Can the scalars of a vector space be of a different field than the entries of the vector space?

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?

• Example: If $F,E$ are fields s.t. $E \subset F$ then $F$ can be considered as $E$-vector space. – flawr Oct 17 '14 at 17:06
• Ok in Friedberg et al the definition goes like so: A vector space V over a field F consists of a set on which two operations are defined so that for each of elements x, y, in V there is a unique x + y in V, and for each element a in F and each element x in V there is a unique element ax in V, such that the following conditions hold. (The conditions are assumed to be know) By that definition it seems a scalar $a$ belongs to the same $F$ that $V$ is over. Further more, they go on to say "The elements of the field $F$ are called scalars and the elements of the vector space are called vectors. – vajra78 Oct 17 '14 at 18:02
• So I think it's been said but scalars can only be from fields that are subsets of the field on which a particular vector space is defined to be over? So if $V$ is the set of all $M_{m\times n}(Q)$ you cannot have scalars from any element of $R$. – vajra78 Oct 17 '14 at 18:06

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.

First, note that a vector space is not necessarily a field. For example, $V$ (as you defined above) is not a field because not every matrix is invertible. However $V$ is certainly a vector space over $R$ AND over $Q$. If you accept the fact that $V$ is a vector space over $R$, then the same is true over $Q$. This is simply because $Q$ is a field with $Q \subset R$. Nothing else has to be verified. Note that if you take matrices with entries from $Q$, then this will be a vector space over $Q$ but not over $R$. This is because of the reason you stated above: if you multiply an irrational by a rational the result is an irrational (unless the rational is zero). Thus if you multiply an irrational scalar by a matrix with entries from $Q$, you will not in general get a matrix with entries from $Q$, hence it is not a vector space over $Q$

There is a little abuse of notation in books. Usually when talking about general vector spaces they will say let V be a vector space without mentioning the field. Always keep in mind that if V is a vector space, then it must be a vector space over some field (perhaps more than one, as in your example).

• I corrected my question. I meant entries of a vector that is an element of a vector space. Clearly a vector space and field are two different things. Although, a field can be a vector space but a vector space is not a field. – vajra78 Oct 17 '14 at 17:27

Yes, the scalars can be from a different field than the entries of the vectors. This is simply because we know how to define, in this case, rational numbers acting on real numbers by multiplication, and this satisfies the vector space axioms. If you review the axioms, you'll see there's nothing there about entries of vectors at all-so there's no way they could rule this out.

• But only if the field of scalars is a subset of the field that the vector space is defined on, right? – vajra78 Oct 17 '14 at 18:04
• That's pretty close to true. It's possible to go in the other direction in certain cases: for example, we can make the complex numbers act on an even-dimensional real vector space, though not in a uniquely defined way. – Kevin Arlin Oct 17 '14 at 19:10
• Is that the other way? Aren't the real numbers a subset of the complex numbers? – vajra78 Oct 18 '14 at 5:23
• Yes, they are, so the real numbers can naturally be the scalars for any complex vector space. I'm saying in some cases you can actually extend the scalars in the other direction. But this isn't an important point-feel free to ignore it. – Kevin Arlin Oct 18 '14 at 16:23

I think you're misinterpreting the question. Let's take a look at it:

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?

Recall that a vector space is a set of indexed arrays (vectors) whose entries come from a field (scalars) that satisfy the vector space axioms. The field is required in the definition of the vector space just as much as the operations. It should not be taken for granted that addition is generally just to mean addition in the elementary sense. Certain fields can be paired with very odd operations that still satisfy the vector space axioms, and so we call the operation "addition" in the context of discussion of the vector space. Really, one might consider a vector space itself as a tuple of three elements($\mathbb{F},[\;]^{m \times n},+)$ with three entries that are first a field of scalars, an array structure to hold the scalars, and third an operation to combine the arrays that behaves like addition in certain ways.

In the first sentence of the question, the author is establishing that the set of $m \times n$ arrays of real entries with Euclidean vector addition satisfies the axioms of a vector space. In the tuple notation, $(\mathbb{R},[\;]^{m \times n},+)$ is a vector space.

In asking "Is V a vector space over F?", the author cannot possibly mean "Is the set of $m \times n$ matrices with real entries a vector space over $F$?" because to say "a vector space over $F$" means that the entries come from $F$, so the answer is a trivial no. It is trivial that $\mathbb{R}^{m \times n}$ is not a vector space over $F$, because by it's very definition it has as its field of scalars the real numbers.

I don't believe that the author intended to ask a trivial question. Instead, the task is to determine if in the definition of $V$ one were to replace "real" with "rational", would the resulting $V$ given $(F,[\;]^{m \times n},+)$ also be a vector space. The question is non-trivial because $F$ is a subset of $\mathbb{R}$, so in this case if $(F,[\;]^{m \times n},+)$ is a vector space, then it is a subspace of $(\mathbb{R},[\;]^{m \times n},+)$ so the result can be demonstrated by showing that the set is closed under linear combinations.

• A vector does not have "entries"; it's an abstract algebraic object. Real matrices are closed under addition, and closed under rational scalar multiplication; and the addition and multiplication satisfy all axioms, so the $\mathbb R$ matrices are a vector space over $\mathbb Q$ . Consider also that $\mathbb R$ itself is a vector space over $\mathbb Q$ . – mr_e_man Oct 3 '18 at 3:53
• @mr_e_man, was that comment directed at my answer or at the OP? I feel like my answer addresses your concern directly. – Andrew Oct 15 '18 at 1:55
• "Is the set of $m\times n$ matrices with real entries a vector space over $F$ [ $=\mathbb Q$ ] ?" The answer is yes. The matrices do not need to be restricted to rational entries. The problem is that "entries" is not well-defined for abstract vector spaces. – mr_e_man Oct 17 '18 at 5:56
• I know it’s yes- I gave the OP advice on how to demonstrate that. I can’t tell if your comment is a criticism of my answer or of the question, because you seem to object to the OP’s use of “entries”. Your objection is frivolous since the question was asked in a linear algebraic context, not an abstract one, and also because the author of the textbook from which he got the question uses the word “entries” himself. It’s clear that “entries” means “the subset of the field that are used as scalars in the set of vectors described; for instance to say “two by two real matrices with even entries” tha – Andrew Oct 21 '18 at 19:52
• Your final paragraph suggests the answer is no: "if in the definition of V one were to replace 'real' with 'rational' ", this would produce "Let V denote the set of all $m\times n$ matrices with rational entries", which is the restriction I criticize. ...What do you mean by "the subset of the field that are used as scalars"? Do the phrases associate left or right, $(AB)C$ or $A(BC)$ ? In the first case, which field? In the second case, which subset? ...And I object not to "entries" of matrices (defined as arrays of numbers), but to "entries" of vectors (defined by algebraic axioms). – mr_e_man Oct 23 '18 at 2:13