Normalization of an Eigenvector in a Matrix

[Dwye]

TL;DR Summary

I am working through some linear algebra questions in the Griffith's Book "Introduction to Quantum Mechanics", and I am unsure why a constant of 1/sqrt(2) is added into the answer.
I understand that the question is detailing a rotation about axis x & y, and that 1/sqrt(2) is the value of 45 degrees for both Sin and Cos, is this the reason for the addition; a generalization?
In fact I have seen this number quite a lot in Quantum Mechanics, is there something more to this number?

 

[vanhees71]

Do you know, how to get the norm of a vector in $\mathbb{C}^2$ or, more generally, how to define a scalar product on a complex vector space? It's very important to get these concepts right, before starting to study quantum theory, for which you need the "infinite-dimensional version" of these ideas, the socalled (separable) Hilbert space (more precisely what physicists do with this is rather the extension to a "rigged Hilbert space").

 

[PeroK]

Summary:: I am working through some linear algebra questions in the Griffith's Book "Introduction to Quantum Mechanics", and I am unsure why a constant of 1/sqrt(2) is added into the answer.
I understand that the question is detailing a rotation about axis x & y, and that 1/sqrt(2) is the value of 45 degrees for both Sin and Cos, is this the reason for the addition; a generalization?
In fact I have seen this number quite a lot in Quantum Mechanics, is there something more to this number?

The vector $(1, i)$ has magnitude $\sqrt 2$, so Griffiths decided to normalise it. The question doesn't actually ask for a normalised eigenvector, so $(1, i)$ would be just as valid an answer.

In QM it's generally a good idea to normalise vectors. The factor here has nothing directly to do with being the sine of $45°$. If the eigenvector were $(1, 2i)$, then the normalisation factor would be $\frac 1 {\sqrt{5}}$.

 

[Dwye]

The vector $(1, i)$ has magnitude $\sqrt 2$, so Griffiths decided to normalise it. The question doesn't actually ask for a normalised eigenvector, so $(1, i)$ would be just as valid an answer.

In QM it's generally a good idea to normalise vectors. The factor here has nothing directly to do with being the sine of $45°$. If the eigenvector were $(1, 2i)$, then the normalisation factor would be $\frac 1 {\sqrt{5}}$.

The vector $(1, i)$ has magnitude $\sqrt 2$, so Griffiths decided to normalise it. The question doesn't actually ask for a normalised eigenvector, so $(1, i)$ would be just as valid an answer.

In QM it's generally a good idea to normalise vectors. The factor here has nothing directly to do with being the sine of $45°$. If the eigenvector were $(1, 2i)$, then the normalisation factor would be $\frac 1 {\sqrt{5}}$.

Thank you very much!