Can the field of complex numbers be regarded as a vector space?

Question

This is from Huffman & Kunze.

"The field C of complex numbers may be regarded as a vector space over the field ℝ of real numbers. More generally, let F be the field of real numbers and let V be the set of n-tuples $\ α=(x_1,..,x_n)$ where $\ x_1,..,x_n$ are complex numbers. Define addition of vectors and scalar multiplication by:

a) let $\ α=(x_1,...,x_n)$ and $\ β=(y_1,...,y_n)$ then $\ α+β=(x_1+y_n,...,x_1+y_n)$

b) let $\ α=(x_1,...,x_n)$ then $\ cα=(cx_1,...,cx_n)$

In this way we obtain a vector space over the field ℝ which is quite different from $\ ℂ^n$ and $\ ℝ^n$"

1) How can the field of complex numbers be considered a vector space? I am asking because I don't think that the field of complex numbers can have n-tuples s.t. n>1, right? Since I believe that in order for an object to be interpreted as a vector, it must have at least an ordered pair of numbers.

2) When it was stated that the constructed vector space is quite different, shouldn't it be "entirely different" because the vector space has n-tuples s.t. n>1.

I am really confused.

Answer 1

It's not necesarily the case that a vector space $V$ over a field $K$ consists of tuples. It's an entirely abstract notion: we only need some addition with $0$ and $-$ (an Abelian group in abstract algebra) and a scalar multiplication which is just a function $K \times V \to V (k, v ) \to k\cdot v$ that satisfies some consistency axioms like

$(kk')\cdot v = k(k'\cdot )v$, tying multiplication in the field $K$ to scalar multiplication, as well as $1\cdot v = v$, $k\cdot(v+w) = k \cdot v + k\cdot w$ and $(k+k')\cdot v = k\cdot v + k'\cdot v$.

But if $K \subset V$ and $V$ is a field too, we can just use the field multiplication in $V$ restricted to $K$ on the left to get a scalar multiplication for free, and the field axioms garantuee all the vector space axioms. So $V$ is a vector space over $K$.

This is also sometimes used to see $\mathbb{R}$ as a vector space over $\mathbb{Q}$ (which has huge dimension), and in a finite field $K$ we always have a subfield of the form $F_p$ for $p$ prime, and the vector space argument then gives us that $|K| = p^n$ for some $n$ ($n$ is the dimension of $K$ as a vector space over $F_p$...)

It's just abstract algebra, not tuples/matrices etc. Later you'll probably see "vector spaces", consisting of functions or equivalence classes of functions, and linear maps that are not "matrices".

Answer 2

It's because it fulfils the requirements of a vector space $V$ over $F$. They are:

• $V$ is an abelian group
• $\alpha(\beta u) = (\alpha\beta)u$
• $1 u = u$
• $\alpha (u+v) = \alpha u+\alpha v$
• $(\alpha+\beta)u = \alpha u + \beta u$

Where $u, v\in V$ and $\alpha, \beta\in F$ and $1$ being the multiplicative unity of $F$.

$\mathbb C$ fulfils all these requirements (with $\mathbb R$ being the field). The complex numbers kind of contains ordered pair, for example $z=x+iy$ contains the ordered pair $(x,y)$. The requirement to contain that doesn't mean that they must do that that explicitely (besides there's no such requirement - the definition allows for a vector space to be one or zero dimensional which means there would be no proper tuples at all in any way).

What is the distinction between quite different and entirely different? I don't know, but of course they are different. The constructed space has dimension $2n$ while $\mathbb C^n$ and $\mathbb R^n$ only has dimension $n$. The difference to $\mathbb R^{2n}$ on the other hand is not within the scope of the linear space part - regarded as linear space only they are isomorphic.