What is TIFR and how to prepare for it?

About TIFR

Tata Institute of Fundamental Research, TIFR is the foremost institution for advanced research in foundational sciences based in Mumbai, Maharashtra, India. The institute offers a master's course, an integrated M.Sc and Ph.D. course and a Ph.D. degree in different science fields. One can get admission into this institute by clearing the TIFR Entrance Exam. It is a perfect place for pioneering scientists to commence their careers.

TIFR Entrance for Integrated M.Sc-Ph.D. Programme in Mathematics

Eligibility

The eligibility criterion to be considered for admission to the Integrated Ph.D. programme is a Bachelor's degree in any of Mathematics, Statistics, Science or Technology (B.A. / B.Sc. / B.Math. / B.Stat. / B.E. / B.Tech.) from any recognized university.

Duration of the course

The duration of integrated courses is 6 years.

Application Procedure

Candidates can fill the application form online through the university website. After filling the form, candidates can pay the application fee online through internet banking or Debit/Credit Card.

Generally, the admission process takes place in the month of December.

Candidates can secure the admission in the following colleges:

School of Mathematics, TIFR Mumbai
Centre for Applicable Mathematics (CAM), BengaluruÂ

Selection Procedure

The selection of the candidates will be done on the basis of the Graduate School (GS) exam. A list of selected candidates will be published on the official website, on the basis of which, interview round will take place. This interview will be conducted by the university.

TIFR GS Entrance Exam Pattern

Part	Type of Questions	Number of Questions	Duration
I	Objective (True/ False)	30	1.5 hrs
II	Subjective	10	1.5 hrs
	Total	40	3 hrs

Marking System

Part I: Two marks will be awarded for each correct answer and one mark is deducted for each incorrect answer.

Part II: Full marks will be given to correct answers and partial marks will be given to partially correct answers in the exam.

Syllabus

The syllabus includes:

Algebra
Analysis
Geometry/ Topology
General

Click here to get the full syllabus:- http://univ.tifr.res.in/gs2020/Files/Maths_Syllabus.pdf

How to prepare for TIFR Entrance

Books for TIFR Entrance

Books for Linear Algebra

Chapter 0 of Serge Lang’s Introduction Linear Algebra
Introduction to Linear Algebra by Gilbert Strang
Linear Algebra: A geometric Approach by S Kumaresan
Linear Algebra by Hoffman & Kunze
Algebra by Artin
Topics in Algebra by Herstein
Linear Algebra Done Right by Axler

Books for Abstract Algebra

Contemporary Abstract Algebra by Gallian
SAGE Math (computation software)
Abstract Algebra by Dummit and Foote

Books for Real Analysis

Introduction to Real Analysis by Bartle Sherbet
Problems in Real Analysis by Kaczor
A basic course in Real Analysis by Kumar & Kumaresan
Analysis 1 & 2 by Terence Tao
Principles of Mathematical Analysis by Rudin

Books for Topology

Topology of Metric Spaces by Kumaresan
Topology by Munkres

Books for Vector Calculus and Differential Equation

Calculus, Early Transcendentals by James Stewart
Basic Multivariable Calculus by Marden
Calculus Vol 1 and 2 by Apostle

For more information, visit the website:- http://univ.tifr.res.in/gs2020/

Preparation Tips

Go through the syllabus:-

Learning about the syllabus is an essential part of preparation. This helps to have a better understanding of the preparation strategy and what topics to focus on. It saves you time as well.

Work on your concept and solve different questions:-

Try to clear your concepts on each topic and practice questions based on that topic from different books. This way, you will be able to understand your understanding of the topic.

Solve Previous Papers:-

This will help you know the exact difficulty level of the exam and prepare accordingly. Try to solve them in a time-bound manner.

Take Mock Tests:-

Take Mock Tests to check your ability of problem-solving. If it is an online test, then it can help you know your preparation status in the competition. You will be able to know where you stand and what should be improved.

Track your progress:-

You should also track your progress. If something unimportant is stealing your time, recognise it and try to avoid it. Keep yourself surrounded by people who encourage you. Discuss your dreams and goals with them.

