To prove whether something is a vector space, my understand is to prove the following properties: commutativity, associativity, additive identity, additive inverse, multiplicative identity, distributive properties. However, in the answer, it tries to prove closed under addition, close under scalar multiplication.

enter image description here

are those conditions for a subspace?

  • $\begingroup$ You are right that in the proof, they do not check commutativity, associativity etc. The picture you included shows that the vector space operations of addition and scalar multiplication $$L(V,W) \times L(V,W) \rightarrow L(V,W), (T,S) \mapsto T+S$$ $$\mathbb F \times L(V,W) \rightarrow L(V,W), (\lambda,T) \mapsto \lambda T$$ are well defined functions. This is the hardest part of showing that $L(V,W)$ is a vector space. You should check the other properties yourself. $\endgroup$
    – D_S
    Jan 13 '19 at 22:20

You are 100% correct. It's an easy trap to fall into, to only verify the subspace conditions instead of all the many conditions for a full vector space, but just verifying subspace conditions is definitely not enough! I've seen countless students (as well as a few teachers) fall into this trap.

Now, there is one potential saving grace here: perhaps it was proven that the set $W^V$ of functions from $V$ to $W$ (linear or not) is a vector space. That is, $W^V$ under the given operations, all of the 8 or so properties of vector spaces were proven previously. Then, proving $\mathcal{L}(V, W)$ is a subspace $W^V$ is a valid way of proving $\mathcal{L}(V, W)$ is a vector space in its own right, as it will inherit almost all the properties from $W^V$.

But otherwise, the proof is incorrect.

  • 2
    $\begingroup$ Note also that the alleged "proof" above didn't even check all the subspace conditions. The existence of zero vector is also important. $\endgroup$
    – BigbearZzz
    Jan 13 '19 at 23:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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