MHB Solve Vector Space Question: Get the Solution Now

LearnerJr
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How do you solve this question I just need a solution
 

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Hello LearnerJr and welcome to MHB! :D

We ask that our users show their progress (work thus far or thoughts on how to begin) when posting questions. This way our helpers can see where you are stuck or may be going astray and will be able to post the best help possible without potentially making a suggestion which you have already tried, which would waste your time and that of the helper.

Can you post what you have done so far?
 
greg1313 said:
Hello LearnerJr and welcome to MHB! :D

We ask that our users show their progress (work thus far or thoughts on how to begin) when posting questions. This way our helpers can see where you are stuck or may be going astray and will be able to post the best help possible without potentially making a suggestion which you have already tried, which would waste your time and that of the helper.

Can you post what you have done so far?

Yes of course. For 2a I know a bit of indirect proof and assume the result
Is not true. But not sure still how to perceive this question in the long run.
2b I know if it's a basis it's vectors have to be linearly independent, they span V.but I still can't solve it.
 
Consider the equation $af(v_1)+ bf(v_2)+ cf(v_3)= 0$. In order to show that $f(v_1)$, $f(v_2)$, and $f(v_3)$ are independent we must show that a= b= c= 0.

Since f is a linear transformation, $af(v_1)+ bf(v_2)+ cf(v_3)= f(av_1+ bv_2+ cv_3)= 0$. Since the kernel of f is only the 0 vector, we must have $av_1+ bv_2+ cv_3= 0$. But we were given that $v_1$, $v_2$, and $v_3$ are independent so a= b= c= 0 as we wished.
 
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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...
##\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...
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