Mathematics Asked by Abhijeet Vats on August 12, 2020
Here’s what I’m trying to prove right now:
Let $V$ be a vector space over $mathbb{F}$. Let $M$ be a linear subspace of $V$ and $N$ be a linear subspace of $M$. Prove that the mapping $x+N mapsto x+M$ between the quotient spaces $V/N to V/M$ is linear with kernel $M/N$. Then, deduce that:
$$frac{V/N}{M/N} cong V/M$$
Proof Attempt:
We have to show that the given relation $T: V/N to V/M$ is a linear mapping. We define:
$$forall x in V: T(x+N) = x+M$$
This is totally-defined. To show well-definedness, let $x+N = y+N$ where $x,y in V$. Then, $x-y in N$. So, $x-y in M$. Hence:
$$x + M = y+M$$
$$iff T(x+N) = T(y+N)$$
To prove that it is linear, we need to show additivity and homogeneity.
Let $u,v in V/N$. Then, $u = x+N$ and $v = y+N$ for some $x,y in V$. Then:
$$T(u+v) = T((x+N)+(y+N)) = T((x+y)+N) = (x+y)+M = (x+M)+(y+M) = T(u) + T(v)$$
This proves additivity.
Let $alpha in mathbb{F}$ and $u in V/N$. Then, $u = x+N$ for some $x in V$. So:
$$T(alpha u) = T(alpha(x+N)) = T(alpha x +N) = alpha x + M = alpha (x+M) = alpha T(u)$$
This proves homogeneity. Hence, $T$ is a linear map. To show that the kernel of $T$ is $M/N$, we have:
$$T(x+N) = x+M = theta_V+M$$
$$iff x in M$$
$$iff x+N in M/N$$
$$iff ker(T) = M/N$$
Now, we notice that $T$ is surjective. By the first isomorphism theorem, it follows that:
$$frac{V/N}{M/N} cong V/M$$
That proves the desired result.
Does the proof above work? If it doesn’t, why? How can I fix it?
Your approach is absolutely right!
I would change the part concerning $ker T$, writing: begin{align} x + N in ker T & iff T(x+N) = 0 in V/M \ & iff x+M = 0 in V/M \ & iff x in M \ & iff x + N in M/N end{align} which means $ker T = M/N$.
Correct answer by Rodrigo Dias on August 12, 2020
Get help from others!
Recent Answers
Recent Questions
© 2024 TransWikia.com. All rights reserved. Sites we Love: PCI Database, UKBizDB, Menu Kuliner, Sharing RPP