When denoting a particular vector space can the field that denotes the entries in the vectors of a vector space be different than the field that denotes the scalars that can be used on that vector space. For example if $V$ is a vector space denoted by $M_{m\times n}(R)$ do the scalars $c$ that act on that vector space have to be elements of $R$? I was reviewing the text "Linear Algebra" by Friedberg, Insel and Spence, and was a little confused by the following question:
Let $V$ denote the set of all $m\times n$ matrices with real entries; so $V$ is a vector space over $R$ by example 2 (note this was the general definition of matrices over $R$). Let $F$ be the field of rational numbers. Is $V$ a vector space over $F$ with the usual definitions of matrix addition and scalar multiplication?
If $c\in R$ then no because you could do scalar multiplication with an irrational number and that would take you out of the vector space over $Q$ because an irrational number times a rational number is an irrational number. Now if $c$ is in $Q$ then yes it is a vector space. Without $c$ being explicitly denoted do I assume it belongs to the field that was defined with the vector space? Also, can $c$ and $V$ be different fields when denoting a vector space?
Best Answer
Yes, it is a vector space over F.
A vector space over F is a set V (in your case $m\times n$ matrices with real entries) with operations $+:V\times V \to V$ and $\cdot:F\times V \to V$ satisfying certain axioms, as found for example on the wikipedia vector space page.
That these axioms are satisfied is easily verified. The first four amount to V being an abelian group, and that last four must be satisfied by F since it is subfield of R.
It also follows easily from these observations that if V is a vector space over a field K, then it is a vector space over any subfield F of K (i.e., the example above generalizes).
It is important to note, though, that V is NOT a finite dimensional vector space over F. The dimensional isn't even countable, in fact.