MATH2601 Higher Linear Algebra (1 Viewer)

leehuan · Mar 1, 2017

$\text{Define an operation}*\text{over }\mathbb{Z}_n\text{ where }x*y=xy \mod n\\ \text{Proven earlier: This operation is binary, commutative and associative}$

$\text{Let }p\in \mathbb{Z}\text{ be prime. Show that }\mathbb{U}_p=\mathbb{Z}_p - \{ 0\}\text{ is a group under the operation }*\text{ defined above.}$

Hint was to use the Bezout property but I have no idea to use it. I assume it's related to proving the existence of an inverse because that was the only bit I had trouble proving. (Associativity is just a repeat proof and the identity element is obviously 1)
____________

Side note - Am not sure why my threads had to be moved here.

seanieg89 · Mar 1, 2017

Re: Linear Algebra

Suppose the integer a is a representative of a nonzero equivalence class in Z_p.

Then a is coprime to p and hence am+pn=1 for some integers m and n.

Projecting to equivalence classes we get [a][m]=1 (mod p). ([z] denotes the equivalence class in Z_p of the integer z.)

I.e. every element of U_p has an inverse.

seanieg89 · Mar 1, 2017

Re: Linear Algebra

In fact this argument straightforwardly generalises to tell us that the subset of Z_n consisting of only the equivalence classes coprime to n form a group w.r.t. multiplication.

This yields Euler's theorem just as the version originally posted will yield Fermat's little theorem.

seanieg89 · Mar 1, 2017

Re: Linear Algebra

Ps this isn't really linear algebra at all.

dan964 · Mar 1, 2017

Re: Linear Algebra

seanieg89 said:
Ps this isn't really linear algebra at all.

thread title updated to "Group Theory" so no longer misleading

leehuan · Mar 1, 2017

Re: Group Theory

I am well aware that group theory may not fall under the umbrella of 'linear algebra'. However, in my course (MATH2601), it is the first thing that they choose to teach, hence I set the thread up with that title. I would like my thread name back so I can reserve this thread for all of my math2601 questions. Thanks.

Green Yoda · Mar 1, 2017

Re: Group Theory

leehuan said:
I am well aware that group theory may not fall under the umbrella of 'linear algebra'. However, in my course (MATH2601), it is the first thing that they choose to teach, hence I set the thread up with that title. I would like my thread name back so I can reserve this thread for all of my math2601 questions. Thanks.

When you double click on your thread when viewing from "whats new" you can change the name yourself.

leehuan · Mar 1, 2017

Re: Group Theory

Rathin said:
When you double click on your thread when viewing from "whats new" you can change the name yourself.

I know that. I just don't want to appear as though I am undoing a moderator action without permission.

dan964 · Mar 2, 2017

Re: Group Theory

leehuan said:
I know that. I just don't want to appear as though I am undoing a moderator action without permission.

its totally fine to edit the thread yourself

I have renamed it again so that it is useful for other users of the site.

leehuan · Mar 7, 2017

Re: MATH2601 Linear Algebra/Group Theory Questions

Just out of curiosity, what's an example of a vector space and/or a field which is defined in a way, that does not use the conventional means of addition and (scalar) multiplication?

Drsoccerball · Mar 7, 2017

Re: MATH2601 Linear Algebra/Group Theory Questions

Prove: $(a^{-1})^{-1} = a$

InteGrand · Mar 7, 2017

Re: MATH2601 Linear Algebra/Group Theory Questions

Drsoccerball said:
Prove: $(a^{-1})^{-1} = a$

$\noindent What's the context? Assuming it's in a group say, note that by definition, $aa^{-1} = a^{-1}a = e$, where $e$ is the identity element. Hence $a$ is an inverse of $a^{-1}$ (ask yourself what it really means for something to be an inverse of $a^{-1}$ and you'll see from above that $a$ satisfies the requirements). By uniqueness of inverses in a group, $a$ is \emph{the} inverse of $a^{-1}$, of course.$

Shadowdude · Mar 7, 2017

Re: MATH2601 Linear Algebra/Group Theory Questions

leehuan said:
Just out of curiosity, what's an example of a vector space and/or a field which is defined in a way, that does not use the conventional means of addition and (scalar) multiplication?

there are probably some real valued matrix examples

Drsoccerball · Mar 7, 2017

Re: MATH2601 Linear Algebra/Group Theory Questions

Let G be a group with identity e. Prove that if x^2 = e for all x in G then G is abelian.

InteGrand · Mar 7, 2017

Re: MATH2601 Linear Algebra/Group Theory Questions

Drsoccerball said:
Let G be a group with identity e. Prove that if x^2 = e for all x in G then G is abelian.

$\noindent Let $a,b \in G$. We need to show that $ab = ba$. We know from the question's assumption that $x = x^{-1}$ for all $x\in G$. Thus$

$$\begin{align*} ab &= \left(ab\right)^{-1} \\ &= b^{-1}a^{-1} \\ &= ba.\end{align*}$

$\noindent This completes the proof.$

Drsoccerball · Mar 7, 2017

Re: MATH2601 Linear Algebra/Group Theory Questions

InteGrand said:
$\noindent Let $a,b \in G$. We need to show that $ab = ba$. We know from the question's assumption that $x = x^{-1}$ for all $x\in G$. Thus$

$$\begin{align*} ab &= \left(ab\right)^{-1} \\ &= b^{-1}a^{-1} \\ &= ba.\end{align*}$

$\noindent This completes the proof.$

Didn't notice that an element is it's own inverse thanks

leehuan · Mar 8, 2017

Re: MATH2601 Linear Algebra/Group Theory Questions

Drsoccerball said:
Didn't notice that an element is it's own inverse thanks

It's basically this

$\begin{align*}x^2&=e\\ \implies (x^{-1}x)x&=x^{-1}e\\ \implies ex&=x^{-1}\\ \implies x&=x^{-1}\end{align*}$

Having used the associativity axiom and the definition of the identity element.

Right-operating also works

Paradoxica · Mar 8, 2017

Re: MATH2601 Linear Algebra/Group Theory Questions

This is strictly for Leehuan's understanding, but here's a concrete example to flesh out for the non-standard addition operation question:

$a \oplus b = \sinh{(\sinh^{-1}{a} + \sinh^{-1}{b})}$

and more generally:

for a bijective function φ(x):

$a \oplus b = \varphi(\varphi^{-1}(a)+\varphi^{-1}(b))$

leehuan · Mar 17, 2017

Re: MATH2601 Linear Algebra/Group Theory Questions

$\text{Prove that if }V\text{ is a finite dimensional vector space and }W\text{ is a subspace of }V,\text{ then }W\text{ is also finite dimensional.}$

InteGrand · Mar 17, 2017

Re: MATH2601 Linear Algebra/Group Theory Questions

leehuan said:
$\text{Prove that if }V\text{ is a finite dimensional vector space and }W\text{ is a subspace of }V,\text{ then }W\text{ is also finite dimensional.}$

Since V is finite dimensional, by definition V has a finite spanning set S, and this set S also spans W, so W is also finite dimensional.

MATH2601 Higher Linear Algebra (1 Viewer)

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