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    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 Caley-Hamilton theorem; the easy one just a special case which is much more straightforward. Getting boring just seeing number theory and integrals.
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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 A : V \to  V where V is a finite dimensional vector space over some algebraically closed field, call it K. If we do not assume the field algebraically closed, we simply choose a larger algebraically closed field which contains our field.
    Now view V as a module over K[t] under the map f(t)v = f(A)v, where v \in V. Trivial observation shows that V_{A} is a torsion module. Hence, we can apply the structure theorem for modules over principal rings. Thus, we obtain the Jordan canonical form V = V_{1} \oplus V_{2} \oplus \cdots \oplus V_{q}, where V_{i} are A-invariant which can be written as W_{V_{i}}_{1} \oplus \cdots \oplus W_{V_{i}}_{r(V_{i})} where W_{V_{i}}_{j} are subspaces isomorphic to K[t]v_{i}_{j} for some v_{i}_{j} \in V_{i}.
    Let P_{A} = (t-\alpha_{1})^{m_{1}}\cdots (t-\alpha_{q})^{m_{q}} be the characteristic polynomial. Then the kernel of the map f(t) \mapsto f(A)W_{V_{i}}_{j} is (t-\alpha_{i})^{s_{i}_{j}}, where s_{i}_{j} depends on i and j. Also, s_{i}_{j} \le m_{i} and \displaystyle \sum_{j} s_{i}_{j} = m_{i}. By (A-\alpha_{i}I)^{m_{i}}v_{i}_{j}=0 and the fact that V is generated by the elements of the form f(A)v_{i}_{j}, for which P_{A}(A)f(A)v_{i}_{j}=f(A)P_{A}(  A)v_{i}_{j}=0, we get for all v \in V, P_{A}(A)v=0 and consequently P_{A}(A)=0.

    We can think of this proof as follows: firstly, we decompose V as a direct sum of torsion modules over the principal ring K[t] 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.
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    This 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).
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    Why is everyone boycotting my question 364?

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    (Original post by Arieisit)
    Why is everyone boycotting my question 364?

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    I have just seen the problem. I personally am not going to do it for it requires too much calculations (for me), and I dislike them passionately.
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    Problem 370**/***

    Prove that in the real number system that the inverse of addition is represented by x + (-x) = 0

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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 x + (-x) = 0

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    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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    (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.
    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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    (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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    My problem is that I could say that -x is the additive inverse of x (by definition of -x) and so x+(-x)=0 as 0 is defined as the additive identity. This is the definition of the reals that I was taught.
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    Not really a problem, but

    \large \sum_{n\ge1}\frac{1}{\binom{n+m+  1}{m+1}}

    can be evaluated using Beta functions. I'm not sure how well known this is, so apologies if this is a little mundane.
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    (Original post by henpen)
    Not really a problem, but

    \large \sum_{n\ge1}\frac{1}{\binom{n+m+  1}{m+1}}

    can be evaluated using Beta functions. I'm not sure how well known this is, so apologies if this is a little mundane.
    \displaystyle \binom{n+m+1}{m+1}^{-1} = \frac{\Gamma (m+2) \Gamma (n+1)}{\Gamma (m+n+2)}.

    Now, \Gamma (z+1) = z \Gamma (z) and \displaystyle \frac{\Gamma (x) \Gamma (y) }{\Gamma (x+y)} = \text{B} (x,y).

    Solution

    \displaystyle\begin{aligned} \sum_{n\geq 1} \binom{n+m+1}{m+1}^{-1} & = \sum_{n\geq 1} \frac{(m+1) \Gamma (m+1) \Gamma (n+1)}{\Gamma (n+m+2)} \\ & = (m+1) \sum_{n\geq 1} \text{B} (m+1, n+1) \\ & = (m+1) \int_0^1 \sum_{n\geq 1} x^n (1-x)^m \ dx \\ & = (m+1) \int_0^1 x (1-x)^{m-1} \ dx \\ & = \frac{1}{m}
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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).
    Go for it.
    How's the self-study coming along now? Also, did you finish with Lang's "Algebra" yet?
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    (Original post by Felix Felicis)
    \displaystyle \binom{n+m+1}{m+1}^{-1} = \frac{\Gamma (m+2) \Gamma (n+1)}{\Gamma (m+n+2)}.

    Now, \Gamma (z+1) = z \Gamma (z) and \displaystyle \frac{\Gamma (x) \Gamma (y) }{\Gamma (x+y)} = \text{B} (x,y).

    Solution

    \displaystyle\begin{aligned} \sum_{n\geq 1} \binom{n+m+1}{m+1}^{-1} & = \sum_{n\geq 1} \frac{(m+1) \Gamma (m+1) \Gamma (n+1)}{\Gamma (n+m+2)} \\ & = (m+1) \sum_{n\geq 1} \text{B} (m+1, n+1) \\ & = (m+1) \int_0^1 \sum_{n\geq 1} x^n (1-x)^m \ dx \\ & = (m+1) \int_0^1 x (1-x)^{m-1} \ dx \\ & = \frac{1}{m}
    Indeed. Any other nice problems solvable with Beta?
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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 Caley-Hamilton 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(A-xI). So p(A) = det(A-AI) = det(A-A) = det(0) = 0.

