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MoreDrinks
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Homework Statement
Is the following set a subspace of R2?
Homework Equations
(a,b2)
The Attempt at a Solution
I'm exhausted and stumped.
MoreDrinks said:Homework Statement
Is the following set a subspace of R2?
Homework Equations
(a,b2)
The Attempt at a Solution
I'm exhausted and stumped.
No, this represents a set of vectors of this form.Zondrina said:(a, b2) is a set? Not a vector?
Zondrina said:(a, b2) is a set? Not a vector?
MoreDrinks said:Homework Statement
Is the following set a subspace of R2?
Homework Equations
(a,b2)
The Attempt at a Solution
I'm exhausted and stumped.
Mark44 said:You still need to show what you have tried. This problem is similar to the other one you posted today - you need to show that for vectors in the set, addition is closed, and scalar multiplication is closed.
Mark44 said:Are there any restrictions on a and b? IOW, are they restricted to integers or can they be any real numbers?
Clearly (1, 4) is in the set, since it could be written as (1, 22). What about (2, -9)?
Mark44 said:Are there any restrictions on a and b? IOW, are they restricted to integers or can they be any real numbers?
Clearly (1, 4) is in the set, since it could be written as (1, 22). What about (2, -9)?
No, a and b are real numbers. A vector <a, b> with these as its components would be in R2.MoreDrinks said:I had thought that we had to begin by assuming a and b are within R2?
More accurately, it's not in R (i.e., it's not real).MoreDrinks said:I mean, I see where you're going - the square root of negative nine is not within R2
So the set of vectors does not form a subspace (of R2).MoreDrinks said:, so the vector is not a subspace.
MoreDrinks said:There just weren't any ground rules laid down in this class for how we could go about this. Do you, perhaps, have a link to a webpage that really goes over vector spaces and subspaces?
Mark44 said:No, a and b are real numbers. A vector <a, b> with these as its components would be in R2.
More accurately, it's not in R (i.e., it's not real).So the set of vectors does not form a subspace (of R2).
vela said:You're given the set S=##\{(a,b^2)\in \mathbb{R}^2: a, b \in \mathbb{R}\}##. It's a collection of ordered pairs, right? So any element in S is also in ##\mathbb{R}^2##, so it's a subset of ##\mathbb{R}^2##. The vector (1,4) = (1,22) is in the set. The vector (2,-9), however, isn't, because you can't express it in the form (a,b2) where a and b are real numbers. S is a set of vectors, but it doesn't contain every vector in ##\mathbb{R}^2##.
To prove S is a subspace, you have to show the three things Mark44 listed in his post.
You've already shown that (0,0) is an element of the set because a=0, b=0 works.
Now take two elements of S, (a,b2) and (c,d2). When you add them together, you get (a+c, b2+d2). Can this be written in the form (e, f2) for some real numbers e and f? You need to explain why you know there are such e and f, or you need to find a counterexample to prove you can't.
Now let k be a real number. Is k(a,b2) = (ka, kb2) in S? In other words, can you always find real numbers e and f such that e=ka and f=kb2? Remember k can be any real number.
mfb said:It is not closed under scalar multiplication. Try a factor of -1 for any vector with non-zero second component.
mfb said:(1,4) is element of your set of vectors (coming from a=1, b=2), but -1(1,4)=(-1,-4) is not an element of your set (as there is no real b where b^2=-4).
Mark44 said:Clearly (1, 4) is in the set, since it could be written as (1, 22). What about (2, -9)?
mfb said:It is not closed under scalar multiplication. Try a factor of -1 for any vector with non-zero second component.
mfb said:(1,4) is element of your set of vectors (coming from a=1, b=2), but -1(1,4)=(-1,-4) is not an element of your set (as there is no real b where b^2=-4).
Mark44 said:MoreDrinks, now do you see where I was going with my question?
vela said:It would help if, instead of simply claiming you get it, you demonstrate your understanding by explaining your solution. For example, you said S is closed under vector addition. How do you know? I wonder if you truly understand the concepts because you later claimed S was closed under scalar multiplication.
OK.MoreDrinks said:Well, we start in R2 with the form (a,b2).