Take proper rest and believe in yourself:-

It is the biggest myth that studying for long hours brings productivity at your work. There should be a routine with a proper work-rest balance and then you should ritually follow it. Also having low self-esteem can ruin your performance in exams.

All the best!

Order of rings: TIFR GS 2018 Part B Problem 12

[et_pb_section fb_built="1" _builder_version="3.22.4" fb_built="1" _i="0" _address="0"][et_pb_row _builder_version="3.25" _i="0" _address="0.0"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||" _i="0" _address="0.0.0"][et_pb_text _builder_version="3.22.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_padding="20px|20px|20px|20px" _i="0" _address="0.0.0.0"]

Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.27" text_font="Raleway||||||||" background_color="#f4f4f4" box_shadow_style="preset2" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" _i="1" _address="0.0.0.1"]The number of rings of order 4, up to isomorphism, is:
(a) 1
(b) 2
(c) 3
(d) 4.[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25" _i="1" _address="0.1"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||" _i="0" _address="0.1.0"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.27" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px" _i="0" _address="0.1.0.0"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="3.27" hover_enabled="0" _i="0" _address="0.1.0.0.0"]TIFR GS 2018 Part B Problem 12[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="3.27" hover_enabled="0" _i="1" _address="0.1.0.0.1" open="off"]Abstract Algebra[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.27" hover_enabled="0" _i="2" _address="0.1.0.0.2" open="off"]Easy[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="3.27" hover_enabled="0" _i="3" _address="0.1.0.0.3" open="off"]Dummit and Foote[/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.23.3" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" _i="1" _address="0.1.0.1"]

Start with hints

[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="3.27" tab_text_color="#ffffff" tab_font="||||||||" background_color="#ffffff" _i="2" _address="0.1.0.2"][et_pb_tab title="Hint 0" _builder_version="3.22.4" _i="0" _address="0.1.0.2.0"]Do you really need a hint? Try it first!

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.27" hover_enabled="0" _i="1" _address="0.1.0.2.1"]

First, ask yourself how many groups are there of order 4. 
the answer is simple => Z/4Z and Klein’s four group (K).

[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.27" hover_enabled="0" _i="2" _address="0.1.0.2.2"]

So intuitively there should be two rings with 4 elements.

Okay, I will give you two rings or same order=> (R,+,.), (R,+,*) . see that order of the rings are same but I have changed the multiplication, and

I define a*b=0 for all a,b in R. [(R,+,*) is called zero ring]

[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.27" hover_enabled="0" _i="3" _address="0.1.0.2.3"]

Question: Prove that (R,+,.) and (R,+,*) are not isomorphic! (easy)

So, we had (Z/4Z,+,.) and (K,+,.) as our answers. But if you change the multiplication 
to “*” then there will be 4 different rings 
*upto isomorphism* right?

Hence the answer is 4.

[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.27" hover_enabled="0" _i="4" _address="0.1.0.2.4"]

Bonus Problem:
Prove that there are only two non-ismorphic p-rings(ring with p elements) upto isomorphism.

[/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.23.3" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" _i="3" _address="0.1.0.3"]

Watch the video

[/et_pb_text][et_pb_code _builder_version="3.26.4" _i="4" _address="0.1.0.4"]

[/et_pb_code][et_pb_text _builder_version="3.23.3" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" min_height="12px" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" _i="5" _address="0.1.0.5"]

Connected Program at Cheenta

The higher mathematics program caters to advanced college and university students. It is useful for I.S.I. M.Math Entrance, GRE Math Subject Test, TIFR Ph.D. Entrance, I.I.T. JAM. The program is problem driven. We work with candidates who have a deep love for mathematics. This program is also useful for adults continuing who wish to rediscover the world of mathematics.[/et_pb_blurb][et_pb_button button_url="https://www.cheenta.com/collegeprogram/" button_text="Learn More" button_alignment="center" _builder_version="3.23.3" custom_button="on" button_bg_color="#0c71c3" button_border_color="#0c71c3" button_border_radius="0px" button_font="Raleway||||||||" button_icon="%%3%%" button_text_shadow_style="preset1" box_shadow_style="preset1" box_shadow_color="#0c71c3" background_layout="dark" _i="7" _address="0.1.0.7"][/et_pb_button][et_pb_text _builder_version="3.22.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" _i="8" _address="0.1.0.8"]

