Author Topic: TT's Maths Thread (Read 134036 times) Tweet Share

TrueTears · « **Reply #1230 on:** May 16, 2011, 06:17:12 am »

[IMG]http://img713.imageshack.us/img713/8/ps9s.jpg[/img]

how do i do part a) and b)?

Cheers

Mao · « **Reply #1231 on:** May 16, 2011, 11:02:22 am »

a) the set of all solutions is a vector space if it is closed under addition and scalar multiplication (taking into account the n-degrees of freedom from constants of integration)

that is, if $\textbf{y}$ is a solution, then $a\textbf{y}$ must also be a solution.

$\frac{d(a\textbf{y})}{dt} = a\frac{d\textbf{y}}{dt} = a(A\textbf{y}) = A(a\textbf{y})$ , satisfied

if $\textbf{y}_1$ and $\textbf{y}_2$ are solutions, then $\textbf{y}_1+\textbf{y}_2$ must also be a solution.

$\frac{d(\textbf{y}_1+\textbf{y}_2)}{dt} = \frac{d\textbf{y}_1}{dt} + \frac{d\textbf{y}_2}{dt} = A\textbf{y}_1 + A \textbf{y}_2 = A(\textbf{y}_1+\textbf{y}_2)$ , satisfied

$\textbf{y}$ is a subspace.

b) $A\textbf{y} = c_1 \lambda_1 e^{\lambda_1 t} \textbf{v}_1 + c_2 \lambda_2 e^{\lambda_2 t} \textbf{v}_2 \cdots$ , since v1, v2,... are eigenvectors of A.
differentiating normally, $\frac{d}{dt} \textbf{y} = c_1 \lambda_1 e^{\lambda_1 t} \textbf{v}_1 + c_2 \lambda_2 e^{\lambda_2 t} \textbf{v}_2 \cdots$

They are the same, verified.

/0 · « **Reply #1232 on:** May 16, 2011, 11:07:35 am »

a)
You need to show that $\mathbf{0}$ solves the DE.
You need to show that if $\mathbf{y}_1$ and $\mathbf{y}_2$ solve the DE, then so does $a\mathbf{y}_1+b\mathbf{y}_2$ (adding equations etc.)

b)
Suppose that $\mathbf{y}=\mathbf{e}^{\lambda_1 t}\mathbf{v}_1$ solves the DE.

Then $\lambda_1 \mathbf{e}^{\lambda_1 t}\mathbf{v}_1= A\mathbf{e}^{\lambda_1 t}\mathbf{v}_1$

$(A-\lambda_1 I)\mathbf{e}^{\lambda_1 t}\mathbf{v}_1=0$ .

$(A-\lambda_1 I)\mathbf{v}_1=0$ .

So $\lambda_1$ must be an eigenvalue and $\mathbf{v}_1$ must be an eigenvector of $A$ . Then by part a), you take linear combinations of them.

(Beaten!)

TrueTears · « **Reply #1233 on:** May 23, 2011, 05:29:52 pm »

[IMG]http://img860.imageshack.us/img860/9325/18467514.jpg[/img]

How would you do these 2 questions?

Cheers

kamil9876 · « **Reply #1234 on:** May 23, 2011, 05:44:38 pm »

In fact every homomorphism from $\mathbb{Z}$ to any group is uniquely determined by the image of $1$ .

Prove that $\phi(x)=6x$ (where by $6x$ I of course mean that element of $\mathbb{Z}/10\mathbb{Z}$ ).

You can prove it for x>0 by noticing that $x=1+1...+1$ $x$ times and so $\phi(x)=\phi(1+1+...+1)$ , now use the homomorphism property. For x<0 the proof is similair you just have to notice that $\phi(-1)=-6$ . Now it should be clear.

As for the first question notice that $48=8*2*3*1$ So you should take $\mathbb{Z}_8 \times G_2 \times G_3 \times G_4$ where $G_2$ is a subgroup of $\mathbb{Z}_4$ with 2 elements, $G_3$ is a subgrpup of $\mathbb{Z}_9$ with three elements and $G_4$ is a subgroup of $\mathhbb{Z}_7$ with 1 element. Can you find such groups?