Here you're using the symbols a and b to represent three different quantities because you're talking about three different vectors. A symbol should represent only one thing at a time.Anything of the form (a,b2) that we can add to our original (a,b2) is going to leave us with a (a,b2).
The phrase in red doesn't make sense because b is just a regular old number while R2 is a set of order pairs, so b can't be in R2. You mean b is in R, not R2.Any b in R2 that is squared is going to be positive, which doesn't leave us with a second element b2 that is negative and, subsequently, possessing a non-real root outside of R2.
You seem to be on the right track, and there's no need to apologize as long as you're trying, which you are. I'll mention that a lot of students struggle with learning how to write proofs, so you're not alone. It's a skill you can pick up as you go along, though.Am I barking up the wrong tree? I'm sorry if the way I'm going about this is totally non-standard - there has never been any sort of "how to prove it" instruction in my educational history, so I'm sort of just jumping into this. It's not even required for my major: I just want to know, and I've enjoyed the matrix algebra in the course hitherto. Thank you for your help thus far.
vela said:It looks like you have the basic idea, but you need to be more careful about how you write it out.
OK.
Here you're using the symbols a and b to represent three different quantities because you're talking about three different vectors. A symbol should represent only one thing at a time.
You're also just restating what it is you're trying to prove, that the sum of two vectors is in the right form.
The phrase in red doesn't make sense because b is just a regular old number while R2 is a set of order pairs, so b can't be in R2. You mean b is in R, not R2.
The rest of the sentence isn't very clear because, again, you're using b to mean different things.
You seem to be on the right track, and there's no need to apologize as long as you're trying, which you are. I'll mention that a lot of students struggle with learning how to write proofs, so you're not alone. It's a skill you can pick up as you go along, though.
There is always some field (this is a precise mathematical term) from which components of a vector come. The field is usually the real numbers or the complex numbers, but finite fields such as {0, 1} can be used. In vector spaces such as R2 and R3, the field is the real numbers. In vector spaces such as C2, the field is the complex numbers, so in Cartesian form, each vector in C2 looks like this: <a1 + b1i, a2 + b2i>.MoreDrinks said:Surely, Mark. Follow-up question: when you're checking to see if a subset of vectors is closed under scalar multiplication, must a scalar be within the original vector space from which the subset is derived?
As vela said, many students struggle when it comes to doing proofs, as it isn't taught much or at all in courses before you get to linear algebra. IMO, it's a useful skill to have, though, as it fosters logical thinking that's not limited just to mathematics.MoreDrinks said:Am I barking up the wrong tree? I'm sorry if the way I'm going about this is totally non-standard - there has never been any sort of "how to prove it" instruction in my educational history, so I'm sort of just jumping into this. It's not even required for my major: I just want to know, and I've enjoyed the matrix algebra in the course hitherto. Thank you for your help thus far.
A subspace in R^2 is a set of vectors that satisfies two conditions: closure under vector addition and closure under scalar multiplication. This means that if you take any two vectors in the subspace and add them together, the resulting vector will also be in the subspace. Similarly, if you multiply any vector in the subspace by a scalar, the resulting vector will also be in the subspace.
A vector is determined to be in a subspace in R^2 if it satisfies the two conditions of closure under vector addition and scalar multiplication. This means that if the vector can be written as a linear combination of other vectors in the subspace, then it is considered to be in the subspace.
Yes, a vector can be in multiple subspaces in R^2. This is because a subspace is defined by its conditions of closure under vector addition and scalar multiplication, and a vector can satisfy these conditions for multiple subspaces at the same time.
A subspace and a span are similar concepts in R^2, but there are some key differences. A subspace is a set of vectors that satisfies the conditions of closure under vector addition and scalar multiplication, while a span is the set of all possible linear combinations of a given set of vectors. In other words, a subspace is a specific set of vectors, while a span is a broader concept that can encompass multiple subspaces.
The subspace status of a vector can affect its properties in R^2 in a few ways. For example, if a vector is in a subspace, it will have certain properties such as closure under vector addition and scalar multiplication. This can also affect how the vector can be manipulated and used in calculations involving other vectors in the subspace.