The condition = 0 to form a subspace of is that f ∈ S is that otherwise, S wouldn't be closed under addition and thus would not be a subspace of .
<h3>What is additive identity?</h3>
An identity element (like 0 in the collective of whole numbers under operation of addition) that leaves unaffected any element that it is added in a provided mathematical system.
Now, for the given question.
I'd just like to comment on the paragraph about intuition as well as the additive identity: That instinct is incorrect.
First, we could perhaps define a object that we need to confirm is a subspace of in a more precise manner than the translation in which it is initially presented. We want to look at a set, S.
Where,
We are specifically tasked with having to prove the restriction just on derivative of f, which is best accomplished in the manner made reference.
The fact that S, as described above, describes a functional space explains why my initial intuition about problem is ultimately incorrect. I was hindering myself by viewing it solely as a geometric space.
As a result, I missed the fact that zero function (the additive identity in such a functional space) is indeed a natural member of S.
To know more about the additive identity, here
brainly.com/question/10737691
#SPJ4
The complete question is-
I'm working my way through Axler's "Linear Algebra Done Right", and I've run into a proposition that I don't understand. Namely:
The set of differentiable real-valued functions f on the interval (0, 3) such that = b is a subspace of if and only if b = 0.
My intution is that this has something to do with the fact that a subspace requires an additive identity, but that because the domain of the differentiable functions is open and exludes 0, we have to make some concessions (i.e. = 0), though I'm unsure of how to work up a proof that backs this.
Why must = 0 to form a subspace of ?