Normalising Imaginary Eigenvector

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Discussion Overview

The discussion revolves around the normalization of an imaginary eigenvector encountered while solving a system of coupled differential equations. Participants explore the properties of complex vectors and the appropriate methods for normalization within a complex vector space.

Discussion Character

  • Exploratory, Technical explanation, Conceptual clarification

Main Points Raised

  • One participant presents an eigenvector ##\vec{e_{1}} = (^{1}_{i})## and questions how to normalize it, noting an issue with the sum of squares yielding zero.
  • Another participant suggests that in a complex vector space, an inner product must be introduced that satisfies the property ##\langle x, y\rangle = \overline{\langle y,x\rangle}##.
  • A different participant states that the squared length of a complex vector can be defined using the product of the vector and its complex conjugate, which would yield a length of ##\sqrt{2}##.
  • One participant acknowledges the relationship between complex numbers and vectors, indicating a conceptual understanding of complex numbers as vectors.
  • A later reply agrees with the previous statement but clarifies the terminology, emphasizing that a complex number is an element in a complex one-dimensional vector space.

Areas of Agreement / Disagreement

Participants express agreement on the conceptual understanding of complex numbers as vectors, but there is no consensus on the normalization process or the implications of the eigenvector's properties.

Contextual Notes

The discussion does not resolve the normalization issue and leaves open questions regarding the definitions and properties of inner products in complex vector spaces.

BOAS
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Hello,

whilst solving a system of coupled differential equations I came across an eigen vector of ##\vec{e_{1}} = (^{1}_{i})##.

Assuming that this is a correct eigenvector, how do I normalise it? I want to say that ##\vec{e_{1}} = \frac{1}{\sqrt{2}} (^{1}_{i})## but if I sum ##1^{2} + i^{2}## I get zero.

It seems sensible to me that the vector's length is root two, but how do I justify this, if at all?

Thank you.
 
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In a complex vector space, you have to introduce an inner product which satisfies ##\langle x, y\rangle = \overline{\langle y,x\rangle}##.
 
The squared length of a complex vector v is defined by v.v(bar) where v(bar) is the complex conjugate, i believe. That will give you sqrt2
 
Ah of course. A complex number is essentially a vector.

Thank you.
 
BOAS said:
Ah of course. A complex number is essentially a vector.

Thank you.

I agree if you take away the "essentially". It is an element in a complex one-dimensional vector space. :)
 
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