \sqrt{ab}i \end{array} \right)\left( \begin{array}{ccc} v_1 \\ v_2 \end{array} \right)=0$

So, for getting the values of $u$ and $v$ we have to solve the equations written above.

$-\sqrt{ab}i·u_1-b·u_2=0$

If we let $u_1$ be $1$ (we just care about de direction and not the magnitude) we can find the value of $u_2$:

$-\sqrt{ab}i-b·u_2=0$

$u_2=\frac{\sqrt{ab}}{b}i=\sqrt{\frac{ab}{b^2}}i=\sqrt{\frac{a}{b}}i$

This meaning that $u=(1,\sqrt{\frac{a}{b}}i)$

Following the same steps for $v$ leads to the result $v=(1,-\sqrt{\frac{a}{b}}i)$

Make an ansatz $v=\begin{pmatrix}1\\y\end{pmatrix}$ and solve $Jv=\lambda v$. You immediately get $-by=\pm\sqrt{ab}i$ (and $\pm\sqrt{ab}iy=a$), so $y=\mp\sqrt{\frac ab}i$.

• The $x$ and $y$ component of the eigenvectors in wolfram alpha are the opposite of what you have here - wolframalpha.com/input/… - Or are they just scaled? – sonicboom Oct 14 '14 at 21:47
• Note that if you multiply @Hagen answer $(1,\pm\sqrt{\frac{a}{b}}i)$ by $\pm\sqrt{\frac{b}{a}}i$ (with opposite signs +|-,-|+) you get wolfram's same result. – Ioannes Oct 14 '14 at 22:13

If $\lambda$ is an eigenvalue of a $2\times2$ matrix $A$, an eigenvector $v$ is a nonzero solution of $(A-\lambda I)v=0$. In your case, if $v=\begin{bmatrix}x_1\\x_2\end{bmatrix}$, we get the linear system $$\begin{bmatrix} -\lambda & -b \\ a & -\lambda \end{bmatrix} \begin{bmatrix}x_1\\x_2\end{bmatrix} =0$$ If we assume $b\ne0$ (or the matrix wouldn't be diagonalizable unless also $a=0$), the second equation is linearly dependent on the first (because we do know $\lambda$ is an eigenvalue), so the system becomes simply one equation: $$\lambda x_1 + bx_2 = 0$$ and an eigenvector will be $$\begin{bmatrix} b \\ -\lambda \end{bmatrix}$$