Indian Statistical Institute, ISI MMATH 2021 PMB Solutions & Discussions

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ISI MMath PMB 2021 Subjective Questions, solutions and discussions

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Problem 1.

Let $M$ be a real $n \times n$ matrix with all diagonal entries equal to $r$ and all non-diagonal entries equal to s. Compute the determinant of $M$ .

Topic: Linear Algebra, Difficulty level: Medium

Solution:

Fix $s.$ Let $M_n(r)$ be the $n\times n$ matrix whose diagonal elements are equal to $r$ and non-diagonal elements are equal to $s.$ Denote $P_n(r)=\det M_n(r).$ Then $P_n$ is a polynomial of degree $n$ in $r.$ We observe that $P_n(s)=0.$ The sum of rows of the matrix $M_n(r)$ is equal to

$\begin{pmatrix} r+(n-1)s & r+(n-1)s & \ldots & r+(n-1)s\end{pmatrix}.$

If $r=-(n-1)s$ then this row is zero, $P_n(-(n-1)s)=0.$ Now we compute the derivative of $P_n:$

$P'_n(r)=\frac{d}{dt}\det M_n(r)=\det \begin{pmatrix} \frac{d}{dr}r & \frac{d}{dr}s & \frac{d}{dr}s & \ldots & \frac{d}{dr}s \\ s& r & s & \ldots & s\\ s & s & r & \ldots & s \\ \ldots & \ldots & \ldots & \ldots & \ldots \\ s & s & s & \ldots & r \\ \end{pmatrix}+$

$+\det \begin{pmatrix} r & s & s & \ldots & s \\ \frac{d}{dr}s& \frac{d}{dr}r & \frac{d}{dr}s & \ldots & \frac{d}{dr}s\\ s & s & r & \ldots & s \\ \ldots & \ldots & \ldots & \ldots & \ldots \\ s & s & s & \ldots & r \\ \end{pmatrix}+\ldots +\det \begin{pmatrix} r & s & s & \ldots & s \\ s& r & s & \ldots & s\\ s & s & r & \ldots & s \\ \ldots & \ldots & \ldots & \ldots & \ldots \\ \frac{d}{dr}s & \frac{d}{dr}s & \frac{d}{dr}s & \ldots & \frac{d}{dr}r \\ \end{pmatrix}=$

$=\det \begin{pmatrix} 1 & 0 & 0 & \ldots & 0 \\ s& r & s & \ldots & s\\ s & s & r & \ldots & s \\ \ldots & \ldots & \ldots & \ldots & \ldots \\ s & s & s & \ldots & r \\ \end{pmatrix}+\det \begin{pmatrix} r & s & s & \ldots & s \\ 0& 1 & 0 & \ldots & 0\\ s & s & r & \ldots & s \\ \ldots & \ldots & \ldots & \ldots & \ldots \\ s & s & s & \ldots & r \\ \end{pmatrix}+\ldots +$

$+\det \begin{pmatrix} r & s & s & \ldots & s \\ s& r & s & \ldots & s\\ s & s & r & \ldots & s \\ \ldots & \ldots & \ldots & \ldots & \ldots \\ 0 & 0 & 0 & \ldots & 1 \\ \end{pmatrix}=$

$=n\det M_{n-1}(r)=nP_{n-1}(r).$

By induction, $P^{(k)}_n(r)=\frac{n!}{(n-k)!}P_{n-k}(r),$ $1\leq k\leq n-1.$ It follows that $r=s$ is a root of $P^{(k)}_n(r)$ . Hence, $r=s$ is a root of order $n-1$ of $P_n(r),$ and $r=-(n-1)s$ is a root of order $1.$ Finally,

$\det M_n(r)=P_n(r)=(r-s)^{n-1}(r+(n-1)s).$

Problem 2.

Let $F[X]$ be the polynomial ring over a field $F$ . Prove that the rings $F[X] /\left\langle X^{2}\right\rangle$ and $F[X] /\left(X^{2}-1\right)$ are isomorphic if and only if the characteristic of $F$ is $2$ .

