Home > Algebra, Analysis, Undergraduate > SEEMOUS 2013 + Solutions

SEEMOUS 2013 + Solutions

Here are some of the problems of SEEMOUS 2013. Update: the 4th problem has arrived; it is number 3 below.

1. Let {f:[1,8] \rightarrow \Bbb{R}} be a continuous mapping, such that

\displaystyle \int_1^2 f^2(t^3)dt+2\int_1^2f(t^3)dt=\frac{2}{3}\int_1^8 f(t)dt-\int_1^2 (t^2-1)^2 dt.

Find the form of the map {f}.

Solution: Change the variable from {t} to {t^3} in the RHS integral and DO NOT calculate the last integral in the RHS. Get all the terms in the left and find that it is in fact the integral of a square equal to zero.

2. Let {M,N \in \mathcal{M}_2(\Bbb{C})} be nonzero matrices such that {M^2=N^2=0} and {MN+NM=I_2}. Prove that there is an invertible matrix {A \in \mathcal{M}_2(\Bbb{C})} such that {M=A\begin{pmatrix} 0&1\\ 0&0\end{pmatrix}A^{-1}} and {N=A\begin{pmatrix} 0&0 \\ 1&0\end{pmatrix}A^{-1}}.

Solution: One solution can be given using the fact that {M,N} can be written in that form, but for different matrices {A}.

Another way to do it is to consider applications {f,g: \Bbb{C}^2 \rightarrow \Bbb{C}^2,\ f(x)=Mx,\ g(x)=Nx}. We get at once {f^2=0,g^2=0,fg+gf=Id} and from these we deduce that {(fg)^2=fg} and {(gf)^2=gf}. First note that {fg} is not the zero application. Then there exists {u \in Im(fg) \setminus\{0\}}, i.e. there exists {w (\neq 0)} such that {f(g(w))=v}. We have {fg(u)=(fg)^2(w)=fg(w)=u}. Consider {v=g(u)}.

Then {u,v} are not collinear, {f(u)=0,f(v)=u, g(u)=v,g(v)=0}. Consider now the basis formed by {u,v} and take {A} to be the change of base matrix from the canonical base to {\{u,v\}}.

3. Find the maximum possible value of

\displaystyle \int_0^1 |f'(x)|^2|f(x)|\frac{1}{\sqrt{x}}dx

over all continuously differentiable functions {f:[0,1] \rightarrow \Bbb{R}} with {f(0)=0} and {\int_0^1|f'(x)|^2 dx\leq 1}.

4. Let {A \in \mathcal{M}_2(\Bbb{Q})} such that there is {n \in \Bbb{N},\ n\neq 0}, with {A^n=-I_2}. Prove that either {A^2=-I_2} or {A^3=-I_2}.

Solution: Consider {p \in \Bbb{Q}[X]} the minimal polynomial of {A}, which has degree at most {2}. The eigenvalues of {A} satisfy {\lambda_1^n=\lambda_2^n=-1}. We have two cases: either the eigenvalues are real and therefore they are both equal to {-1} either they are complex and conjugate of modulus one. In both cases the determinant of {A} is equal to {1}. Therefore, by Cayley Hamiltoh theorem {A} satisfies an equation of the type {A^2-qA+I_2=0}.

By hypothesis the minimal polynomial {p} divides {X^n+1}. If {p} has degree one then {A=\lambda I_2} and {\lambda \in \Bbb{Q},\lambda^n=-1} so {A=-I_2}.

If not, then the minimal polynomial is {X^2-qX+1} and we must have

\displaystyle X^2-qX+1 | X^n+1.

It can be proved that {q} is in fact an integer. (using the fact that the product of two primitive polynomials is primitive)

Since {q=2\cos \theta} it follows that {q \in \{0,\pm 1,\pm 2\}}. The rest is just casework.

Categories: Algebra, Analysis, Undergraduate Tags:
  1. pedro
    March 24, 2013 at 7:41 pm

    SEEMOUS is comprised of four problem or not?
    Thank you by posting these problems.

    • March 24, 2013 at 7:56 pm

      Yes, SEEMOUS contest subject usually has four problems, but I didn’t receive the 4th problem yet. I’ve searched the internet to see if I can find the problems on other forums, but so far no success. I will post the 4th problem as soon as I find it.

  2. pedro
    March 24, 2013 at 11:47 pm

    Ok very nice. Beni Bogoşel I didn’t understand when you used to solve the problem 3, q = 2cos (x)…


    • March 24, 2013 at 11:50 pm

      The trace of the matrix is q = \lambda_1+\lambda_2=\cos \theta+i\sin \theta+\cos \theta-i\sin\theta=2\cos\theta. I forgot a minus sign in the proof.

  3. March 25, 2013 at 12:22 am

    Pedro, the 4th problem has arrived. It was number 3 who was missing.

  4. pedro
    March 25, 2013 at 5:06 am

    The solution of the problem 3 is perfect, but the original statement says:

    • pedro
      March 25, 2013 at 5:06 am

      pedro :
      The solution of the problem 3 is perfect, but the original statement says:

      • March 25, 2013 at 2:19 pm

        I thought that something was too easy there.

  5. GTz
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