Indian Statistical Institute, ISI MMATH 2021 PMB Solutions & Discussions
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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
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,
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
Assume that the rings and are isomorphic, and let be an isomoprhism. Then
Since is an isomorphism, necessarily or
It follows that
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,
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
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.
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
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
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
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
Function attains a local maximum at
For we have
Function attains a local minimum at
On each interval function attains a local maximum and a local minimum.
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
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
holds for all and all Events are mutually independent.
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
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.
Let be a group having exactly three subgroups. Prove that is cyclic of order for some prime .
Topic: Group Theory, Difficulty Level: Easy
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 .