Eigenvectors of a complex matrix

Question

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!

You need to solve $MV = \lambda V$, or $(M - \lambda I)V = 0$, which requires you to know how to compute null space basis. — DanielV, Mar 23 '15 at 23:41
I know that, but I dont know how to deal with the complex numbers, e.g. how do they vanish in Gaussian elimination? — , Mar 23 '15 at 23:42

abel · Accepted Answer · 2015-03-24T00:27:35.253

0

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 $$\pmatrix{1-i\\2\\1+i} $$

edited Mar 24 '15 at 00:27

answered Mar 23 '15 at 23:56

abel

29,170

1

Thanks for your answer, but I had already found the eigenvalues. I asked for a strategy of finding the eigenvecorts. Did you just see the eigenvector for 0 or did you use Gaussian Elimination? I checked that it is one, but I couldnt find it. Could you give me any first step or hint how to proceed, then I would try myself. Thank you! – Mar 23 '15 at 23:59
@Ratatouille, eigenvector corresponding to the zero eigenvalue was made easier because $y = 0.$ i computed the eigenvalues because you said you could not find any eigenvectors. if you can't find the eigenvector for $2$ or $-2$ i can do it. it is just like with the real numbers. now you need to complex number arithmetic. – abel Mar 24 '15 at 00:02
Thank you! In this situation, $(i\pm 1)$-factors, this is an efficient way to 'see' the eigenvectors. Great! – Mar 24 '15 at 00:28
@Ratatouille, you are welcome. once we are sure that we have the right eigenvalue, you can make this kind of simplifications. – abel Mar 24 '15 at 00:29

score 0 · Answer 2 · answered Mar 23 '15 at 23:57

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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 $1/(1+i) = (1-i)/2$, ...

answered Mar 23 '15 at 23:57

Robert Israel

448,999

Mnifldz · Answer 3 · 2015-03-24T00:10:33.320

0

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.

edited Mar 24 '15 at 00:10

answered Mar 23 '15 at 23:59

Mnifldz

12,880

somehow you have got that $(1-i)/(1+i) = 1-i...$ I get $v_1=-iv_3$ – danimal Mar 24 '15 at 00:02
@Mnifidz, please check your work. – abel Mar 24 '15 at 00:04
@abel Yes, you're right. Sorry for the mistake. – Mnifldz Mar 24 '15 at 00:08
@Mnifldz, that is fine. we all make mistakes. – abel Mar 24 '15 at 00:09

score 0 · Answer 4 · answered Mar 24 '15 at 00:08

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 $$ \begin{bmatrix} -i\\ i-1\\ 1 \end{bmatrix} $$

Thank you, this was very helpful! Especially the step $R_2-(1+i)R_1$ didnt come to my mind. But now I see! :) — , Mar 24 '15 at 00:26

Eigenvectors of a complex matrix

4 Answers4