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.

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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
    Commented Jan 13, 2019 at 22:20

1 Answer 1


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
    Commented Jan 13, 2019 at 23:02

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