Normalising Imaginary Eigenvector

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To normalize the eigenvector ##\vec{e_{1}} = (^{1}_{i})##, one must use the inner product defined for complex vector spaces, which involves the complex conjugate. The squared length of the vector is calculated as ##\vec{e_{1}} \cdot \overline{\vec{e_{1}}}##, leading to a length of ##\sqrt{2}##. Normalization can then be achieved by dividing the vector by its length, resulting in ##\vec{e_{1}} = \frac{1}{\sqrt{2}} (^{1}_{i})##. The discussion emphasizes that complex numbers can be viewed as vectors in a one-dimensional complex vector space. Understanding these concepts is crucial for working with eigenvectors in complex systems.
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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