# ISI MMath Solutions and Discussion 2017 : PMB

## ISI MMath PMB 2017 Subjective Questions along with hints and solutions and discussions

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## Group A

**Problem 1.**

Let , such that and as . Let . Show that .

### Topic: Real Analysis, Difficulty level: Medium

**Indian Statistical Institute, ISI MMath 2017 Problem 1 Hints along with Solution & Discussion**

#### Hint 1

Use the representation

Fix any There exists such that for all

#### Hint 2

#### Hint 3

Then for all we have

#### Full Solution

We use the representation

Fix any There exists such that for all

Then for all we have

For any there exists such that for all

In other words,

**Problem 2.**

Let . Prove that there exists such that for all .

### Topic: Real Analysis, Difficulty Level: Medium

**Indian Statistical Institute, ISI MMath 2017 Problem 2 Hints along with Solution & Discussion**

#### Hint 1

Apply the arithmetic mean-geometric mean inequality:

for all and

#### Hint 2

Let where Then

#### Hint 3

Take

#### Full Solution

We apply the arithmetic mean-geometric mean inequality:

for all and

Let where Then

So, for any we have

Take

Then

for any The answer is

**Problem 3.**

Let be an increasing function. Suppose there are sequences and such that for all and as . Prove that is continuous at .

### Topic: Real Analysis, Difficulty Level: Medium

**Indian Statistical Institute, ISI MMath 2017 Problem 3 Hints along with Solution & Discussion**

#### Hint 1

Try to form the mathematical expression with and from this given statement:

“Suppose there are sequences and such that for all and as “.

#### Hint 2

Let There exists such that for all

#### Hint 3

Fix such and take Consider arbitrary If then

and

If then

#### Full Solution

Let There exists such that for all

Fix such and take Consider arbitrary If then

and

If then

and

In any case,

as soon as is continuous at

## Problem 4.

Do there exist continuous functions and on [0,1] such that is a solution to on for all justify your answer.

### Topic: Elements of Ordinary Differential Equations, Difficulty Level: Medium

**Indian Statistical Institute, ISI MMath 2017 Problem 4 Hints along with Solution & Discussion**

#### Hint 1

Assume that such functions exist and then proceed.

#### Hint 2

Derivatives of are equal to

#### Hint 3

In particular,

#### Full Solution

Assume that such functions exist. Derivatives of are equal to

In particular,

If for all the equality

then taking the limit as we get

which is impossible. So, such functions and do not exist.

**Problem 5.**

Let be defined by

for all . Prove that is monotone.

### Topic: Real Analysis, Difficulty Level: Easy

**Indian Statistical Institute, ISI MMath 2017 Problem 5 Hints along with Solution & Discussion**

#### Hint 1

Observe the relation

where

#### Hint 2

We have

#### Hint 3

#### Full Solution

Observe the relation

where We have

In particular, for Since we have and

The function is strictly increasing.

**Problem 6.**

Let be an symmetric matrix with non-negative real entries such that if and only if . Show that is a metric on .

### Topic: Metric Space, Difficulty Level: High

**Indian Statistical Institute, ISI MMath 2017 Problem 6 Hints along with Solution & Discussion**

#### Hint 1

At first we verify that the minimum is achieved. Consider a path between and of length Among elements

there are at least two identical elements: for some Then

is a path of length Also, and

It means that it is enough to consider only paths of length There are only finitely many such paths, hence the minimum in the definition of is achieved. We check axioms of a metric.

#### Hint 2

For any path between and we have

Hence, Assume Since the minimum in the definition of is achieved, we have

for some path between and Since all terms in the left-hand side of (*) are non-negative, we deduce that

By definition of it follows that

In particular, We’ve proved that and

#### Hint 3

Consider a path Then But hence

This proves that

#### Full Solution

At first we verify that the minimum is achieved. Consider a path between and of length Among elements

there are at least two identical elements: for some Then

is a path of length Also, and

It means that it is enough to consider only paths of length There are only finitely many such paths, hence the minimum in the definition of is achieved. We check axioms of a metric.

- For any path between and we have

Hence, Assume Since the minimum in the definition of is achieved, we have

for some path between and Since all terms in the left-hand side of (*) are non-negative, we deduce that

By definition of it follows that

In particular, We’ve proved that and

- Consider a path Then But hence

This proves that

- Consider a path between and such that

Observe that is a path between and So,

We’ve proved that

Since this is true for all we also get

Finally,

- Consider two paths such that

Observe that is a path between and Hence

use that

The triangle inequality is proved.

is a metric on

## Group B

## Problem 7.

Factory produces bad watch in and factory produces bad watch in . You are given two watches from one of the factories and you don’t know which one.

- What is the probability that the second watch works?
- Given that the first watch works, what is the probability that the second watch works

### Topic: Probability Theory, Difficulty Level: Medium

**Indian Statistical Institute, ISI MMath 2017 Problem 7 Hints along with Solution & Discussion**

#### Hint 1

Introduce two hypothesis

#### Hint 2

Then Consider events

It is given that

#### Hint 3

By the law of total probability

#### Full Solution

Introduce two hypothesis

Then Consider events

It is given that

- By the law of total probability

- We compute the probability by the definition:

From the previous part, we know By the law of total probability

So,

## Problem 8.

Let be a commutative ring containing a field as a sub-ring. Assume that is a finite dimensional -vector space. Show that every prime ideal of is maximal.

