TIFR 2015 Problem 7 Solution -Increasing Function and Continuity


TIFR 2015 Problem 7 Solution is a part of TIFR entrance preparation series. The Tata Institute of Fundamental Research is India's premier institution for advanced research in Mathematics. The Institute runs a graduate program leading to the award of Ph.D., Integrated M.Sc.-Ph.D. as well as M.Sc. degree in certain subjects.
The image is a front cover of a book named Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert. This book is very useful for the preparation of TIFR Entrance.

Also Visit: College Mathematics Program of Cheenta

Problem:

Let $f$ and $g$ be two functions from $([0,1])$ to $([0,1])$ with $f$ strictly increasing. Which of the following statements is always correct?

A. If (g) is continuous, then $(f\circ g)$ is continuous

B. If  (f) is continuous, then $(f \circ g)$ is continuous

C. If (f) and $(f \circ g)$ is continuous, then (g) is continuous

D. If (g) and $(f \circ g)$ are continuous, then (f) is continuous

Discussion:

A: Let (g(x)=x) for all $(x\in [0,1])$.

(f(x)=x) for $(x \in [0,\frac{1}{2}])$ and (f(x)=5+x) for $(x\in (\frac{1}{2},1])$.

Then $(f \circ g=f)$ and (f) is not continuous.

So A is False.

B: Reverse (f) and (g) in A to show that B is False.

C: If (f) and $(f \circ g)$ are continuous then (f) is 1-1 (increasing), continuous map ([0,1]to [0,1]).

(A subset [0,1] ) be closed. Then (A) is compact. (Closed subsets of compact spaces are compact).

Therefore (f(A)) is compact. (continuous image of compact set is compact).

We have that (f(A)) is a compact subset of ([0,1]). Therefore (f(A)) is closed in ([0,1]). (compact subspace of Hausdorff space is closed).

Therefore, (f) is a closed map. So $(f^{-1})$ is continuous.

Hence $(f^{-1} \circ f \circ g=g)$ is continuous.

So, C is True.

D: Let $(g(x)=\frac{x}{4})$ for all $(x\in [0,1])$.

(f(x)=x) for $(x \in [0,\frac{1}{2}])$ and (f(x)=5+x) for $(x\in (\frac{1}{2},1])$.

Then $(f \circ g(x)=f(\frac{x}{4})=\frac{x}{4})$ for all $(x \in [0,1])$.

So $(f \circ g)$ is continuous but (f) is not continuous.

So, D is False.

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TIFR 2015 Problem 4 Solution - Groups without Commuting Elements


TIFR 2015 Problem 4 Solution is a part of TIFR entrance preparation series. The Tata Institute of Fundamental Research is India's premier institution for advanced research in Mathematics. The Institute runs a graduate programme leading to the award of Ph.D., Integrated M.Sc.-Ph.D. as well as M.Sc. degree in certain subjects.
The image is a front cover of a book named Topics in Algebra by I.N.Herstein. This book is very useful for the preparation of TIFR Entrance.

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PROBLEM:

Let (S) be the collection of isomorphism classes of groups (G) such that every element of G commutes with only the identity element and itself. Then what is (|S|)?

Discussion:

Given any $g\in G$, it commutes with few obvious elements: $e,g,g^2,...$ , i.e, all integral powers of (g).

So by given condition, this whole set ${(e,g,g^2,...)}$ is same as the set ${(e,g)}$. So any element in (G) must have order ( \le 2).

Now let us look at (e). The identity commutes with every element. But by given condition, (e) commutes with (e) only. That implies there is no other element in (G).

So, (G=(e)).

So, (|S|=1).

HELPDESK

TIFR 2015 Problem 3 Solution - Trace of Product of Matrix

TIFR 2015 Problem 3 Solution is a part of TIFR entrance preparation series. The Tata Institute of Fundamental Research is India's premier institution for advanced research in Mathematics. The Institute runs a graduate programme leading to the award of Ph.D., Integrated M.Sc.-Ph.D. as well as M.Sc. degree in certain subjects.
The image is a front cover of a book named Linear Algebra Done Right by Sheldon Axler. This book is very useful for the preparation of TIFR Entrance.

Also Visit: College Mathematics Program of Cheenta

Problem Type:

Let (A) be a (10\times 10) matrix with complex entries such that all its eigenvalues are non-negative real numbers, and at least one eigenvalue is positive. Which of the following statements is always false?

A. There exists a matrix B such that (AB-BA=B)

B. There exists a matrix B such that (AB-BA=A)

C. There exists a matrix B such that (AB+BA=A)

D. There exists a matrix B such that (AB+BA=B)

Discussion:

We know that for two square matrix (A) and (B) of same size, (Tr(AB)=Tr(BA)) ((TrM)=Trace of (M) ).