TrueTears · « **Reply #1235 on:** May 23, 2011, 06:00:06 pm »

Quote from: kamil9876 on May 23, 2011, 05:44:38 pm

In fact every homomorphism from $\mathbb{Z}$ to any group is uniquely determined by the image of $1$ .

Prove that $\phi(x)=6x$ (where by $6x$ I of course mean that element of $\mathbb{Z}/10\mathbb{Z}$ ).

You can prove it for x>0 by noticing that $x=1+1...+1$ $x$ times and so $\phi(x)=\phi(1+1+...+1)$ , now use the homomorphism property. For x<0 the proof is similair you just have to notice that $\phi(-1)=-6$ . Now it should be clear.

As for the first question notice that $48=8*2*3*1$ So you should take $\mathbb{Z}_8 \times G_2 \times G_3 \times G_4$ where $G_2$ is a subgroup of $\mathbb{Z}_4$ with 2 elements, $G_3$ is a subgrpup of $\mathbb{Z}_9$ with three elements and $G_4$ is a subgroup of $\mathhbb{Z}_7$ with 1 element. Can you find such groups?

Not really sure on that homomorphism question, I'm a bit vague in this area, can you finish off the gaps in your working?

Also for the subgroup question for order 48, LCM(8*2*3*1) is not 48 so you can't really do that?

Also what about the other question? (the isomorphic one?)

Thanks.

kamil9876 · « **Reply #1236 on:** May 23, 2011, 06:50:26 pm »

You can take G_2={0,2}, G_3={0,3,6}, G_4={0} Viola.

What's not clear? The point is $\phi(x)=6x$ is what you have to really prove and then you can answer i and ii immediately.

As for the question I missed I think you should use the chineese remainder theorem to decompose it into a product of prime power groups and then it may be easier to analyse.

TrueTears · « **Reply #1237 on:** May 23, 2011, 07:01:02 pm »

Thanks.

Well what I mean is how did you get $\phi(x)=6x$ from the $\phi(1) = 6$ ? And how do you prove it for the case x<0?

Quote

As for the question I missed I think you should use the chineese remainder theorem to decompose it into a product of prime power groups and then it may be easier to analyse.

Can you show me? I've never used or heard of this technique of CRT to decompose...

kamil9876 · « **Reply #1238 on:** May 23, 2011, 07:17:19 pm »

for $x>0$ , we have $\phi(x)=\phi(1+1+1...+1)=\phi(1)+\phi(1)....+\phi(1)=6+6+6...+6=6x$

$0=\phi(0)=\phi(-x)+\phi(x)$ so we have $\phi(-x)=\phi(x)$ , hence $\phi(x)=6x$ is verified for negative $x$ as well.

As for the other one:

$\mathbb{Z}_{126}=\mathbb{Z}_{2*3^2*7}=\mathbb{Z}_2\mathbb{Z}_{3^2} \times \mathhbb{Z}_{7}$

TrueTears · « **Reply #1239 on:** May 23, 2011, 08:38:44 pm »

Quote from: kamil9876 on May 23, 2011, 07:17:19 pm

for $x>0$ , we have $\phi(x)=\phi(1+1+1...+1)=\phi(1)+\phi(1)....+\phi(1)=6+6+6...+6=6x$

$0=\phi(0)=\phi(-x)+\phi(x)$ so we have $\phi(-x)=\phi(x)$ , hence $\phi(x)=6x$ is verified for negative $x$ as well.

As for the other one:

$\mathbb{Z}_{126}=\mathbb{Z}_{3^2*7^2}=\mathbb{Z}_{3^2} \times \mathhbb{Z}_{7^2}$ (because $3^2$ and $7^2$ have no common factor) decompose the others in a similair fashion.

Thanks for that, but it should be $\phi(-x) = -\phi(x)$ right?

Also is this right for ii?