Topic: Abstract Algebra, Difficulty Level: Medium

Solution:

Assume that the rings $F[X]/\langle X^2\rangle$ and $F[X]/\langle X^2-1\rangle$ are isomorphic, and let $\varphi:F[X]/\langle X^2\rangle\to F[X]/\langle X^2-1\rangle$ be an isomoprhism. Then

$\varphi(X+\langle X^2\rangle)=b_0+b_1 X+\langle X^2-1\rangle.$

Since $\varphi$ is an isomorphism, necessarily $b_0\ne 0$ or $b_1\ne 0.$

It follows that

$\langle X^2-1\rangle=0+\langle X^2-1\rangle=\varphi((X+\langle X^2\rangle)^2)=$

$=\varphi(X+\langle X^2\rangle)^2=b^2_0+(b_0b_1+b_0b_1)X+b^2_1 X^2+\langle X^2-1\rangle=$

$=b^2_0+b^2_1+(b_0b_1+b_0b_1)X+\langle X^2-1\rangle.$

That is,

$b^2_0+b^2_1=0, \ (b_0b_1+b_0b_1)=0.$

Then $(b_0-b_1)^2=0,$ hence $b_0=b_1$ and $b^2_0+b^2_0=0.$ If $b_0=0,$ then $b_1=0$ and $\varphi$ can’t be an isomorphism. So, $b_0\ne 0$ and we find that $F$ is of characteristic $2.$

Conversely, assume that $F$ is of characteristic $2.$ Consider the mapping

$\varphi: F[X]/\langle X^2\rangle\to F[X]/\langle X^2-1\rangle, \varphi(a_0+a_1X+\langle X^2\rangle )=a_0+a_1(1+X)+\langle X^2-1\rangle.$

The mapping $\varphi$ is an isomorphism. Indeed,

$\varphi(a_0+a_1X+c_0+c_1X+\langle X^2\rangle )=a_0+c_0+(a_1+c_1)(1+X)+\langle X^2-1\rangle=$

$=\varphi(a_0+a_1X+\langle X^2\rangle )+\varphi(c_0+c_1X+\langle X^2\rangle );$

$\varphi((a_0+a_1X+\langle X^2\rangle)(c_0+c_1X+\langle X^2\rangle ) )=\varphi(a_0c_0+(a_0c_1+a_1c_0)X+\langle X^2\rangle )=$

$=a_0c_0+(a_0c_1+a_1c_0)(1+X)+\langle X^2-1\rangle=(a_0+a_1(1+X)+\langle X^2-1\rangle)(c_0+c_1(1+X)+\langle X^2-1\rangle)=$

$=\varphi(a_0+a_1X+\langle X^2\rangle )\varphi(c_0+c_1X+\langle X^2\rangle ),$

$(1+X+\langle X^2-1\rangle)^2=(1+X+X+X^2+\langle X^2-1\rangle)=(1+X+X+1+\langle X^2-1\rangle)=\langle X^2-1\rangle$

Problem 3.

Let $C$ be a subset of $\mathbb{R}$ endowed with the subspace topology. If every continuous real-valued function on $C$ is bounded, then prove that $C$ is compact.

Topic: Topology, Difficulty Level: Easy

Solution:

The function $f:C\to \mathbb{R},$ $f(x)=x,$ is continuous. Hence it is bounded on $C:$

$\sup_{x\in C}|x|<\infty.$

The set $C$ is bounded.

Let $x_0\in \overline{C}$ (i.e. $x_0$ is in the closure of $C$ ). If $x_0\not\in C,$ then the function

$g(x)=\frac{1}{|x-x_0|}$

is continuous on $C.$ For every $\epsilon>0$ there exists a point $x\in C$ such that $|x-x_0|<\epsilon.$ Then $g(x)>\frac{1}{\epsilon}.$ As $\epsilon>0$ is arbitrary, this shows that $g$ is unbounded on $C$ . The obtained contradiction proves that $C=\overline{C}.$ $C$ is bounded and closed, hence compact.

Problem 4.

Let $A=\left(a_{i j}\right)$ be a nonzero real $n \times n$ matrix such that $a_{i j}=0$ for $i \geq j$ . If $\sum_{i=0}^{k} c_{i} A^{i}=0$ for some $c_{i} \in \mathbb{R}$ , then prove that $c_{0}=c_{1}=0$ . Here $A^{i}$ is the $i$ -th power of $A$ .

Topic: Linear Algebra, Difficulty Level: Medium

Solution:

Let $r\in \{1,\ldots,n-1\}$ be the first number such that $a_{i,i+r}\ne 0$ for some $i$ (such $r$ exists as $A\ne 0$ and $a_{ij}=0$ for $j\leq i$ ) Denote $A^d=(a^{(d)}_{ij})_{i\leq j}.$ We check by induction that for all $d\geq 1$

$j\leq i+r+d-2\Rightarrow a^{(d)}_{ij}=0.$

For $d=1$ the statement is true by the choice of $r.$ Assume the statement is proved for $d.$ Let $j<i+r+d.$ Then

$a^{(d+1)}_{ij}=\sum^n_{l=1} a^{(d)}_{il}a_{lj}=\sum_{l: l\geq i+r+d-1, j\geq l+r}a^{(d)}_{il}a_{lj}$

the sum is taken over non-empty set of indices when $j-r\geq l\geq i+r+d-1,$ i.e. when $j\geq i+2r+d-1\geq i+r+d=i+r+(d+1)-1.$ The statement is proved.

