If $\mathbb{V}$ is a vector space, is it always true that $\mathbb{V} = \text{span}(\mathbb{V})$?

Question

This question have just came to my mind after starting to study some linear algebra. I tried to write the following proof:

Let $\mathbb{V}$ be a vector space.

($\subseteq$) Let $v \in \mathbb{V}$. We can write $v$ as the linear combination $1v$. Hence, $v \in \text{span}(\mathbb{V})$.

($\supseteq$) If we take a linear combination of vectors in $\mathbb{V}$, that linear combination shall also be in $\mathbb{V}$ because $\mathbb{V}$ is a vector space and by definition it is closed under the sum of vectors and the scalar product. Hence, $\text{span}(\mathbb{V}$) is contained in $\mathbb{V}$.$\blacksquare$

I'd like to ask if my proof is correct. I've seen a few times the spanning set written as $A \subset \text{span}(A)$ but why isn't it equal?

Thank you! :)

Answer 1

Yes, it is correct. Reworded a little,

• $v \in \mathbb{V} \implies v \in \text{span}(\mathbb{V})$, since $v$ is a (trivial) linear combination of elements in $\mathbb{V}$.

• If we take $z \in \text{span}(\mathbb{V})$, then we can write $z = \sum_{i=1}^n \alpha_i v_i$ where, for all $i$, $\alpha_i$ are scalars in $\mathbb{V}$'s underlying field and $v_i \in \mathbb{V}$. It is easy to show from the axioms defining a vector space that they are closed under finite linear combinations, and hence $z \in \mathbb{V}$.

It is enough to define a spanning set with the condition $\text{span}(A) \subseteq A$, since $A \subseteq \text{span}(A)$ is always true for any $A$ by its very definition. (Hence, if $\text{span}(A) \subseteq A$ then $\text{span}(A) = A$ immediately!) Many mathematicians like to boil down definitions to their barest essence, i.e. to show an object satisfies a definition, we verify as few axioms as we truly need to. Sometimes this results in confusion to novices, of course.

Answer 2

It is true that Span(V) = V.

Because 1.v + zero times any other vectors = v.

But this is a non-profound statement. If a computer key board has a key for all words of the language which words of the dictionary can be typed? Obviously all of English language.

But 26 letters of alphabet suffices, these letters make a "spanning set'.

These are something like small handles to help us carry a much bigger objects. So a handle would be useful only when it much smaller than the object it is attached to.

Answer 3

Yes, it is always true that $V = span(V)$ for any vector space $V$.

By definition, the span of a set of vectors is the set of all possible linear combinations of those vectors. In other words, $span(V)$ consists of all possible linear combinations of vectors from $V$.

Since $V$ is a vector space, it contains the zero vector, and it is closed under scalar multiplication and vector addition. Therefore, any vector in $V$ can be expressed as a linear combination of vectors from $V$.

This implies that every vector in $V$ is already in the $span(V)$ because $span(V)$ includes all possible linear combinations of vectors from $V$.

Hence, we can conclude that $V = span(V)$ for any vector space V.