Given the following matrix
$\begin{pmatrix} 0 & 1-i & 0\\ 1+i & 0 &1-i\\ 0& 1+i &0\\ \end{pmatrix}$
I have found the Eigenvalues $0, 2,-2$. But I have no idea how to calculate the corresponding Eigenvectors and I failed with Gaussian method. What could you recommend?
Thanks in advance!
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$\begingroup$the eigenvalues $\lambda$ are given by $$det\, A = det \pmatrix{-\lambda&1-i&0\\1+i& -\lambda&1-i\\0&1+i&-\lambda} = 0$$ expanding by the first row, we have $$0=-\lambda\left(\lambda^2 -(1-i)(1+i)\right)+(1-i)\lambda(1+i)=-\lambda^3 + 4\lambda = -\lambda(\lambda - 2)(\lambda + 2)$$ the eigenvalues of $A$ are $0, 2, -2.$
the eigenvector corresponding to the eigenvalue $0.$ we need to solve $$ \pmatrix{0&1-i&0\\1+i& 0&1-i\\0&1+i&0}\pmatrix{x\\y\\z} = \pmatrix{0\\0\\0}.$$
we can take an eigenvector to be $\pmatrix{1 - i\\0\\ -1 - i}.$
i will leave you the task of finding the other eigenvectors.
$\bf edit:$ now that egreg has done the eigenvalue $-2,$ i will finish it by finding the eigenvector corresponding to $2.$ if you look at the first equation $$-2x + (1-i)y = 0$$ so we will choose $$y = 2, x = 1-i. $$ now look at the last equation: that give $$(1+i)y -2z = 0 \to z = 1 + i $$ therefore an eigenvector corresponding to the eigenvalue $2$ is $$\pmatrix{1-i\\2\\1+i} $$
$\endgroup$ 4 $\begingroup$It's exactly the same as with real numbers, except that you have to do arithmetic with complex numbers. Thus for the eigenvalue $0$, start by interchanging the first and second rows, then multiply the first row by $1/(1+i) = (1-i)/2$, ...
$\endgroup$ $\begingroup$For the eigenvalue $-2$: \begin{align} \begin{bmatrix} 2 & 1-i & 0\\ 1+i & 2 &1-i\\ 0& 1+i &2\\ \end{bmatrix} &\to \begin{bmatrix} 1 & (1-i)/2 & 0\\ 1+i & 2 &1-i\\ 0& 1+i &2\\ \end{bmatrix} && R_1\gets \tfrac{1}{2}R_1 \\[6px]&\to \begin{bmatrix} 1 & (1-i)/2 & 0\\ 0 & 1 &1-i\\ 0& 1+i &2\\ \end{bmatrix} && R_2\gets R_2-(1+i)R_1 \\[6px]&\to \begin{bmatrix} 1 & (1-i)/2 & 0\\ 0 & 1 &1-i\\ 0& 0 &0\\ \end{bmatrix} && R_3\gets R_3-(1+i)R_2 \\[6px]&\to \begin{bmatrix} 1 & 0 & i\\ 0 & 1 &1-i\\ 0& 0 &0\\ \end{bmatrix} && R_1\gets R_1-\tfrac{1-i}{2}R_2 \end{align} Thus the equations can be written $$ \begin{cases} x_1=-ix_3\\ x_2=(i-1)x_3 \end{cases} $$ so an eigenvector is $$ \begin{bmatrix} -i\\ i-1\\ 1 \end{bmatrix} $$
$\endgroup$ 1 $\begingroup$I'll find an eigenvector corresponding to $\lambda = 0$. We want to find a vector $v = \left [ \begin{array}{ccc} v_1 & v_2 & v_3 \\ \end{array} \right ]^T$ such that
$$ \left [ \begin{array}{ccc} 0 & 1 -i & 0 \\ 1 + i & 0 & 1 - i \\ 0 & 1+i & 0 \\ \end{array} \right ] \left [ \begin{array}{c} v_1\\ v_2\\ v_3\\ \end{array} \right ] \;\; =\;\; \textbf{0}_{3\times 3}. $$
It should be clear from multiplication that $v_2 = 0$. The only other constraint is $(1+i)v_1 = (1-i)v_3$ or rather
$$ v_1 \;\; = \;\; \frac{1-i}{1+i} v_3 \;\; =\;\; \frac{1-i}{1+i}\cdot \frac{1-i}{1-i}v_3 \;\; =\;\; \frac{-2i}{2} v_3 $$
Or $v_1 = -iv_3$. The technique used above can always be used to get rid of complex numbers in the denominator of a fraction. Take $v_3 = 1$ and we have
$$ v \;\; =\;\; \left [ \begin{array}{c} -i \\ 0 \\ 1 \\ \end{array} \right] $$
Is one of the eigenvectors. The others can be found similarly, except you need to compute $(A - \lambda I)w=0$ where $A$ is the matrix given, $\lambda$ is an eigenvalue, and $w$ is a proposed eigenvector.
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