# Why is the linear space V = $\mathbb{C}$ a real linear space when scalars are from $\mathbb{R}$

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?

• 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. – Randy Marsh Nov 10 '20 at 14:50
• @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? – reveance Nov 10 '20 at 15:20
• 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 $v\neq 0$. – Randy Marsh Nov 10 '20 at 16:15
• That's very insightful, thanks! – reveance Nov 10 '20 at 19:09

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 $$\mathbb{R}.$$

• 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 $\mathbb{R}$? – reveance Nov 10 '20 at 15:12
• 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 – Air Mike Nov 10 '20 at 15:17
• If you still have doubts please tell me :) – Air Mike Nov 10 '20 at 15:19
• "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? – reveance Nov 10 '20 at 15:33
• 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 :) – Air Mike Nov 10 '20 at 15:35

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.

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 $$\bf Q$$.