# The role of linear combination in definition of a subspace.

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.

• How does Kunze defines subspace spanned by a subset? Commented May 25, 2022 at 22:24
• Let me post pictures of the text. One moment. Commented May 25, 2022 at 22:25
• I have posted a picture of Kunze's definition of subspace spanned by a subset as well as theorem 3 and its proof. Commented May 25, 2022 at 22:26

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 $$c\textbf a + \textbf b$$.

• Just to check my understanding of your answer: Your first sentence refers to the proof that variants (the ones you stated) 4 and 5 are equivalent? Thanks! Commented May 25, 2022 at 21:52
• Also, is Kunze's use of variant 4 just a matter of conciseness? Thinking about variant 2 makes the most sense to me, and I am wondering if thinking about subspaces in this way will hamper my ability to understand other concepts in linear algebra or follow proofs in Kunze. I feel it makes most sense because it seems the simplest since if a, b are vectors in V and c, d are scalars in F, closure under s. multiplication => ca is in V and db is in V; and if closure under v. addition the previous implication => ca + db is in V. Commented May 25, 2022 at 22:04
• The first line of Kunze's proof of theorem 3 seems to invoke the definition of a subspace. In particular, that a subspace W spanned by S which is a subset of V is itself a vector space under the operations of V. This invocation leads Kunze to state that all linear combinations of vectors in S are "clearly" in W. If all that is being invoked is the fact that W is a subspace, then mustn't Kunze solely be working with the fact that vectors of the form ca + b are also in W to come to the conclusion that linear combinations ca + db are also in W? Commented May 25, 2022 at 22:16
• But, I must be wrong since your counter example makes sense. Commented May 25, 2022 at 22:17
• Yes, the first line is about your "we can set b to be the 0 vector". For use variant 4 as definition - yes, I think just because it's the most compact. Commented May 25, 2022 at 22:23