# Is this a Linear subspace?

LCSphysicist
W = {f(t) | f(0) = 2f(1)}
The answer say yes, but i don't know how to prove the neutral element.

Mentor
2022 Award
W = {f(t) | f(0) = 2f(1)}
The answer say yes, but i don't know how to prove the neutral element.
What would be the zero?

LCSphysicist
What would be the zero?
i have no idea :|, the answer is literally "yes", just it.

You mention the word subspace. Of what vector space is this a subspace? What is the zero in the larger vector space? Can you conclude something about the zero of the subspace as well?

Mentor
2022 Award
i have no idea :|, the answer is literally "yes", just it.
What does ##f(t)## stand for? A certain number at a point ##t##, or what does it mean?

What does ##f(t)## stand for? A certain number at a point ##t##, or what does it mean?

My guess is that it is the usual abuse of notation.

Mentor
2022 Award
My guess is that it is the usual abuse of notation.
I know. I just want to get the OP think about it. The question is trivial once it is understood, so it is all about understanding, not answering.

• member 587159
LCSphysicist
Oh i forgot, subspace of the P(R) space of the real polynomials. You know:
The conjunt of reals Polynomials n deegre or smaller more the nule polynomial.
Yes, indeed is a trivial question, actually it's in a book introductory to linear algebra, but i still don't understand how to prove this item.

Mentor
2022 Award
Oh i forgot, subspace of the P(R) space of the real polynomials. You know:
The conjunt of reals Polynomials n deegre or smaller more the nule polynomial.
Yes, indeed is a trivial question, actually it's in a book introductory to linear algebra, but i still don't understand how to prove this item.
In order to prove this property, you only have to show that zero is in that space. Now what is zero in this context?

Another property is to show that if ##f(t) \in W## and ##\lambda \in \mathbb{R}##, then ##\lambda \cdot f(t)## must be in ##W##. If you had shown this, then what if ##\lambda =0##?

Staff Emeritus
Homework Helper
The conjunt of reals Polynomials n deegre or smaller more the nule polynomial.
What's a conjunt? What does nule mean? I'm trying to figure out what that sentence was intended to mean.

• Delta2
Gold Member
What's a conjunt? What does nule mean? I'm trying to figure out what that sentence was intended to mean.
I assume the OP may be a native Spanish spraker. "Conjunto" is Spanish for "Set" and "Nulo" is Spanish for null. And for others, I assume, per abuse of notation, f(t) is a function. I assume the 0 vector here would be the 0 function/polynomial.

• Delta2
Staff Emeritus
Homework Helper
Ah, now it makes more sense.

Homework Helper
Gold Member
Yes, the first mistake was not including this in the original post: ##\{f(t)\in ? | f(0) = 2f(1)\} ##. Given that, the properties of a subspace are inherited from the superspace. You can easily list the required properties one-by-one saying that each one is satisfied. The only thing left is to show that the subspace is closed under the operations of addition and of multiplication by a real.

zinq
It's always helpful to say what space you are dealing with. If it's a vector space V that contains the functions f in the expression

{f(t) | f(0) = 2f(1)}

then this should be written

{f ∈ V | f(0) = 2f(1)}

and V ought to be defined in terms of what functions it contains (domain, codomain, properties) and what is the field F of scalars (the reals or the complexes).

(Note that I rewrote your "f(t)" as just "f", because f means the function itself that is a vector in V, but f(t) means the value of that function after it has been evaluated at some input t.)

But to decide whether your set is a subspace of V, the things to check is whether a) the sum f + g of two vectors f and g in your set also belongs to the set, and b) whether the product αf of an f in your set by a scalar α ∈ F is also in your set.

• Delta2
W = {f(t) | f(0) = 2f(1)}
The answer say yes, but i don't know how to prove the neutral element.

If you are dealing with a vector space whose elements are functions, the zero vector ( neutral element) must be a function. What function is it?

Perhaps you aren't remembering that there can be constant functions. For example, the function f(x) = 3 is a constant function. It is a function even though its value isn't different for different values of x.

Eclair_de_XII
Question: Would the nature of the answer change if the superspace of ##W## were instead just the space of continuous functions, or maybe just another vector space consisting of real-valued functions defined at ##x=0## and ##x=1##?

Question: Would the nature of the answer change if the superspace of ##W## were instead just the space of continuous functions, or maybe just another vector space consisting of real-valued functions defined at ##x=0## and ##x=1##?

No. As long as the superspaces have the same "operations".

• Eclair_de_XII
Homework Helper
A subspace of a vector space has two properties:
It is "closed under scalar multiplication" and "closed under vector addition".

Here the subset is the set of all functions, f, such that f(0)= 2f(1).

"Closed under scalar multiplication". If a is any number then does af satisfy af(0)= 2af(1)?

"Closed under vector addition". If f satisfies f(0)= 2f(`1) and g satisfies g(0)= 2g(1) does f+ g satisfy f(0)+ g(0)= 2(f(1)+ g(1))?

• roam, Delta2 and atyy
LCSphysicist
The problem in this question is, actually, it say more nothing, just it. So we can't give a answer with just this informations, right?

I assume the OP may be a native Spanish spraker. "Conjunto" is Spanish for "Set" and "Nulo" is Spanish for null. And for others, I assume, per abuse of notation, f(t) is a function. I assume the 0 vector here would be the 0 function/polynomial.

Voce acertou, mas eu sou brasileiro XD
You're right, but i am brazillian XD

sorry by my "portuenglish"

• Delta2
Homework Helper
No, it is very easy to answer the questions I asked before and that is sufficient to answer the question.

• atyy
MidgetDwarf
I assume the OP may be a native Spanish spraker. "Conjunto" is Spanish for "Set" and "Nulo" is Spanish for null. And for others, I assume, per abuse of notation, f(t) is a function. I assume the 0 vector here would be the 0 function/polynomial.
Yes. I also speak Spanish and this is true.

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