I understand there is a duplicate: Show subset $w$ of a vector space $V$ is a subspace of $V$ if and only if $span(w)=w$.

The explanation given in that page does not satisfy me, what actually does it mean when $\text{span}(W)=W$? I think it means that any vector in W can be made from a linear combination from elements in W. I do not really understand subspace. I do understand that I have to confirm 3 criterion:

  • Show it is closed under addition.
  • Show it is closed under scalar multiplication.
  • Show that the vector 0 is in the subset.

I do not know how to do these! What does it even mean, closed under addition or multiplication?

All I know is that if $\text{span}(W)=W$ then for a scalar $\alpha_n$ any vector $w \in W$ can be written as $$w=w_1\alpha_1+w_2\alpha_2+w_3\alpha_3+\cdots+w_n\alpha_n$$ and that for a scalar $k\in \mathbb{R}$ we can write $$kw=k(w_1\alpha_1+w_2\alpha_2+w_3\alpha_3+\cdots+w_n\alpha_n)$$ $$kw=kw_1\alpha_1+kw_2\alpha_2+kw_3\alpha_3+\cdots+kw_n\alpha_n$$

  • $\begingroup$ A set $S$ with a binary operation $\star$ defined on it is closed under that operation if for all $a,b \in S$, the object $a \star b \in S$ as well. For example, $\Bbb{R}$ is closed under $+$, whereas say the set of odd integers, is not closed under the operation of $+$. Perhaps look into brilliant.org/wiki/subspace $\endgroup$
    – Anurag A
    May 26 '21 at 19:49

To begin with, the statement $W = \text{span}(W)$ means every linear combination of elements of $W$ is also in $W.$ Your interpretation is also a part of two sets being equal (that would be $W \subseteq \text{span}(W)$) but this part is trivially true because you can simply say each vector in $W$ is equal to itself, which is a linear combination of vectors from $W.$

As for the closures, the closures under addition and multiplication are two of the axioms which any set must fulfill in order to be a vector space. A subspace of a vector space is also a vector space, so it must fulfill these axioms.

Closure under addition means that adding any two members of the vector space will always give you another member of the vector space. A good example would be the set of integers, since whenever I add two integers I always get another integer. An example of a set not closed on addition is odd whole numbers: $3$ and $5$ are odd and $3 + 5 = 8,$ but $8$ is not odd, so the set is not closed under addition.

Closure under scalar multiplication is similar, except now when we multiply a member of a set by a scalar from some other field, we get another member of the set. As an example, let's let the candidate space be irrational numbers and the field of scalars can be the integers. The irrational numbers are closed under multiplication by an integer because if I multiply any irrational number by an integer, I must get another irrational number, since the product being representable by a rational fraction would give me a way to represent the original number as a rational fraction. An example of a non-closed set could be the set of odd integers under multiplication by the integers: $3$ is odd and $2 \cdot 3 = 6,$ but $6$ is not odd.

Notice also that if your field contains a $0,$ then your closure under multiplication also guarantees that the zero vector is in your subspace.

Now do you see how you can proceed?


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