Meaning of, and how to verify, a vector space *over* $\mathbb{R}$

In Axler's book on Linear Algebra he writes ($\mathbb{F}$ is here either $\mathbb{R}$ or $\mathbb{C}$):

The scalar multiplication in a vector space depends upon $\mathbb{F}$. Thus when we need to be precise, we will say that $V$ is a vector space over $\mathbb{F}$ instead of saying simply that $V$ is a vector space. For example, $\mathbb{R}^n$ is a vector space over $\mathbb{R}$, and $\mathbb{C}^n$ is a vector space over $\mathbb{C}$.

What is actually meant by $\mathbb{R}^n$ being a vector space over $\mathbb{R}$ and how can one verify that it is? What are all the implications of this? Is it just that any scalar multiplication in $\mathbb{R}^n$ depends upon numbers in $\mathbb{R}$?

To me, $\mathbb{R}^n$ feels larger than $\mathbb{R}$, so I find the wording, that one is over the other, hard to digest...

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A vector space must satisfy the condition that if $v,w$ are elements in the vector space, so is $av + bw$ for any $a,b$ elements of the base field. So the "linear combination" condition depends on what you specify as a base field. A vector space over $\mathbb{R}$ just means that you are looking at the base field being $\mathbb{R}$. – Willie Wong Aug 31 '11 at 11:19
For example, taking complex linear combinations of vectors in $\mathbb{R}^n$ will easily make you end up "outside" $\mathbb{R}^n$... – Willie Wong Aug 31 '11 at 11:20
@Willie: That's the answer, why not post it as such? – Rahul Aug 31 '11 at 11:26
@Rahul: I fear that I may be just re-stating what Jodles could've read in a textbook (since that feels like what I would write if I wrote my own textbook). I feel like I don't quite understand his confusion in the last sentence, so I want to wait and see before posting an answer. – Willie Wong Aug 31 '11 at 11:34
@Jodles: have you seen anything on vector spaces over finite fields, e.g., $\mathbb Z/p$; $p$ a prime? that will be radically different from working over $\mathbb R$ or $\mathbb C$; for one thing, you have finite sums that are zero, which does not happen for either of $\mathbb R$ ,nor for $\mathbb C$ – gary Aug 31 '11 at 11:56

When you give the axioms of a vector space $V$, you have to talk about a certain field: for example, in the axioms that talk about scalar multiplication, you'll say something like "for every $x \in \mathbb{F}$ and every $v \in V$ there's an element $xv$ of $V$....". In this case we say that "$V$ is a vector space over $\mathbb{F}$".

So you're right: $V$ is a vector space over $\mathbb{F}$ means that the scalars you can multiply by are elements of $\mathbb{F}$. It's just a conventional use of the word over.

The reason to mention it is that sometimes which field you are working over is not obvious. Take $\mathbb{C}^n$. It is a vector space over $\mathbb{C}$ in the obvious way, but it is also a vector space over $\mathbb{R}$ (if you can use elements of $\mathbb{C}$ as scalars then you can certainly use $\mathbb{R}$ as $\mathbb{R} \subset \mathbb{C}$). What's more it has dimension $n$ over $\mathbb{C}$ but $2n$ over $\mathbb{R}$ so it really matters which field you work over.

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+1 for viewing $\mathbb C^n$ as a vector space over $\mathbb R$ and $\mathbb C$ (it's also perhaps implicit in Willie Wong's comment). – Srivatsan Aug 31 '11 at 11:32
But an interesting question is whether we can make $\mathbb R^n$ as a vector space over $\mathbb R$ (same for $\mathbb C^n$)in more than one way, i.e., can we define an action of the scalars on the vectors (different from usual multiplication) that will give us a vector space too? – gary Aug 31 '11 at 12:15
@gary, the answer to your question is yes. Post it as a new question, and someone will answer it. Alternatively, go through the exercises in a good linear algebra textbook, and you'll probably find an example. – Gerry Myerson Aug 31 '11 at 13:44

In fact, $\mathbb{C}^n$ is an $n$-dimensional vector space over $\mathbb{C}$, but $\mathbb{C}^n$ is also a $2n$-dimensional vector space over $\mathbb{R}$. When you understand why that is so, then you will know the answer to your question.

You can find a set $\{\mathbf{v}_1,\ldots,\mathbf{v}_n\}$ of $n$ vectors such that every vector $v\in\mathbb{C}^n$ can be expressed as a linear combination $c_1\mathbf{v}_1+\cdots+c_n\mathbf{v}_n$, where $c_1,\ldots,c_n$ are complex numbers.

You can also find a set $\{\mathbf{v}_1,\ldots,\mathbf{v}_{2n}\}$ of $2n$ vectors such that every vector $v\in\mathbb{C}^n$ can be expressed as a linear combination $c_1\mathbf{v}_1+\cdots+c_{2n}\mathbf{v}_{2n}$, where $c_1,\ldots,c_{2n}$ are real numbers.

Thus any point in $\mathbb{C}^n$ can be specified by $n$ complex numbers or by $2n$ real numbers.

The set $n$ vectors $(1,0,\ldots,0,0)$, $(0,1,0,\dots,0)$,$(0,0,1,0,\ldots,0),\ldots,(0,0,\ldots,0,1)$ can serve in the first role above, where the scalars by which they will be multiplied are complex numbers.

The $2n$ vectors $(1,0,\ldots,0,0)$, $(0,1,0,\dots,0)$,$(0,0,1,0,\ldots,0),\ldots,(0,0,\ldots,0,1)$, $(i,0,\ldots,0,0)$, $(0,i,0,\dots,0)$,$(0,0,i,0,\ldots,0),\ldots,(0,0,\ldots,0,i)$ can serve in the second role above, where the scalars by which they will be multiplied are real numbers.

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You can also find a set $\{\mathbf{v}_1,\ldots,\mathbf{v}_{2n}\}$ of $2n$ vectors such that every vector $v\in\mathbb{C}^n$ can be expressed as a linear combination $c_1\mathbf{v}_1+\cdots+c_{2n}\mathbf{v}_{2n}$, where $c_1,\ldots,c_{2n}$ are complex numbers. – Rasmus Aug 31 '11 at 12:33
What I'm trying to say is that linear independence should be mentioned here. – Rasmus Aug 31 '11 at 12:46
+1 Thank you for the detailed explanation, Michael! – Jodles Aug 31 '11 at 14:49