In other words, (Tr(AB-BA)=0) for any two square matrices of the same size.

Since trace of a square matrix is also the sum of its eigenvalues, and (A) has all eigenvalues non-negative with at least one positive eigenvalue, we have (Tr(A) > 0). Taking trace of both sides of (AB-BA=A) we get a contradiction. So there does not exist any (B) such that (AB-BA=A).

 

Take (B=0) the 10x10 zero-matrix. Then (AB-BA=B) is satisfied. So is (AB+BA=B).

Take (B=\frac{1}{2}I), where (I) is the 10x10 identity matrix. Then (AB+BA=A) is satisfied.

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TIFR 2015 Problem 2 Solution -Image of continuous function


TIFR 2015 Problem 2 Solution is a part of TIFR entrance preparation series. The Tata Institute of Fundamental Research is India's premier institution for advanced research in Mathematics. The Institute runs a graduate programme leading to the award of Ph.D., Integrated M.Sc.-Ph.D. as well as M.Sc. degree in certain subjects.
The image is a front cover of a book named Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert. This book is very useful for the preparation of TIFR Entrance.

Also Visit: College Mathematics Program of Cheenta

Problem:

Let (f: \mathbb{R} \to \mathbb{R} ) be a continuous function. Which of the following can not be the image of ((0,1]) under (f)?

A. {0}

B. ((0,1))

C. ([0,1))

D. ([0,1])

Discussion:

If f is the constant function constantly mapping to 0, which is continuous, then the image set is {0}.

Suppose that (f((0,1])=(0,1)) . Then (f((0,1))=(0,1)- {f(1)} ). Now since (f(1)\in (0,1) ) the set ( (0,1)- {f(1)} ) is not connected. But ((0,1)) is connected, and we know that continuous image of a connected set is connected. This gives a contradiction. So ((0,1)) can not be the image of ((0,1]) under f.

Define (f(x)=1-x).  Then (f((0,1])= [0,1)).

Define (f(x)=0) for (x\in [0,\frac{1}{2}] ) and (f(x)= 2(x-\frac{1}{2}) ) for (x\in [\frac{1}{2} ,1 ] ). (f) is continuous on  ((0,\frac{1}{2}] ) and ( [\frac{1}{2} ,1 ] ) and (f) agrees on the common points, by pasting lemma (f) is continuous on ( [0,1] ) . And image of ((0,1] ) is ([0,1]).

TIFR 2015 Problem 2 Solution is concluded.

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TIFR 2015 Problem 1 Solution -Invertible Matrix with Sum of Each Row 1


TIFR 2015 Problem 1 Solution is a part of TIFR entrance preparation series. The Tata Institute of Fundamental Research is India's premier institution for advanced research in Mathematics. The Institute runs a graduate programe leading to the award of Ph.D., Integrated M.Sc.-Ph.D. as well as M.Sc. degree in certain subjects.
The image is a front cover of a book named Introduction to Linear Algebra by Gilbert Strang. This book is very useful for the preparation of TIFR Entrance.

Also Visit: College Mathematics Program of Cheenta

Problem:

Let (A) be an invertible (10 \times 10) matrix with real entries such that the sum of each row is 1. Then

A. The sum of the entries of each row of the inverse of A is 1

B. The sum of the entries of each column of the inverse of A is 1

C. The trace of the inverse of A is non-zero.

D. None of the above.

Discussion:

The sum of each row of (A) is 1, means that the sum of the columns of A is the vector ((1,1,...,1)^T ) .

Note that i-th column of (A) is given by (Ae_i ). Therefore, (\sum_{i=1}^{10} Ae_i = (1,1,...,1)^T ).

Since left multiplication by (A) is a linear transformation, the left-hand side of the last expression can be written as (A(\sum_{i=1}^{10}e_i)).

Now, (\sum_{i=1}^{10}e_i = (1,1,...,1)^T ).

Hence we get (A(1,1...,1)^T = (1,1,...,1)^T ).

Another way of saying the last expression is that the vector ( (1,1...,1)^T ) is fixed by A.

Since A is invertible, applying (A^{-1}) on both sides of the last expression we get ((1,...,1)^T = A^{-1}(1,1,...,1)^T ).

By the linearity argument as above, this gives ( (1,1...,1)^T = \sum_{i=1}^{10} A^{-1} (e_i) ). And the i-th term in right-hand side expression is the i-th column of (A^{-1}). Therefore, sum of columns of (A^{-1}) is the vector ((1,1,...,1)^T ). This is same as saying that sum of entries of each row of (A^{-1}) is 1.

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