Normalising Imaginary Eigenvector

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SUMMARY

The discussion focuses on normalizing the imaginary eigenvector ##\vec{e_{1}} = (^{1}_{i})## encountered while solving coupled differential equations. The correct normalization involves using the inner product defined in complex vector spaces, specifically ##\langle x, y\rangle = \overline{\langle y,x\rangle}##. The squared length of the vector is calculated using the formula v.v(bar), leading to a normalization factor of ##\frac{1}{\sqrt{2}} (^{1}_{i})##. This confirms that the length of the vector is indeed root two.

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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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