Normalizing eigenvector with complex entries

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Homework Help Overview

The discussion revolves around the normalization of eigenvectors that contain complex entries, specifically focusing on the eigenvector \(\vec{v}=\left(\begin{array}{c}-2+i\\1\end{array}\right)\). Participants are exploring the correct method to normalize such vectors and the calculations involved.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to normalize the complex eigenvector using a method similar to that used for real eigenvectors, leading to confusion about the normalization factor. Some participants suggest calculating the magnitude of the complex vector directly, while others question the algebraic steps taken in the calculations.

Discussion Status

Participants are actively engaging with the problem, with some providing clarifications on the calculation of the magnitude of complex numbers. There is a recognition of differing interpretations of the algebra involved, but no explicit consensus has been reached.

Contextual Notes

The original poster expresses uncertainty about their calculations and seeks assistance in identifying errors. There is an acknowledgment of the complexities involved in handling complex numbers in the context of eigenvector normalization.

R.Harmon
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Homework Statement


Hi, I'm having a bit of a problem normalizing eigenvectors with complex entries. Currently the eigenvector I'm looking at is \[\vec{v}=<br /> \left(\begin{array}{c}<br /> -2+i\\<br /> 1<br /> \end{array}\right)\]


Homework Equations





The Attempt at a Solution



If the eigenvectors don't have complex elements I can do this, for example if i have \[\vec{v}=<br /> \left(\begin{array}{c}<br /> 3\\<br /> 1<br /> \end{array}\right)\] and I want to normalize I know that this is the same as \[\vec{v}=\left(\begin{array}{c}<br /> 3a\\<br /> 1a<br /> \end{array}\right)\] and (3a)^2+a^2=1 so the normalized eigenvector is \vec{v}=\frac{1}{\sqrt{10}}<br /> \left(\begin{array}{c}<br /> 3\\<br /> 1<br /> \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}}<br /> \left(\begin{array}{c}<br /> -2+i\\<br /> 1<br /> \end{array}\right) where as the solution should be \vec{v}=\frac{1}{\sqrt{6}}<br /> \left(\begin{array}{c}<br /> -2+i\\<br /> 1<br /> \end{array}\right). Could someone please point out where I'm going wrong? Any help is appreciated.
 
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I believe you should just divide by

sqrt( |-2 + i|^2 + |1|^2)
=
sqrt( 5 + 1) = sqrt(6)
 
Unless my algebras gone out the window:
(-2+i)^2=(-2+i)(-2+i)=4-4i+(i^2)=3-4i
not 5?
 
|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
 
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Likes   Reactions: debjit625
Ahh ok makes sense. Thanks a lot, that was bugging me for hours!
 

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