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This started with a question in the book Linear Algebra by S. Friedberg, A. Insel and L. Spence in which they define a vector space $\text{V} = \{(a_1, a_2, ...,a_n):a_i \in \mathbb{C}\}$, so a vector space over $\mathbb{C}$, and ask if it is a vector space over the field of real numbers.

I'm inclined to say this is false because $a_i$ could be any complex number, and multiplying by a scalar from $\mathbb{R}$ could still keep it in $\mathbb{C}$ right? In other words, if $\text{V}$ is defined over $\mathbb{C}$ and an element of $\text{V}$ with complex components is multiplied by a real scalar, its components could still be complex and non-real. In which case it is not an element in the field of $\mathbb{R}$.

The problem is that I've checked community solutions of the book online which say that it is in fact a vector field over the field of $\mathbb{R}$ and also consulted Apostol's Calculus: Vol 2, which says "Let $\text{V}= \mathbb{C}$, the set of all complex numbers, and define $ax$ to be multiplication of the complex number $x$ by the real number $a$. Even though the elements of $\text{V}$ are complex numbers, this is a real linear space because the scalars are real."

So I'm clearly wrong, but I fail to see why. How can a space be over a subset of a certain field by only limiting its scalar multiplication to that subset?

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    $\begingroup$ A real vector space is any vector space over the reals, it is a definition. In general, for a vector space $V$ over a field $\mathbb F$, the requirement is that $\{a\cdot v\mid a\in\mathbb F, v\in\mathbb V\}$ is a subset of $V$, not a subset of $\mathbb F$. In general $V$ need not be a field, for example the vector space of polynomials in one indeterminate. $\endgroup$ – Randy Marsh Nov 10 '20 at 14:50
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    $\begingroup$ @RandyMarsh So is it in essence a definition that when you change the field from which you take the scalar $a$, you change over which field the vector space is defined? Meaning "over" says nothing about the actual field of the components of the elements of V? $\endgroup$ – reveance Nov 10 '20 at 15:20
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    $\begingroup$ Yes to the first question, and not quite to the second. You can infer some information about the components of the elements of $V$ because the multiplication with the elements of $\mathbb F$ has to be compatible. For example, the components of the elements of $V$ can not be in a field that is properly contained in $\mathbb F$. Take $V=\mathbb Q$ and $\mathbb F=\mathbb R$. Then e.g. $\pi\cdot v\notin V$ whenever $v\neq 0$. $\endgroup$ – Randy Marsh Nov 10 '20 at 16:15
  • $\begingroup$ That's very insightful, thanks! $\endgroup$ – reveance Nov 10 '20 at 19:09
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If we have a vector space $V = \mathbb{C}^n$ over the field $\mathbb{R}$ you have indeed a vector space. Note that the operation that you have are the usual addition of vectors and the multiplication by aclarar is defined to be

$$\lambda \cdot (x_1,\dots,x_n) = (\lambda \cdot x_1,\dots, \lambda \cdot x_n) \quad \quad \forall (x_1,\dots,x_n) \in \mathbb{C}^n, \quad\forall \lambda \in \mathbb{R}.$$

Note that his operation if well-defined, since when you multiply $\lambda \in \mathbb{R}$ by each $x_i \in \mathbb{C}$ you get $\lambda \cdot x_i \in \mathbb{C}$ for all $i \in \{1,\dots,n\}.$ So the RHS of the equation above is in $\mathbb{C}^n.$

After realising this it is easy to prove that $V$ in addition with the usual sum of vectors in $\mathbb{C}^n$ and the multiplication by a scalar $\cdot$ as defined above is a vector space over $\mathbb{R}.$

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    $\begingroup$ But I fail to see why it then becomes a vector space over $\mathbb{R}$. How can a vector space be over $\mathbb{R}$ when it is defined as being over $\mathbb{C}$ if the scalar doesn't do anything that causes all of the components of the elements of V to become $\mathbb{R}$? $\endgroup$ – reveance Nov 10 '20 at 15:12
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    $\begingroup$ Note that this vector space is $\mathbb{C}^n$ over $\mathbb{R}.$ It just mean that the scalars that you will multiply are all real numbers $\endgroup$ – Air Mike Nov 10 '20 at 15:17
  • $\begingroup$ If you still have doubts please tell me :) $\endgroup$ – Air Mike Nov 10 '20 at 15:19
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    $\begingroup$ "Note that this vector space is $\mathbb{C}^n$ over $\mathbb{R}$" really made it a lot more clear to me :). So "over" really only says something about the scalars that you multiply the elements of V with? Nothing about the actual space that consists of all elements of V? $\endgroup$ – reveance Nov 10 '20 at 15:33
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    $\begingroup$ Exactly, $V$ is a vector space over a filed $\mathbb{K}$ means that the multiplication is between a scalar in $\mathbb{K}$ and a vector in $V$. Very good :) $\endgroup$ – Air Mike Nov 10 '20 at 15:35
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Well if you have an ambient vector space $V= K^n = \{(a_1,\ldots,a_n)\mid a\in K\}$, where $K$ is a field, then $V$ is also a vector space over any subfield $L$ of $K$.

For this, consider the scalar multiplication

$\kappa (a_1,\ldots a_n) = (\kappa a_1,\ldots,\kappa a_n)$

where $\kappa\in K$. This multiplication is also defined if $\kappa$ is from a subfield $L$. The addition of vectors doesn't change.

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It is an operation known as ‘restriction of scalars’. Any $\mathbf C$-vector space $V$ is also an $\mathbf R$-vector space just by asking the scalar multiplication to be restricted to the real numbers, and we have a relation between the dimensions as a $\bf C$- and $\bf R$-vector space if $V$ is finite-dimensional: $$\dim_{\bf R}V=\dim_{\bf C}V\dim_{\bf R}\mathbf C=2\dim_{\bf C}V.$$ Likewise, the same vector space cab be seem as a $\bf Q$-vectorspace, and in this case, it has infinite dimension as a vector space over $\bf Q$, because $\bf C$ has infinite dimension over $\bf Q$.

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