# ISI MMath Solutions and Discussion 2016 : PMB

## Problem 1.

Let be a sequence of real numbers defined as follows: and for all .

(a)Show that there exists such that for all ,

(b) Prove that exists and find its value.

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

#### Full Solution

(a) By induction we check that Indeed, for this is true by definition. Assuming that we have hence

Next, we estimate

The needed estimate holds with

(b) We have By induction we check

For it is true:

Assuming that we deduce from (a) that

For we estimate

Since we deduce that

i.e. is a Cauchy sequence and the limit exists. Passing to the limit in the equation

we find that

But for all So, and

## Problem 2.

Examine, with justification, whether the following limit exists:

.
If the limit exists, then find its value.

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

#### Full Solution

Consider the elementary estimate

Applying it to the integral we get

Answer: The limit exists and is equal to zero.

## Problem 3.

Does there exist a continuous function that takes every real value exactly twice Justify your answer.

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

#### Full Solution

Assume that such a function exists and let be any of its values. By assumption, there exist such that

On the interval the function never takes the value Consequently, either for all or for all Without loss of generality we can assume that for all (in the opposite case we just consider the function is also continuous and takes each value exactly twice). reaches maximum on the segment at some point Let For each we have so and we necessarily have By assumption, there exists such that Observe that Cconsider two possibilities:

• From the Intermediate Value Theorem, any value is reached at least three times: once between and an end-point of once between and and once between and This is impossible.
• Then If is constant between and then the value is taken infinitely many times, which is impossible. If is not constant, there exists between and such that Fix The value is taken at least four times: between and the end-point of between and the end-point of between and and between and

In any case we get a contradiction, so there is no continuous function that takes every real value exactly twice.

## Problem 4.

Suppose is a bounded function such that is Riemann integrable on for every . Is Riemann integrable on ? justify your answer.

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

#### Full Solution

Let We will prove that is Riemann integrable by showing that for every there exists a partition of such that for the upper Darboux sum and the lower Darboux sum the inequality

holds.

Let be such that Since is Riemann integrable on there exists a partition of such that

Set for so that Observe that

So,

is Riemann integrable on

## Problem 5.

Let be a surjective function such that

for all . Here denotes the Euclidean norm on . Show that the image of every open set (in ) under the map is an open set (in ).

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

#### Full Solution

Let be open. Fix for some There exists such that

Let By assumption, there exists such that We estimate

so,

and We have proved that

which means that is open.

## Problem 6.

Suppose that is a continuous function and

Define a new function by

Now define another function by

### Topic: Partial Differential, Difficulty Level: High

#### Full Solution

At first we check that is continuous. We need two consequences of continuity of

• (boundedness) there exists such that for all
• (uniform continuity) for any there exists such that

Given choose such that (*) holds. Let be such that For any relation (*) implies

So, we have

Continuity of is proved.

For fixed the function is continuous on By the Fundamental Theorem of Calculus, the integral

is differentiable at and

So, we`ve proved that

## Problem 7.

Let be a matrix with complex entries. Suppose that and trace. Show the following:

(a) Kernel(A) Range(A).

(b) span(Kernel(A) Range(A)).

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

#### Full Solution

(a) The characteristic polynomial of is

So, has two eigenvalues and Let and be corresponding eigenvectors. is a basis of Assume Then for some and Write Then

Since we deduce and

(b) so On the other hand and It follows that

From relation it follows that the range of is the one-dimensional subspace generated by

Finally,

## Problem 8.

Suppose that is a nonzero matrix with complex entries. Prove that if and only if the and the matrix are similar.

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

#### Full Solution

Assume that Consider the Jordan normal form of the matrix there exists an invertibale matrix such that

and either or In the first case

Then and This is not the case.

Consequently,

In this case

and We have proved that is similar to the matrix Observe that for we have

So,

Conversely, if is similar to the matrix then is similar to the matrix

i.e.

## Problem 9.

Let be group of all permutation of distinct symbols. How many subgroups of order does have? Justify your answer.

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

#### Full Solution

A subgroup of prime order is cyclic. Let
be a subgroup of of order Every element of distinct from the identity is the generator of The order of a permutation is the least common mulitple of orders of permutations in its cyckle decomposition. If is a generator, then it is of order and its cycle decomposition consists of a single cycle of length

Replacing with for (if needed), we may assume that

There are such elements and they generate distinct subgroups of order in

Answer: there are subgroups of order

## Problem 10.

Suppose that and are two subgroups of a group . Assume that and is not a subgroup of . Show that

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

#### Full Solution

Take element The cosets and are disjoint, so and

## Problem 11.

For any ring , let denote the ring of all polynomials with indeterminate and coefficients from . Examine, with justification, whether the following pairs of rings are isomorphic:

(a) and .

(b) and

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

#### Full Solution

(a) Assume that there exists an isomorphism Observe that in we have

Let for some Then

However, the free term on the left-hand side is

(b) Consider the mapping given by

The mapping is a surjective homeomoprhism. Indeed, for any we have

We find the kernel of

and are roots of i.e. is divisible by

The kernel of coincides with the ideal generated by By the isomoprhism theorem,

are isomorphic.

## Problem 12.

For any , let be the smallest subfield of containing . Find a basis for the vector space over

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

#### Full Solution

Set Then

is a root of the polynomial

If we prove that is irreducible in , we will verify that

and

is a basis for the vector space over

To prove that is irreducible in it is enough to show that the roots do not belong to Assume that

Then

which is impossible as So, is irreducible in

Answer: a basis for the vector space over