Homework Help: Normalizing eigenvector with complex entries

1. Jan 7, 2008

R.Harmon

1. The problem statement, all variables and given/known data
Hi, I'm having a bit of a problem normalizing eigenvectors with complex entries. Currently the eigenvector I'm looking at is $$$\vec{v}= \left(\begin{array}{c} -2+i\\ 1 \end{array}\right)$$$

2. Relevant equations

3. The attempt at a solution

If the eigenvectors don't have complex elements I can do this, for example if i have $$$\vec{v}= \left(\begin{array}{c} 3\\ 1 \end{array}\right)$$$ and I want to normalize I know that this is the same as $$$\vec{v}=\left(\begin{array}{c} 3a\\ 1a \end{array}\right)$$$ and $$(3a)^2+a^2=1$$ so the normalized eigenvector is $$\vec{v}=\frac{1}{\sqrt{10}} \left(\begin{array}{c} 3\\ 1 \end{array}\right)$$. However with the first eigenvector using the same method I get $$(a(-2+i))^2+a^2=1$$ or $$a=\frac{1}{\sqrt{4-4i}}$$ giving the normalized eigenvector as $$\vec{v}=\frac{1}{\sqrt{4-4i}} \left(\begin{array}{c} -2+i\\ 1 \end{array}\right)$$ where as the solution should be $$\vec{v}=\frac{1}{\sqrt{6}} \left(\begin{array}{c} -2+i\\ 1 \end{array}\right)$$. Could someone please point out where I'm going wrong? Any help is appreciated.

2. Jan 7, 2008

nicksauce

I believe you should just divide by

sqrt( |-2 + i|^2 + |1|^2)
=
sqrt( 5 + 1) = sqrt(6)

3. Jan 7, 2008

R.Harmon

Unless my algebras gone out the window:
$$(-2+i)^2=(-2+i)(-2+i)=4-4i+(i^2)=3-4i$$
not 5?

4. Jan 7, 2008

nicksauce

|a + bi|^2 = (a + bi)(a - bi) = a^2 + b^2
or
|z|^2 = zz* (where z* is the complex conjugate)

so

|-2 + i|^2 = 4 + 1 = 5

5. Jan 7, 2008

R.Harmon

Ahh ok makes sense. Thanks alot, that was bugging me for hours!

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