Let $i_0$ be such that $a_{i_0,i_0+r}\ne 0.$ For all $d\geq 2$ we have $i_0+r\leq i_0+r+d-2,$ hence $a^{(d)}_{i_0,i_0+r}=0.$ We find that

$0=\sum^k_{d=0} c_i a^{(d)}_{i_0,i_0+r}=c_1a_{i_0,i_0+r}, \ c_1=0$

( $a^{(0)}_{i_0,i_0+r}=0$ because it is the off-diagonal element of the identity matrix). Also, for all $d\geq 1$ we have $i_0\leq i_0+r+d-2,$ hence $a^{(d)}_{i_0,i_0}=0$ and

$0=\sum^k_{d=0} c_i a^{(d)}_{i_0,i_0}=c_0.$

Problem 5.

Let $g: \mathbb{R} \rightarrow \mathbb{R}$ be the function given by

$g(x)= \begin{cases}x \sin \left(\frac{1}{x}\right), & x \neq 0 \\ 0, & x=0\end{cases}$

Prove that $g(x)$ has a local maximum and a local minimum in the interval $\left(-\frac{1}{m}, \frac{1}{m}\right)$ for any positive integer $m$ .

Topic: Real Analysis, Difficulty Level: Medium

Solution:

On each interval $(-\frac{\pi}{2}+\pi n, \frac{\pi}{2}+\pi n)$ the function $\tan(t)-t$ is strictly increasing. Indeed, its derivative is

$(\tan t-t)'=\frac{1}{\cos^2 t}-1=\frac{\sin^2 t}{\cos^2 t}.$

Let $n\geq 1.$ At $t=\pi n$ we have $\tan t-t=-\pi n < 0,$ and $\lim_{t\to \frac{\pi}{2}+\pi n}(\tan t- t)=\infty.$ Then there is a unique $t_n\in (\pi n, \frac{\pi}{2}+\pi n)$ such that

$t_n<t< \frac{\pi}{2}+\pi n\Rightarrow \tan t>t,$

$\pi n < t< t_n\Rightarrow \tan t<t.$

Consider the derivative of the function $g(x)$ for $x>0:$

$g'(x)=\sin\frac{1}{x}-\frac{1}{x}\cos\frac{1}{x}.$

Let $m\geq 1$ and take $n>\frac{m}{2\pi}.$ Then $(\frac{1}{\frac{\pi}{2}+2\pi n},\frac{1}{2\pi n})\subset (-\frac{1}{m},\frac{1}{m}).$ For $x\in (\frac{1}{\frac{\pi}{2}+2\pi n},\frac{1}{2\pi n})$ we have

$2\pi n<\frac{1}{x}< \frac{\pi}{2}+2\pi n, \cos\frac{1}{x}>0.$

Further,

$\frac{1}{\frac{\pi}{2}+2\pi n}<x< \frac{1}{t_{2n}}\Rightarrow g'(x)=\cos\frac{1}{x}\left(\tan\frac{1}{x}-\frac{1}{x}\right)>0$

and

$\frac{1}{t_{2n}}<x<\frac{1}{2\pi n}\Rightarrow g'(x)=\cos\frac{1}{x}\left(\tan\frac{1}{x}-\frac{1}{x}\right)<0.$

Function $g(x)$ attains a local maximum at $x=\frac{1}{t_{2n}}.$

For $x\in (\frac{1}{\frac{\pi}{2}+(2n+1)\pi },\frac{1}{(2n+1)\pi })$ we have

$(2n+1)\pi <\frac{1}{x}< \frac{\pi}{2}+(2n+1)\pi , \cos\frac{1}{x}<0.$

Further,

$\frac{1}{\frac{\pi}{2}+(2n+1)\pi }<x< \frac{1}{t_{2n+1}}\Rightarrow g'(x)=\cos\frac{1}{x}\left(\tan\frac{1}{x}-\frac{1}{x}\right)<0$

and

$\frac{1}{t_{2n+1}}<x<\frac{1}{(2n+1)\pi }\Rightarrow g'(x)=\cos\frac{1}{x}\left(\tan\frac{1}{x}-\frac{1}{x}\right)>0.$

Function $g(x)$ attains a local minimum at $x=\frac{1}{t_{2n+1}}.$

On each interval $(-\frac{1}{m},\frac{1}{m})$ function $g$ attains a local maximum and a local minimum.