### Topic: Abstract Algebra- Ring and Field Theory, Difficulty Level: Medium

**Indian Statistical Institute, ISI MMath 2017 Problem 8 Hints along with Solution & Discussion**

#### Hint 1

Let be a prime ideal in . It is enough to check that is a field

#### Hint 2

The infinite sequence is linearly dependent and there exists samllest such that

for some non-zero coefficients If

#### Hint 3

Then

#### Full Solution

Let be a prime ideal in . It is enough to check that is a field, i.e. every nonzero element of has a multiplicative inverse. is a finite-dimensional -vector space. Hence the infinite sequence is linearly dependent and there exists smallest such that

for some non-zero coefficients If then

Let be a representative of Then

hence The ideal is prime, so

and

The latter contradicts minimality of So, and

has a multiplicative inverse. is a field and is a maximal ideal.

## Problem 9.

Let be prime numbers and such that then show that .

### Topic: Number Theory, Difficulty Level: Easy

**Indian Statistical Institute, ISI MMath 2017 Problem 9 Hints along with Solution & Discussion**

#### Hint 1

By assumptions of the problem

#### Hint 2

Let be the order of modulo i.e. is the smallest positive integer such that

#### Hint 3

From we deduce that is prime, so

Now use “Fermat’s Little Theorem”.

#### Full Solution

By assumptions of the problem Let be the order of modulo i.e. is the smallest positive integer such that From we deduce that is prime, so By the Fermat Little Theorem,

Since is the order of modulo it follows that

## Problem 10.

Determine all finite groups which have exactly conjugacy classes.

### Topic: Group Theory, Difficulty Level: Hard

**Indian Statistical Institute, ISI MMath 2017 Problem 10 Hints along with Solution & Discussion**

#### Hint 1

Let be a finite group with exactly conjugacy classes. One conjugacy class is always the identity class .

#### Hint 2

Let and be sizes of other two conjugacy classes. Conjugacy classes form a partition of a group, so

#### Hint 3

The size of a conjugacy class of equals to the index of a centralizer of

#### Full Solution

Let be a finite group with exactly conjugacy classes. One conjugacy class is always the identity class Let and be sizes of other two conjugacy classes. Conjugacy classes form a partition of a group, so

The size of a conjugacy class of equals to the index of a centralizer of i.e. it divides So,

Further,

Assume that Then i.e. The only group of order is the cyclic group It is an abelian group, hence it has exactly 3 conjugacy classes.

Assume that Then hence If then and The only groups of order 4 are the Klein 4 group or the cyclic group Both are abelian and thus have 4 conjugacy classes.

If then and There are two groups of order an abelian group and a nonabelian group (permutations on three elements). We check that has three conjugacy classes:

The only finite groups with three conjugacy classes are and

## Problem 11.

Let be a field, a prime integer. Suppose the polynomial is reducible in . Prove that this polynomial has a root in .

### Topic: Field Theory, Difficulty Level: Medium

**Indian Statistical Institute, ISI MMath 2017 Problem 11 Hints along with Solution & Discussion**

#### Hint 1

In the case obviously is the needed root.

#### Hint 2

We assume firther that Let be the splitting field of over

#### Hint 3

Then

for some

#### Full Solution

In the case obviously is the needed root. We assume firther that Let be the splitting field of over Then

for some Since the polynomial is reducible, we may relabel its roots in such a way that for some a polynomial

is a factor of i.e. has coefficients in . Observe that for each Consider the free term of It is of the form Hence, Further,

Since we can find integers such that

We have and

is a root of in .

## Problem 12.

Let be a finite-dimensional vector space over a field and let be a linear transformation. Let be a subspace such that . Suppose is diagonalizable. Is restricted to also diagonalizable?

### Topic: Linear Algebra, Difficulty Level: Hard

**Indian Statistical Institute, ISI MMath 2017 Problem 12 Hints along with Solution & Discussion**

#### Hint 1

By assumption, is a direct sum of subspaces

and on each subspace acts as a multiplication by all are distinct.

#### Hint 2

Consider Then

for some Let be the number of nonzero summands

#### Hint 3

By induction on we will prove that all If then and there is nothing to prove.

#### Full Solution

By assumption, is a direct sum of subspaces

and on each subspace acts as a multiplication by all are distinct.

Consider Then

for some Let be the number of nonzero summands By induction on we will prove that all If then and there is nothing to prove. Assume the result is proved for number of nonzero summands and let Then

By inductive assumption, for all

Eigenvalues are distinct, so for But then

This statement shows that in the decomposition

all summands are in i.e.

The restriction of to is diagonalizable.

Hi! Thanks for your selfless help by putting out all the solutions free of cost on a public domain. I really appreciate it and I’m extremely thankful.

I have a doubt. Can you please explain the reasoning behind the first hint of problem 9?

This is trivially seen from the result . If then it implies that and so .

In problem 10, in the case when m<n, I understood that m+1 should be less than or equal to n.

But how did we immediately get in the next step, that n = m+1 ?

Since then from that it implies and thus the conclusion.

Understood both of them, thanks!

In Q.12. Why have we assumed that all eigenvalues are distinct, when we may have a diagonalizable linear transformation on V which has repeated eigenvalues as long as the sum of dimensions of the eigenspaces is dim(V)?