# Linear Subspace with Neutral Element in Brazilian Portuguese and Spanish

• I
• LCSphysicist
In summary, the conversation focuses on proving whether the set W, consisting of real polynomials of degree n or less with a specific condition, is a subspace of a larger vector space. The main question is how to prove that the neutral element (zero) is also in this set. The conversation also touches on the use of notation and terminology, as well as the properties that a subspace must satisfy.
LCSphysicist
W = {f(t) | f(0) = 2f(1)}
The answer say yes, but i don't know how to prove the neutral element.

LCSphysicist said:
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?

fresh_42 said:
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?

LCSphysicist said:
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?

fresh_42 said:
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.

Math_QED said:
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
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.

LCSphysicist said:
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##?

LCSphysicist said:
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
vela said:
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
Ah, now it makes more sense.

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.

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
LCSphysicist said:
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.

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##?

Eclair_de_XII said:
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
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
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?
WWGD said:
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
No, it is very easy to answer the questions I asked before and that is sufficient to answer the question.

atyy
WWGD said:
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.

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

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

Delta2

## 1. What is a linear subspace?

A linear subspace is a subset of a vector space that satisfies two conditions: closure under vector addition and closure under scalar multiplication. This means that any linear combination of vectors in the subspace will also be in the subspace.

## 2. How do you determine if something is a linear subspace?

To determine if something is a linear subspace, you must check if it satisfies the two conditions of closure under vector addition and scalar multiplication. You can also check if it contains the zero vector and if it is closed under taking linear combinations.

## 3. Can a linear subspace contain only one vector?

Yes, a linear subspace can contain only one vector. This vector would be considered the zero vector, as it is closed under scalar multiplication and vector addition.

## 4. Is the intersection of two linear subspaces always a linear subspace?

Yes, the intersection of two linear subspaces is always a linear subspace. This is because it satisfies the two conditions of closure under vector addition and scalar multiplication.

## 5. Can a linear subspace be infinite?

Yes, a linear subspace can be infinite. For example, the set of all polynomials of degree 2 or less is an infinite linear subspace.

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