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1. Jun 12, 2017

### Buffu

I am confused why is space over field $R$ not over field $C$ ? The entries in each vector is an element of $\Bbb C$ not $\Bbb R$.

2. Jun 12, 2017

### Staff: Mentor

I'm confused, too. Can you sort this out? Where's the quotation from? Is $F$ meant to be a subfield of $\mathbb{R}$? And is the scalar field of $V$ now $F$ or $\mathbb{R}\,$? Or $\mathbb{C}$ as that's where the vector operations are defined for?

I can only guess, that the statement is: Complex and real vector spaces are different and both differ from vector spaces with a scalar field like e.g. $F=\mathbb{Q}(\pi,\sqrt{2},e,\log 2)$ or whatever. To me this statement is a total mess.

3. Jun 12, 2017

### Buffu

It is written $F$ is a field of $R$. I don't think it matters if the scalar field is $R$ or $F$ since both are same. It is certainly not $C$.

4. Jun 12, 2017

### Buffu

Original statement :

This bit might also help :

5. Jun 12, 2017

### Staff: Mentor

So we have $F \subseteq \mathbb{R} \subset \mathbb{C}$. Next we have a vector space $V_\mathbb{K}$ of finite dimension $n$ and a field of scalars $\mathbb{K} \in \{F,\mathbb{R},\mathbb{C}\}$. All versions of $V_\mathbb{K}$ lead to different vector spaces, despite the fact that they all are $n-$dimensional.

I do not understand the remark $\alpha = (x_1,\ldots ,x_n) \in \mathbb{C}^n$.
Essential to the vector space $V_\mathbb{K}$ is where $c$ is from, and that is $c \in \mathbb{K}$.

E.g. let's consider $V := \mathbb{C}^2_\mathbb{R}$ and $W:= \mathbb{C}^2_\mathbb{C}$. Then $(i,2i)$ and $(1,2)$ are two different vectors in both, but in $V$ they point in two different directions (i.e. they are linearly independent), whereas in $W$ there is an equation $-i \cdot (i,2i) = (1,2)$ which means one is a multiple of the other and thus point in the same direction (and they are linearly dependent). All because we have $c= -i$ available for $W$ which is not available for $V$.

Last edited: Jun 12, 2017
6. Jun 12, 2017

### Buffu

Why $K \in \{F, \Bbb R, \Bbb C\}$ ? Should not it be just $K = F$ ?

Neither do I.

7. Jun 12, 2017

### Staff: Mentor

I've chosen the most general case, because the author(s) introduced a field $F$ of real numbers, e.g. $\mathbb{Q}(\pi)$, then also spoke about the entire real number field $\mathbb{R}$ as well as of complex numbers $\mathbb{C}$. I thought a neutral $\mathbb{K}$ would split this Gordian knot of fields.
I guess it is because of the example I added in my previous post.

8. Jun 12, 2017

### Buffu

Nevertheless I think I got it.
I think that $\alpha \in \Bbb C^n$ is to show that there can be different spaces for the same vector set $V$ like $\Bbb C^n$ and vector space of this example.