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    Time for some number theory

    Problem 371 ***


    Prove that for any integer M \geq 1

    \displaystyle \sum_{m=1}^{M} \frac{1}{m} \leq \prod_{p\leq M} \left(1-\frac{1}{p}\right)^{-1}

    where p is a prime number
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    Let p\in \mathbb{P}. Observe that

    RHS=\prod_{p\le M}\left(\frac{1}{1-1/p}\right)=\prod_{p\le M}\left(1+\frac{1}{p}+\frac{1}{p  ^2}+\frac{1}{p^3}+\cdots \right) (infinite geometric series: 1+x^2+x^3+...=1/(1-x), given that |x|<1, which is the case here since all prime are greater than 1)

    Thus, by the fundamental theorem of arithmetic, when p\le M the RHS of the inequality is

    1+\frac{1}{2}+\frac{1}{3}+\cdots,

    where the \cdots indicates that the sum goes on forever, since there are infinitely many numbers that have factors in p\le M. A simple example would be \frac{1}{2^a\times 3^a\times 5^a\times \cdots \times p^a} (p being the largest prime \le M), where a take on infinitely many values.

    Then the problem boils down to showing that

    1+\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{M-1}+\frac{1}{M}\le 1+\frac{1}{2}+\frac{1}{3}+\cdots.

    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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    (Original post by arkanm)
    Let p\in \mathbb{P}. Observe that

    RHS=\prod_{p\le M}\left(\frac{1}{1-1/p}\right)=\prod_{p\le M}\left(1+\frac{1}{p}+\frac{1}{p  ^2}+\frac{1}{p^3}+\cdots \right) (infinite geometric series: 1+x^2+x^3+...=1/(1-x), given that |x|<1, which is the case here since all prime are greater than 1)

    Thus, by the fundamental theorem of arithmetic, when p\le M the RHS of the inequality is

    1+\frac{1}{2}+\frac{1}{3}+\cdots,

    where the \cdots indicates that the sum goes on forever, since there are infinitely many numbers that have factors in p\le M. A simple example would be \frac{1}{2^a\times 3^a\times 5^a\times \cdots \times p^a} (p being the largest prime \le M), where a take on infinitely many values.

    Then the problem boils down to showing that

    1+\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{M-1}+\frac{1}{M}\le 1+\frac{1}{2}+\frac{1}{3}+\cdots.

    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.
    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).
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    (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).
    Yeah I don't think it's an interesting problem. I would rather have people posting interesting problems that are elementary (e.g. olympiad problems) than ones that are easy but require some advanced theory.

    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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    Solution 371

    Note that \displaystyle \prod_{p \le M} \left(\frac{1}{1-p^{-1}} \right) = \sum_{s \in S} \frac{1}{s}, where S consists of all numbers whose prime factors are less than or equal to M. Hence \begin{aligned} \displaystyle \prod_{p \le M} \left(\frac{1}{1-p^{-1}} \right) = \sum_{s: p|s, p \le M} \frac{1}{s} \ge \sum_{s \le M} \frac{1}{s} \end{aligned}.

    (Original post by MW24595)
    Go for it.
    How's the self-study coming along now? Also, did you finish with Lang's "Algebra" yet?
    It goes well except for the last week during which I completed my application to read mathematics at Cambridge . Nowadays, I read from one not-so-well-known book on algebraic topology by Prodanov, Spanier's classic, Hu's Homotopy Theory (I heard that Bott and Tu's Differential Forms in Algebraic Topology is one of the best books on the subject and introduces spectral sequences from the very beginning). I finished Lang's Algebra two months ago, and I dare say that now I find it perfectly clear and not so dry (it has to be supplemented since it is not so heavy on category theory); this is just because I tried Switzer's Homotopy and Homology which is a total dead end (there exists even worse choice - Elements of Homotopy Theory by Whitehead).

    Good luck and enjoy your time at Cambridge.

    Problem 372***

    Let A and B be subsets of the topological space X so that X = A^{\circ} \cup B^{\circ}, where A^{\circ} is the interior of A. Show that there exist natural homomorphisms \Delta : H_{n}(X) \to H_{n-1}(A \cap B) such that \cdots \mathop \to \limits^{\Delta} H_{n}(A \cap B) \mathop \to \limits^{\varphi} H_{n}(A) \oplus H_{n}(B) \mathop \to \limits^{\psi} H_{n}(X) \mathop \to \limits^{\Delta} H_{n-1}(A \cap B) \mathop \to \limits^{\varphi} \cdots is an exact sequence, where \varphi and \psi are defined by the induced homomorphisms, namely \varphi = (i_{\ast},j_{\ast}) and \psi = k_{\ast} - l_{\ast}, i_{\ast} : H_{n}(A \cap B) \to H_{n}(A), j_{\ast} : H_{n}(A \cap B) \to H_{n}(B), k_{\ast} : H_{n}(A) \to H_{n}(X) and l_{\ast} : H_{n}(B) \to H_{n}(X).
    This is a well-known sequence in algebraic topology which plays crucial role in homology theory. Using the same assumptions, show that the homomorphism \Delta : H_{n}(X) \to H_{n-1}(A \cap B) is a composition of the following morphisms H_{n}(X) \mathop \to \limits^{j_{\ast}} H_{n}(X,B) \approx H_{n}(A, A \cap B) \mathop \to \limits^{\delta_{\ast}} H_{n-1}(A \cap B) (now, you might come up with a second proof of the result).

    Problem 373***

    Let X be a subset of S^{n}, homeomorphic to S^{k}, k \in \{0, \cdots, n-1 \}. Prove that \tilde{H}_{n-k-1}(S^{n}-X) = \mathbb{Z} and \tilde{H}_{i}(S^{n}-X) =0, i \not= n-k-1.

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    The order of the above problems is intentional.
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    Problem 372***

    Prove the following:

    If

    f : D \rightarrow \mathbb{C}

    where D is a connected domain, and if f is holomorphic, and

    \forall z \in D, f'(z)=0

    then

     f= constant
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    Some strange glitch. The latest posts aren't visible :/
 
 
 
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