If I have two vector spaces $V$ and $W$, and I collect all the linear maps between the two vector spaces. Whether the collection (set of linear maps between two vector spaces) can be guaranteed to be a vector space?
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3Does this answer your question? [Hom(V,W) is a vector space](https://math.stackexchange.com/questions/3148665/homvwisavectorspace) – DanLewis3264 Nov 10 '21 at 02:42

I do not really think so(probably I miss something?). I'm wondering that if I have two vector spaces $V$ and $W$. (It seems to be a silly question) Can I just say that there "must" be a third vector space which is $L(V, W) $? I'm not very sure about that. Even though I know the linear maps between two vector spaces can form a vector space, does it happen all time? Whatever two vector spaces I choose, there must be a vector space that contains all the maps between them. Is this true? Thanks – Hubert Nov 10 '21 at 02:49

1Yes, that's correct. I appreciate it may seem a leap at first. But I don't think it requires a new post here, when Hom(V,W) is welldocumented online, including already on this site. – DanLewis3264 Nov 10 '21 at 02:57

Yes, this is true **all the time**. The only sort of caveat is that $V$ and $W$ have to be over the same field, in order for a linear map between $V$ and $W$ to even make sense. But, other than that, given any two such vector spaces $V$ and $W$, the set of linear maps from $V$ to $W$ forms a vector space under the usual definition of addition and scalar multiplication of functions. I don't know if a proof is written out in full on this site somewhere (I'd suspect not), but check out here: https://math.stackexchange.com/questions/2381942/thesetofalllinearmapstvwisavectorspace/ – Theo Bendit Nov 10 '21 at 03:41

@TheoBendit I prove that the set of all linear transformations from $V$ to $W$ form a vector space under pointwise operations at the top of [this post](https://math.stackexchange.com/a/111383/742); I do the sum explicitly, and leave the scalar multiplication as an exercise for the reader. Some of the details are only stated and not proven, though. – Arturo Magidin Nov 10 '21 at 03:48

@ArturoMagidin I too, in the link in my comment above, prove several axioms in detail, and leave the the rest to the reader. I suspect that nobody here has put the time into writing the whole proof, beginning to end, without holes. Perhaps we shouldn't? Proving something like this is a vector space is a linear algebra rite of passage! – Theo Bendit Nov 10 '21 at 03:51
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Go through the requirements.
Requires a 0 element. The 0 function is such an element.
Must be closed under addition. The Sun of two linear maps is a linear map.
Must be closed under scalar multiplication. This is also true.
Can you take it from here?
NicNic8
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It bugs me slightly that these are the three conditions for a subspace, given that these three conditions are so often illegitimately used as a shortcut to prove all the vector space axioms. Readers beware: you need to prove many other things about this set and its operations before you can conclude that it is a vector space. Or, if you know that the set of functions from $V$ to $W$ is a vector space (via a comparably long proof), you can use these three properties to show $L(V, W)$ is a subspace, and hence a vector space in its own right. – Theo Bendit Nov 10 '21 at 04:13