Complex conjugation in inner product space?

  • #1
ATY
34
1

Main Question or Discussion Point

I got the following derivation for some physical stuff (the derivation itself is just math)
http://thesis.library.caltech.edu/5215/12/12appendixD.pdf
I understand everything until D.8.

So in the eqation ε is a symmetric matrix and δx(t) is just the difference between two points.

After D.7 they get the eigenvalue and eigenvectors from ε. The text says that my δx(t) gets aligned in the same direction as the the eigenvektor of the largest eigenvalue (this would be my explanation for the fact that there is no eigenvector in the scalar product, but this might be wrong, cause I do not know much about eigenvectors).
But what I don't get is why in D.8. my δx(t) is suddenly complex conjugated. I can not find the reason for this.
I would be really happy about any explanation.

have a nice day
ATY

PS: sorry for the weird titel. Had no clue how to describe my problem (mea culpa)
 

Answers and Replies

  • #2
Ssnow
Gold Member
509
149
Hi, I don't know if can help but ##\langle \overline{\delta(x)},\lambda \overline{\delta(x)} \rangle=\overline{\lambda} \langle \overline{\delta(x)}, \overline{\delta(x)} \rangle ## for the properties of the hermitian product and your eigenvector is the eigenvector of the conjugate ##\lambda##, so why the notation with the bar. But this is true only if the matrix has real entries so ##\overline{A}=A## ...
I think you must see how is defined the inner product on your manifold possible there are remarcable properties ...
 
  • #3
Svein
Science Advisor
Insights Author
2,022
648
Just a quick reminder: if you are considering complex inner products, you have [itex]\left\langle x, y \right\rangle = \overline{\left\langle y, x \right\rangle} [/itex].
 

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