Problem 369***(easy)
Prove that any diagonalisable matrix satisfies it's own characteristic polynomial
Problem 369***(hard)
This time, without the assumption of diagonalisability.
The hard version is the CaleyHamilton theorem; the easy one just a special case which is much more straightforward. Getting boring just seeing number theory and integrals.

FireGarden
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Solution 369
What's a matrix? It's a morphism between vector spaces. So we identify our matrix with the endomorphism of vector spaces where is a finite dimensional vector space over some algebraically closed field, call it . If we do not assume the field algebraically closed, we simply choose a larger algebraically closed field which contains our field.
Now view as a module over under the map , where . Trivial observation shows that is a torsion module. Hence, we can apply the structure theorem for modules over principal rings. Thus, we obtain the Jordan canonical form , where are invariant which can be written as where are subspaces isomorphic to for some .
Let be the characteristic polynomial. Then the kernel of the map is , where depends on and . Also, and . By and the fact that is generated by the elements of the form , for which , we get for all , and consequently .
We can think of this proof as follows: firstly, we decompose as a direct sum of torsion modules over the principal ring such that these torsion modules have exponents which are prime powers. Then, we use the structure theorem to decompose each of the summands in a direct sum of cyclic modules. Thus, under the corresponding basis our endomorphism is composed of cells in which all the elements are zero, except these on the diagonal (they are equal) and these on the super diagonal (they are not relevant). So we are done if we show that for an endomorphism composed of a single such cell, the proposition is true. But this is obvious since the vector space viewed as a module over the polynomial ring is cyclic and the ring of polynomials is commutative.
Spoiler:ShowThis gibberish is because I think that the notation obscures the main idea.
Edit: I forgot to note the obvious connection between linear algebra and algebra established by the structure theorems for modules over principal rings. I quite like this link. I can post much more interesting propositions; however, I restrain myself from doing it since I do not know whether people would like them (these would be mainly from my current reading list which includes some category theory and algebraic topology).Last edited by Mladenov; 05102013 at 00:42. 
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Why is everyone boycotting my question 364?
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Problem 370**/***
Prove that in the real number system that the inverse of addition is represented by
Posted from TSR MobileLast edited by Arieisit; 05102013 at 15:12. 
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(Original post by Arieisit)
Problem 370*/**
Prove that in the real number system that the inverse of addition is represented by
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 05102013 03:16
(Original post by james22)
From what starting point? One perfectly valid definition of the reals is as an ordered field (with a few other properties). Under this definition each numeber, x, has an additive inverse which is normally denoted by x.
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 05102013 14:46
(Original post by Arieisit)
There are a few ways I can think of in which this can be proven but its up to you to decide what to do (if you can).
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 05102013 18:55
Not really a problem, but
can be evaluated using Beta functions. I'm not sure how well known this is, so apologies if this is a little mundane. 
Felix Felicis
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(Original post by henpen)
Not really a problem, but
can be evaluated using Beta functions. I'm not sure how well known this is, so apologies if this is a little mundane.
Now, and .
Last edited by Felix Felicis; 05102013 at 19:23. 
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(Original post by Mladenov)
I can post much more interesting propositions; however, I restrain myself from doing it since I do not know whether people would like them (these would be mainly from my current reading list which includes some category theory and algebraic topology).
How's the selfstudy coming along now? Also, did you finish with Lang's "Algebra" yet? 
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(Original post by FireGarden)
Problem 369***(easy)
Prove that any diagonalisable matrix satisfies it's own characteristic polynomial
Problem 369***(hard)
This time, without the assumption of diagonalisability.
The hard version is the CaleyHamilton theorem; the easy one just a special case which is much more straightforward. Getting boring just seeing number theory and integrals.
"Solution" 369:
The characteristic polynomial of a square matrix A is p(x) = det(AxI). So p(A) = det(AAI) = det(AA) = det(0) = 0.

Indeterminate
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Last edited by Indeterminate; 08102013 at 01:49. 
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Let . Observe that
(infinite geometric series: 1+x^2+x^3+...=1/(1x), given that x<1, which is the case here since all prime are greater than 1)
Thus, by the fundamental theorem of arithmetic, when the RHS of the inequality is
,
where the indicates that the sum goes on forever, since there are infinitely many numbers that have factors in . A simple example would be (p being the largest prime ), where take on infinitely many values.
Then the problem boils down to showing that
.
No idea now. This seems more like an analytic number theory problem to me, which I've never studied in my life. (in A2 at the moment; don't even know single variable calculus properly yet lol..)
I can note that the LHS is a finite sum while the RHS is an infinite one. I don't know how to show if it diverges or not though, and how then to deploy that fact on the problem.
EDIT: I also don't know exactly how to deal with the equality case here.Last edited by arkanm; 08102013 at 18:32. 
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 08102013 19:25
(Original post by arkanm)
Let . Observe that
(infinite geometric series: 1+x^2+x^3+...=1/(1x), given that x<1, which is the case here since all prime are greater than 1)
Thus, by the fundamental theorem of arithmetic, when the RHS of the inequality is
,
where the indicates that the sum goes on forever, since there are infinitely many numbers that have factors in . A simple example would be (p being the largest prime ), where take on infinitely many values.
Then the problem boils down to showing that
.
No idea now. This seems more like an analytic number theory problem to me, which I've never studied in my life. (in A2 at the moment; don't even know single variable calculus properly yet lol..)
I can note that the LHS is a finite sum while the RHS is an infinite one. I don't know how to show if it diverges or not though, and how then to deploy that fact on the problem.
EDIT: I also don't know exactly how to deal with the equality case here. 
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 08102013 19:32
(Original post by james22)
In your last inequality, the RHS diverges and the inequality is obvious enough to assume. You would not be able to prove it formally without knowing some analysis though (not a very intersting problem either and just an application of definitions).
I didn't know that the RHS would diverge because it isn't the harmonic series as the numbers that are in the denominators depend on M. But if it does diverge then I guess it's easy to prove as the LHS is a finite sum.
What about the equality case? 
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 09102013 15:03
Solution 371
Note that , where consists of all numbers whose prime factors are less than or equal to . Hence .
(Original post by MW24595)
Go for it.
How's the selfstudy coming along now? Also, did you finish with Lang's "Algebra" yet?
Good luck and enjoy your time at Cambridge.
Problem 372***
Let and be subsets of the topological space so that , where is the interior of . Show that there exist natural homomorphisms such that is an exact sequence, where and are defined by the induced homomorphisms, namely and , , , and .
This is a wellknown sequence in algebraic topology which plays crucial role in homology theory. Using the same assumptions, show that the homomorphism is a composition of the following morphisms (now, you might come up with a second proof of the result).
Problem 373***
Let be a subset of , homeomorphic to , . Prove that and , .
Spoiler:ShowThe order of the above problems is intentional. 
Indeterminate
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 11102013 20:57
Problem 372***
Prove the following:
If
where D is a connected domain, and if f is holomorphic, and
then
Last edited by Indeterminate; 12102013 at 21:22. 
Indeterminate
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Some strange glitch. The latest posts aren't visible :/
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