$Ker(\phi) = \{x \in \mathbb{Z} | \phi(x) = 0\}$ where 0 is the identity element in $\mathbb{Z}_{10}$

Thus we have $6x \equiv 0 \pmod{10} \implies x = 10k$ where $k \in \mathbb{Z}$

Also $\mathbb{Z}_{4} \times \mathbb{Z}_{126}$

The factorisation you got for 126 is wrong right... should be:

$\mathbb{Z}_4 = \mathbb{Z}_{2^2}$ and $\mathbb{Z}_{126} = \mathbb{Z}_{23^27}$ which means it is isomorphic to $\mathbb{Z}_{2} \times \mathbb{Z}_{3^2} \times \mathbb{Z}_{7}$

So $\mathbb{Z}_{4} \times \mathbb{Z}_{126}$ is isomorphic to $\mathbb{Z}_{2^2} \times \mathbb{Z}_{2} \times \mathbb{Z}_{3^2} \times \mathbb{Z}_{7}$

For $\mathbb{Z}_6 \times \mathbb{Z}_{84}$

$\mathbb{Z}_6$ is isomorphic to $\mathbb{Z}_2 \times \mathbb{Z}_3$ and $\mathbb{Z}_{84}$ is isomorphic to $\mathbb{Z}_{2^2} \times \mathbb{Z}_{3} \times \mathbb{Z}_{7}$

So $\mathbb{Z}_6 \times \mathbb{Z}_{84}$ is isomorphic to $\mathbb{Z}_2 \times \mathbb{Z}_3 \times \mathbb{Z}_{2^2} \times \mathbb{Z}_{3} \times \mathbb{Z}_{7}$

Now what?

kamil9876 · « **Reply #1240 on:** May 23, 2011, 08:48:33 pm »

Quote

Thanks for that, but it should be \phi(-x) = -\phi(x) right?

yes sorry.

No it is not x=10k, because x=5 is in the kernel but it is not of that form. Think about it, when is 6x a multiple of 10? (ie when is 2*3*x a multiple of 2*5 ?)

Now the beauty of that decomposition is that you can see that the first group has an element of order 9 (the element (0,0,1,0) in $\mathbb{Z}_{2^2} \times \mathbb{Z}_{2} \times \mathbb{Z}_{3^2} \times \mathbb{Z}_{7}$ ). But you can argue that the second group does not and hence that means they are NOT isomorphic.

TrueTears · « **Reply #1241 on:** May 23, 2011, 08:56:53 pm »

Quote from: kamil9876 on May 23, 2011, 08:48:33 pm

Quote
Thanks for that, but it should be \phi(-x) = -\phi(x) right?

yes sorry.

No it is not x=10k, because x=5 is in the kernel but it is not of that form. Think about it, when is 6x a multiple of 10? (ie when is 2*3*x a multiple of 2*5 ?)

Now the beauty of that decomposition is that you can see that the first group has an element of order 9 (the element (0,0,1,0) in $\mathbb{Z}_{2^2} \times \mathbb{Z}_{2} \times \mathbb{Z}_{3^2} \times \mathbb{Z}_{7}$ ). But you can argue that the second group does not and hence that means they are NOT isomorphic.

oh shit opps, i mean x = 5k since lcm(6, 10) = 30, thats better yeah?

yeah check my edited post, i realised that lol

TrueTears · « **Reply #1242 on:** May 23, 2011, 11:06:15 pm »

[IMG]http://img716.imageshack.us/img716/9218/4bcz.jpg[/img]

Any ideas on these kamil?

I'm a bit lost on the first one however the second one I have a rough starting, not sure where to go with it though.

for a homomorphism $\phi: G \to G'$ , then we know $Ker(\phi)$ is a normal subgroup of G, by Lagrange's Theorem $|Ker(\phi)|$ divides $|G|$ .

This is the only thing I can think of which relates the order of G, but then i'm kinda stuck to prove what's required.

kamil9876 · « **Reply #1243 on:** May 23, 2011, 11:45:20 pm »

To prove the first one it depends on what you know, are you aware of what the Sign of a permutation is?

As for the second one, try to use the first isomorphism theorem.

TrueTears · « **Reply #1244 on:** May 23, 2011, 11:52:08 pm »

Nope, I haven't heard of what a sign of a permutation is... I know most about them, transpositions, even/odd and etc etc

Yeah I've thought about the first isomorphism theorem too but how is it of use here? can you show me by finishing the proof with it? i dont get how to apply it

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