Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Theorem: Let V be a vector space over the field K, and let W be a subset of V. Then W is a subspace if and only if it satisfies the following three conditions:

  • The zero vector, 0, is in W.
  • If u and v are elements of W, then any linear combination of u and v is an element of W
  • If u is an element of W and c is a scalar from K, then the scalar product c u is an element of W

I was just wondering if there was a mathematical way of expressing these facts which are necessary to prove a the existence of a subspace in a subset.

Here is what I've come up with I don't know if it's correct though:

  • $0 \in$ W
  • $u + v = v + u$ $\forall$ $u,v \in$ W
  • Couldn't figure out how to properly express this.
share|cite|improve this question
"$u+v = v+u$ [...]" Excuse me? Everything already takes place inside the vector space $V$, so this is always going to be true. – kahen Nov 5 '11 at 22:10
Also, one generally puts quantifiers before the formula they refer to, not after. – Arturo Magidin Nov 5 '11 at 22:34
One thing to note is that your original statement is a way to mathematically express those conditions. Words have just as big a place in math as symbols. – Riley E Nov 5 '11 at 23:38
up vote 2 down vote accepted

The first one is essentially correct, though you should use $\mathbf{0}$ (boldface) to distinguish it from the scalar $0$, just like the original does.

It is more common to put quantifiers first and the clause to which they apply following them, in parenthesis. So you would write the second one as $$\forall u,v\in W (u+v=v+u).$$ That said, this does not correspond to the second statement above. Instead, what it says is that addition is commutative for elements of $W$ (it says: if $u$ and $v$ are any vectors in $W$, then the result of adding $u$ to $v$ is the same as the result of adding $v$ to $u$, which is not "any linear combination of $u$ and $v$ is an element of $W$).

Instead, you want to say something like (here $K$ is the field of scalars) $$\forall \alpha,\beta\in K\Biggl(\forall \mathbf{u},\mathbf{v} \Bigl( \mathbf{u},\mathbf{v}\in W\longrightarrow \alpha \mathbf{u}+\beta\mathbf{v}\in W\Bigr)\Biggr).$$ This says: for any two scalars $\alpha$ and $\beta$, and any two vectors $\mathbf{u}$ and $\mathbf{v}$, if $\mathbf{u}$ and $\mathbf{v}$ are both in $W$, then (their linear combination) $\alpha\mathbf{u}+\beta\mathbf{v}$ is also in $W$.

For the third, it's very similar to the second (in fact, I'm surprised it is included, since it is a special case of the second one, taking $\mathbf{v}=\mathbf{0}$): $$\forall c\in K\Bigl(\forall \mathbf{u}\bigl( \mathbf{u}\in W\longrightarrow c\mathbf{u}\in W\bigr)\Bigr).$$ This says: for every $c\in K$, for every vector $\mathbf{u}$, if $\mathbf{u}$ is in $W$, then $c\mathbf{u}$ is in $W$.

The above are a bit of "pidgin logical notation" rather than completely formal, because of the use of things like $\forall \alpha,\beta\in K$; but it is a common (ab)use of the notation.

If you really wanted to be more formal, the quantifiers should only have one variable each, and the conditions would be put in the statement. So the third statement could be $$\forall c\forall\mathbf{u}\Bigl( \bigl((c\in K)\land (\mathbf{u}\in W)\bigr) \longrightarrow (c\mathbf{u}\in W)\Bigr),$$ (you can put the universal quantifiers in the other order) and the second could be $$\forall\mathbf{u}\forall\mathbf{v}\forall\alpha\forall\beta\Bigl(\bigl( (\mathbf{u}\in W)\land(\mathbf{v}\in W)\land (\alpha\in K)\land (\beta\in K)\bigr)\longrightarrow (\alpha\mathbf{u}+\beta\mathbf{v}\in W)\Bigr).$$

share|cite|improve this answer
Thanks for the thorough explanation, I am not fully experienced yet with proof/notation based mathematics, so I wasn't sure of the proper expression to use for the axioms. – eWizardII Nov 5 '11 at 23:00

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.