Last three digit of the last year: TIFR GS 2018 Part B Problem 9

Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.27" text_font="Raleway||||||||" background_color="#f4f4f4" box_shadow_style="preset2" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" _i="1" _address="0.0.0.1"]What are the last 3 digits of \(2^{2017}\)?
(a) 072
(b) 472
(c) 512
(d) 912.[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25" _i="1" _address="0.1"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||" _i="0" _address="0.1.0"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.27" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px" _i="0" _address="0.1.0.0"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="3.27" hover_enabled="0" _i="0" _address="0.1.0.0.0"]TIFR GS 2018 Part B Problem 9[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="3.27" hover_enabled="0" _i="1" _address="0.1.0.0.1" open="off"]Number theory[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.27" hover_enabled="0" _i="2" _address="0.1.0.0.2" open="off"]Easy[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="3.27" hover_enabled="0" _i="3" _address="0.1.0.0.3" open="off"]Burton[/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.23.3" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" _i="1" _address="0.1.0.1"]

Start with hints

[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="3.27" tab_text_color="#ffffff" tab_font="||||||||" background_color="#ffffff" hover_enabled="0" _i="2" _address="0.1.0.2"][et_pb_tab title="Hint 0" _builder_version="3.22.4" _i="0" _address="0.1.0.2.0"]Do you really need a hint? Try it first!

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.27" hover_enabled="0" _i="1" _address="0.1.0.2.1"]Pretty convenient problem for number theory lovers. I’m going to give some insight from Group Theory and solve with basic congruence techniques.

Now, an obvious fact is \(2^{2017}\) ≅ \(0(mod8)\)
so my idea is to find modul0 1000. so where does group theory lend you a hand?
see that 2 is a generator of (Z/125Z)* (why?)

[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.27" hover_enabled="0" _i="2" _address="0.1.0.2.2"]

So we get \(2^{2017}\) ≅ \(2^{2017}(mod125)\) ≅ \(2^{17}\) ≅ \(72 (mod 125)\)

4. Now, what is the most demanding step after this?

combining 1 and 3 we get \(2^{2017}\) ≅ \(072(mod 1000)\) [as \(125|2^{2017}-72, 8|2^{2017}-72 \)=> third digit is 0]

Hence the answer is 072.

[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.27" hover_enabled="0" _i="3" _address="0.1.0.2.3"]

Bonus Problems:

Q. find last two digits of 2^2016 like this process.

SOME NUMBER THEORIC PROBLEMS:

Q. Prove Wilson's theorem using basic group theory  

Q. Prove Wilson's theorem by using Sylows theorems.

Watch the video

[/et_pb_text][et_pb_code _builder_version="3.26.4" _i="4" _address="0.1.0.4"]

Connected Program at Cheenta

Graphs in groups or Groups in graphs : TIFR GS 2018

[et_pb_section fb_built="1" _builder_version="3.22.4"][et_pb_row _builder_version="3.25"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]

Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.27.4" text_font="Raleway||||||||" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"]Let G be a finite group and g ∈ G an element of even order. Then we
can colour the elements of G with two colours in such a way that x and
gx have different colours for each x ∈ G.[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.27" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="3.27"]TIFR GS 2018 Part A Problem 24[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="3.27" open="off"]Abstract Algebra[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.27" open="off"]Hard[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="3.27" open="off"]Abstract Algebra, Dummit and Foote[/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]

Start with hints

[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="3.27" tab_text_color="#ffffff" tab_font="||||||||" background_color="#ffffff"][et_pb_tab title="Hint 0" _builder_version="3.22.4"]Do you really need a hint? Try it first!

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.27"]

One needs to know the basics of Graph Theory to understand the solution.

