Are These Linear Algebra Statements True or False?

  • Thread starter Thread starter ammar555
  • Start date Start date
ammar555
Messages
12
Reaction score
0
I have these questions on my studyguide and I know both of them are false. I just don't have a good counterexample or a good explaintion to prove so.

1) two subsets of a vector space V that span the same subspace of V are equal.

False: They don't have to be equal

2) The union of any 2 subspaces of a vector space V is a subspace of V.

False: Adding two subspaces doesn't necessary mean they will stay inside the vector space
 
Physics news on Phys.org
1) +1 and -1 in R
2) x and y axes in R^2. Addition and set union of two vector subspaces are not equal notions, though.
 
Alesak said:
1) +1 and -1 in R
2) x and y axes in R^2. Addition and set union of two vector subspaces are not equal notions, though.

Please I need a little more explaintions, I don't fully understand your examples. And does that mean my answer to number 2 is wrong?

Thanks though!
 
1) real line R is a vector space, and it is generated by both +1 and -1. So these two one-element subsets {1} and {-1} generate R yet are different.

2) it's correct, just unclear. Consider, for example, subspaces of R^2 that coincide with usual x and y axes. Vector (1, 0) is on x axis, vector (0, 1) on y axis, but if you add them they lie out of both subspaces.

So set theoretic union of two vector subspaces doesn't have to be vector subspace. Perhaps you can try to work out operation that makes some kind of "union" of two vector subspaces again a subspace.

You can surely find more information about vector spaces on wikipedia or in your textbook.
 
Alesak said:
1) real line R is a vector space, and it is generated by both +1 and -1. So these two one-element subsets {1} and {-1} generate R yet are different.

2) it's correct, just unclear. Consider, for example, subspaces of R^2 that coincide with usual x and y axes. Vector (1, 0) is on x axis, vector (0, 1) on y axis, but if you add them they lie out of both subspaces.

So set theoretic union of two vector subspaces doesn't have to be vector subspace. Perhaps you can try to work out operation that makes some kind of "union" of two vector subspaces again a subspace.

You can surely find more information about vector spaces on wikipedia or in your textbook.


Much better explanations precisely for number 1, I think I understand it now. Number 2 I'm still shady about it. Maybe if I read more about vectors, I can then understand it more clearly.

Thanks a million times!
 
ammar555 said:
Much better explanations precisely for number 1, I think I understand it now. Number 2 I'm still shady about it. Maybe if I read more about vectors, I can then understand it more clearly.

Thanks a million times!

Sure, get any good textbook on linear algebra, it will be crystal clear to you in no time.
 
I asked online questions about Proposition 2.1.1: The answer I got is the following: I have some questions about the answer I got. When the person answering says: ##1.## Is the map ##\mathfrak{q}\mapsto \mathfrak{q} A _\mathfrak{p}## from ##A\setminus \mathfrak{p}\to A_\mathfrak{p}##? But I don't understand what the author meant for the rest of the sentence in mathematical notation: ##2.## In the next statement where the author says: How is ##A\to...
##\textbf{Exercise 10}:## I came across the following solution online: Questions: 1. When the author states in "that ring (not sure if he is referring to ##R## or ##R/\mathfrak{p}##, but I am guessing the later) ##x_n x_{n+1}=0## for all odd $n$ and ##x_{n+1}## is invertible, so that ##x_n=0##" 2. How does ##x_nx_{n+1}=0## implies that ##x_{n+1}## is invertible and ##x_n=0##. I mean if the quotient ring ##R/\mathfrak{p}## is an integral domain, and ##x_{n+1}## is invertible then...
The following are taken from the two sources, 1) from this online page and the book An Introduction to Module Theory by: Ibrahim Assem, Flavio U. Coelho. In the Abelian Categories chapter in the module theory text on page 157, right after presenting IV.2.21 Definition, the authors states "Image and coimage may or may not exist, but if they do, then they are unique up to isomorphism (because so are kernels and cokernels). Also in the reference url page above, the authors present two...
Back
Top