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.

1. March 24, 2013 at 7:41 pm

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

• 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. 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)…

Pedro.

• 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. March 25, 2013 at 5:06 am

The solution of the problem 3 is perfect, but the original statement says:
\int_{1}^{0}|f'(x)|^{2}dx\leq1.

• March 25, 2013 at 5:06 am

pedro :
The solution of the problem 3 is perfect, but the original statement says:
$$\int_{1}^{0}|f’(x)|^{2}dx\leq1.$$

• March 25, 2013 at 2:19 pm

I thought that something was too easy there.

5. March 25, 2013 at 10:15 am