Problem 6.

Fix an integer $n \geq 1$ . Suppose that $n$ is divisible by distinct natural numbers $k_{1}, k_{2}, k_{3}$ such that

$\operatorname{gcd}\left(k_{1}, k_{2}\right)=\operatorname{gcd}\left(k_{2}, k_{3}\right)=\operatorname{gcd}\left(k_{3}, k_{1}\right)=1$

Pick a random natural number $j$ uniformly from the set $\{1,2,3, \ldots, n\}$ . Let $A_{d}$ be the event that $j$ is divisible by $d$ . Prove that the events $A_{k_{1}}, A_{k_{2}}, A_{k_{3}}$ are mutually independent.

Topic: Probability, Difficulty Level: Medium

Solution:

Let $l$ divides $n.$ There are $\frac{n}{l}$ numbers in $\{1,2,\ldots,n\}$ that are divisible by $l.$ So, $P(A_{l})=\frac{1}{l}.$

Since $k_i$ and $k_j$ are co-prime when $i\ne j,$ a number is divisible by $k_ik_j$ if and only if it is divisible by $k_i$ and by $k_j:$

$A_{k_i}\cap A_{k_j}=A_{k_ik_j}, P(A_{k_i}\cap A_{k_j})=\frac{1}{k_ik_j}=P(A_{k_i})P(A_{k_j}), \ 1\leq i<j\leq 3.$

Similarly, $A_{k_1}\cap A_{k_2}\cap A_{k_3}=A_{k_1k_2k_3}$ and

$P(A_{k_1}\cap A_{k_2}\cap A_{k_3})=\frac{1}{k_1k_2k_3}=P(A_{k_1})P(A_{k_2})P(A_{k_3}).$

The equality

$P(A_{k_{i_1}}\cap \ldots A_{k_{i_m}})=P(A_{k_{i_1}}) \ldots P(A_{k_{i_m}})$

holds for all $m=1,2,3$ and all $1\leq i_1<\ldots<i_m\leq 3.$ Events $A_{k_1},$ $A_{k_2},$ $A_{k_3}$ are mutually independent.

Problem 7.

Let $f:[0,1] \rightarrow[0, \infty)$ be a function. Assume that there exists $M \geq 0$
such that $\sum_{i=1}^{k} f\left(x_{i}\right) \leq M$ for all $k \geq 1$ and for all $x_{1}, \ldots, x_{k} \in[0,1]$
Show that the set $\{x \mid f(x) \neq 0\}$ is countable.

Topic: Real Analysis, Difficulty Level: Easy

Solution:

Remark: I assume that $x_1,\ldots,x_k$ in the statement of the problem are distinct. Otherwise, $f$ must be identically zero.

Denote $A_n=\{x\in [0,1]:f(x)\geq \frac{1}{n}\}.$ Then $\{x:f(x)\ne 0\}=\cup^\infty_{n=1}A_n.$ We will show that each set $A_n$ is finite. Indeed, if $x_1,\ldots,x_k\in A_n$ are distinct, then

$M\geq \sum^k_{i=1}f(x_i)\geq \frac{k}{n}.$

So, $k\leq nM$ and the set $A_n$ has at most $nM$ elements.

Problem 8.

Let $G$ be a group having exactly three subgroups. Prove that $G$ is cyclic of order $p^{2}$ for some prime $p$ .

Topic: Group Theory, Difficulty Level: Easy

Solution:

$G$ has three subgroups: $\{1\},$ $G$ and some other subgroup $H$ . The group $G$ has an element $g,$ such that $g\ne 1.$ Consider a cyclic subgroup $S_g$ generated by $g.$ If this subgroup coincides with $G$ , then $G$ is cyclic. Otherwise, $S_g=H.$ There is an element $h\not\in S_g.$ Let $S_h$ be a cyclic subgroup generated by $h.$ Since $S_h\ne \{1\}$ and $S_h\ne S_g=H,$ then $S_h=G$ and $G$ is cyclic. The subgroup $H$ has no proper subgroups, so its order is prime: $|H|=p.$ Correspondingly, $|G|=pn$ for some integer $n.$ By the fundamental theorem of cyclic groups, if $k$ divides $n,$ then $G$ has a subgroup of order $k.$ It follows that the only possible divisor of $n$ is $p.$ $|G|=p^2$ .