- #1

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## Main Question or Discussion Point

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

The answer say yes, but i dont know how to prove the neutral element.

The answer say yes, but i dont know how to prove the neutral element.

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- #1

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

The answer say yes, but i dont know how to prove the neutral element.

The answer say yes, but i dont know how to prove the neutral element.

- #2

fresh_42

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What would be the zero?W = {f(t) | f(0) = 2f(1)}

The answer say yes, but i dont know how to prove the neutral element.

- #3

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

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Math_QED

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fresh_42

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

- #6

Math_QED

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My guess is that it is the usual abuse of notation.What does ##f(t)## stand for? A certain number at a point ##t##, or what does it mean?

- #7

fresh_42

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

- #8

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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 dont understand how to prove this item.

- #9

fresh_42

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In order to prove this property, you only have to show that zero is in that space. Now what is zero in this context?

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 dont understand how to prove this item.

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##?

- #10

vela

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

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WWGD

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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.What's a conjunt? What does nule mean? I'm trying to figure out what that sentence was intended to mean.

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vela

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Ah, now it makes more sense.

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FactChecker

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{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

(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 ∈

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Stephen Tashi

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

The answer say yes, but i dont know how to prove the neutral element.

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.

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Math_QED

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

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HallsofIvy

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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))?

- #19

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Voce acertou, mas eu sou brasileiro XDI 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.

You're right, but i am brazillian XD

sorry by my "portuenglish"

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HallsofIvy

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Yes. I also speak Spanish and this is true.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.

- #22

WWGD

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But read above where he said he's Brazilian.Yes. I also speak Spanish and this is true.

- #23

Mark44

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Spanish and Portuguese share a lot of words. Brazilian Portuguese is a dialect of Portuguese.But read above where he said he's Brazilian.

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