Indian Statistical Institute, ISI MMATH 2021 PMB Solutions & Discussions
ISI MMath PMB 2021 Subjective Questions, solutions and discussions
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Problem 1.
Let be a real
matrix with all diagonal entries equal to
and all non-diagonal entries equal to s. Compute the determinant of
.
Topic: Linear Algebra, Difficulty level: Medium
Solution:
Fix Let
be the
matrix whose diagonal elements are equal to
and non-diagonal elements are equal to
Denote
Then
is a polynomial of degree
in
We observe that
The sum of rows of the matrix
is equal to
If then this row is zero,
Now we compute the derivative of
By induction,
It follows that
is a root of
. Hence,
is a root of order
of
and
is a root of order
Finally,
Problem 2.
Let be the polynomial ring over a field
. Prove that the rings
and
are isomorphic if and only if the characteristic of
is
.
Topic: Abstract Algebra, Difficulty Level: Medium
Solution:
Assume that the rings and
are isomorphic, and let
be an isomoprhism. Then
Since is an isomorphism, necessarily
or
It follows that
That is,
Then hence
and
If
then
and
can’t be an isomorphism. So,
and we find that
is of characteristic
Conversely, assume that is of characteristic
Consider the mapping
The mapping is an isomorphism. Indeed,
as
Problem 3.
Let be a subset of
endowed with the subspace topology. If every continuous real-valued function on
is bounded, then prove that
is compact.
Topic: Topology, Difficulty Level: Easy
Solution:
The function
is continuous. Hence it is bounded on
The set is bounded.
Let (i.e.
is in the closure of
). If
then the function
is continuous on For every
there exists a point
such that
Then
As
is arbitrary, this shows that
is unbounded on
. The obtained contradiction proves that
is bounded and closed, hence compact.
Problem 4.
Let be a nonzero real
matrix such that
for
. If
for some
, then prove that
. Here
is the
-th power of
.
Topic: Linear Algebra, Difficulty Level: Medium
Solution:
Let be the first number such that
for some
(such
exists as
and
for
) Denote
We check by induction that for all
For the statement is true by the choice of
Assume the statement is proved for
Let
Then
the sum is taken over non-empty set of indices when i.e. when
The statement is proved.
Let be such that
For all
we have
hence
We find that
( because it is the off-diagonal element of the identity matrix). Also, for all
we have
hence
and
Problem 5.
Let be the function given by
Prove that has a local maximum and a local minimum in the interval
for any positive integer
.
Topic: Real Analysis, Difficulty Level: Medium
Solution:
On each interval the function
is strictly increasing. Indeed, its derivative is
Let At
we have
and
Then there is a unique
such that
Consider the derivative of the function for
Let and take
Then
For
we have
Further,
and
Function attains a local maximum at
For we have
Further,
and
Function attains a local minimum at
On each interval function
attains a local maximum and a local minimum.
Problem 6.
Fix an integer . Suppose that
is divisible by distinct natural numbers
such that
Pick a random natural number uniformly from the set
. Let
be the event that
is divisible by
. Prove that the events
are mutually independent.
Topic: Probability, Difficulty Level: Medium
Solution:
Let divides
There are
numbers in
that are divisible by
So,
Since and
are co-prime when
a number is divisible by
if and only if it is divisible by
and by
Similarly, and
The equality
holds for all and all
Events
are mutually independent.
Problem 7.
Let be a function. Assume that there exists
such that for all
and for all
Show that the set is countable.
Topic: Real Analysis, Difficulty Level: Easy
Solution:
Remark: I assume that in the statement of the problem are distinct. Otherwise,
must be identically zero.
Denote Then
We will show that each set
is finite. Indeed, if
are distinct, then
So, and the set
has at most
elements.
Problem 8.
Let be a group having exactly three subgroups. Prove that
is cyclic of order
for some prime
.
Topic: Group Theory, Difficulty Level: Easy
Solution:
has three subgroups:
and some other subgroup
. The group
has an element
such that
Consider a cyclic subgroup
generated by
If this subgroup coincides with
, then
is cyclic. Otherwise,
There is an element
Let
be a cyclic subgroup generated by
Since
and
then
and
is cyclic. The subgroup
has no proper subgroups, so its order is prime:
Correspondingly,
for some integer
By the fundamental theorem of cyclic groups, if
divides
then
has a subgroup of order
It follows that the only possible divisor of
is
.