# 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. Commented Nov 10, 2020 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? Commented Nov 10, 2020 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 $v\neq 0$. Commented Nov 10, 2020 at 16:15
• That's very insightful, thanks! Commented Nov 10, 2020 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 $\mathbb{R}$? Commented Nov 10, 2020 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 Commented Nov 10, 2020 at 15:17
• If you still have doubts please tell me :) Commented Nov 10, 2020 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? Commented Nov 10, 2020 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 :) Commented Nov 10, 2020 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 $$\bf Q$$.