The role of linear combination in definition of a subspace.

Question

I am an undergraduate who just finished taking my first course in Linear Algebra. I am wanting to get a deeper understanding of the subject out of interest and for its wide reaching applications (primarily to physics).

To do so, I have been working through Hoffman and Kunze's 2nd Edition of Linear Algebra. I am on the second chapter, which concerns vector spaces.

Kunze defines a vector space in the way I was taught this past semester: axiomatically. He then defines subspaces as such:

"Let V be a vector space and F be a field. A subspace of V is a subset W of V which is itself a vector space over F with the operations of vector addition and scalar multiplication on V."

The first theorem presented after this definition states:

"A non-empty subset of W of V is a subspace of V iff for each pair of vectors a, b in W and each scalar c in F the vector ca + b is again in W."

Now, the definition of a subspace that I am familiar with is something like "a subspace of a vector space V is a subset of the vectors in V closed under vector addition and scalar multiplication. This translates into closure under linear combination."

I am quite confused as to why Kunze uses "...ca + b is again in W" instead of closure under linear combination which would look like ""...ca + db is again in W."

My thought is that "...ca + b is again in W" actually implies that "...ca + db is again in W". Thus, it is unnecessary to say closure under linear combination. I have thought through a sketch of this proof which goes something like:

Let W be a non-empty subset of vectors in vector space V on field F such that for vectors a, b and for an arbitrary scalar c in the field F ca + b is contained in W. I.e. let W be a subspace as defined by Kunze. Then, any scalar multiple of a is itself a vector in W because we can set b to be the 0 vector which necessarily must be in a subspace. If we then now set b to be the said scalar multiples of a that we just found out are elements of W, we can now input our original arbitrary vectors b in ca and produce linear combinations of vectors a, b as we set out to do.

However, one of the chapter's emphasis is that it's useful to consider a set of objects and linear combinations of said objects. So, why not emphasize it here?

Further adding to my confusion is Kunze's theorem 3:

"The subspace spanned by a non-empty subset S of a vector space V is the set of all linear combinations of vectors in S."

Here, Kunze actually refers to linear combinations of vectors in S. Now, if my hypothesis about "...ca + b is again in W" implying that "...ca + db is again in W" is correct, why wouldn't Kunze just say "...is the set of all combinations of vectors in S of the form ca + b"?

This leads me to believe I am misunderstanding something. Any help would be much appreciated.

Explicit edit: I thought it might be helpful to provide Kunze's proof for the above stated theorem 1:

Below is Kunze's definition of a subspace spanned by a subset, theorem 3, and theorem 3's proof.

Answer 1

You also need to show that $\textbf a, b \in W \Rightarrow c\textbf a + \textbf b \in W$ implies $\textbf 0 \in W$, but it's simple (and essentially proved in the proof of theorem 1).

For non-empty $W \subseteq V$, the following are equivalent:

1. $W$ is vector space itself (with the same operations)

2. $W$ is closed under vector addition and scalar multiplication

3. $W$ is closed under taking linear combinations

4. if $\textbf a, \textbf b \in W$ and $c \in F$ then $c \textbf a + \textbf b \in W$

5. if $\textbf a, \textbf b \in W$ and $c,d \in F$ then $c \textbf a + d \textbf b \in W$

The 4th variant is usually shortest to write, while the 2nd is simplest to check.

For the theorem 3, it's important to consider all linear combinations, not just of form $c\textbf a + \textbf b$, because when you talk about building a closure (not of checking if something is closed) you may need to use this operations more than once. And set $\{c\textbf a + \textbf b | \textbf a, b \in S, c \in F\}$ in general isn't a subspace spanned by $S$.

For example, if $V = \mathbb R^2$ and $S = \{(0, 1), (1, 0)\}$, then subspace spanned by $S$ is entire $V$, but vector $(2, 2)$ isn't of form $c\textbf a + \textbf b$.