As noted Colouring is a fundamental topic in Graph Theory,so we need to convert the problem into a graph theory problem.
Consider the elements of the group G as the vertices and consider edges between two elements say x and y if x=gy or \(x=(g^{-1})x\). Call graph G*.
Check that x---y---z (x is adjacent to y and y is adjacent to z) iff \(y = (g^{-1})x , z = (g^{-1})y = (g^{-2})x\) or \(y = x , z = (g)y = (g^2)x\).We will assume left multiplication by g ( the proof for \(g^{-1}\) is exactly the same.)

[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.27"]

Now observe that we need to answer whether this graph G* is 2-colourable.
There is an elementary result in graph theory characterizing the 2-colourable graphs.

Theorem 1 : A graph is 2-colourable iff it is bipartite.

Theorem 2: A graph is bipartite iff it has no odd-cycle.

[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.27"]

Thus Theorem 1 and Theorem 2 ⇒ G* is 2 -colourable iff it has no odd cycle.
Now what does odd cycle mean in here in terms of group.
A path of odd length means that \(x---gx---(g^2)x---...---(g^k)x\) in odd number of steps i.e. k is odd.

[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.27"]

A cycle of odd length means that \((g^k)x=x ⇒ g^k=1\).
We are given that g is even ordered so it can only happen if k is even.
Hence an odd cycle cannot exist and we can colour G* with 2 colours.

The answer is therefore True.

[/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]

Watch the video

[/et_pb_text][et_pb_code _builder_version="3.26.4"]

[/et_pb_code][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" min_height="12px" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]

Connected Program at Cheenta

[/et_pb_text][et_pb_blurb title="College Mathematics Program" url="https://www.cheenta.com/collegeprogram/" image="https://www.cheenta.com/wp-content/uploads/2018/03/College-1.png" _builder_version="3.23.3" header_font="||||||||" header_text_color="#e02b20" header_font_size="48px" link_option_url="https://www.cheenta.com/collegeprogram/" border_color_all="#e02b20"]The higher mathematics program caters to advanced college and university students. It is useful for I.S.I. M.Math Entrance, GRE Math Subject Test, TIFR Ph.D. Entrance, I.I.T. JAM. The program is problem driven. We work with candidates who have a deep love for mathematics. This program is also useful for adults continuing who wish to rediscover the world of mathematics.[/et_pb_blurb][et_pb_button button_url="https://www.cheenta.com/collegeprogram/" button_text="Learn More" button_alignment="center" _builder_version="3.23.3" custom_button="on" button_bg_color="#0c71c3" button_border_color="#0c71c3" button_border_radius="0px" button_font="Raleway||||||||" button_icon="%%3%%" background_layout="dark" button_text_shadow_style="preset1" box_shadow_style="preset1" box_shadow_color="#0c71c3"][/et_pb_button][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]

Are juniors countable if seniors are?: TIFR GS 2018 Part A Problem 21

Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.27" text_font="Raleway||||||||" background_color="#f4f4f4" box_shadow_style="preset2" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" _i="1" _address="0.0.0.1"]A countable group can have only countably many distinct subgroups.[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25" _i="1" _address="0.1"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||" _i="0" _address="0.1.0"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.27" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px" _i="0" _address="0.1.0.0"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="3.27" hover_enabled="0" _i="0" _address="0.1.0.0.0"]TIFR GS 2018 Part A Problem 21[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="3.27" hover_enabled="0" _i="1" _address="0.1.0.0.1" open="off"]Group Theory[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.27" hover_enabled="0" _i="2" _address="0.1.0.0.2" open="off"]Easy[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="3.27" hover_enabled="0" _i="3" _address="0.1.0.0.3" open="off"]Abstract Algebra, Dummit and Foote[/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.23.3" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" _i="1" _address="0.1.0.1"]

Start with hints

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.27" hover_enabled="0" _i="1" _address="0.1.0.2.1"]

This is a really artistic problem.

Pre-Solution Thoughts:

Lemma: A group is finite iff the number of subgroups of a group is finite.(Check!)

The lemma is a way to understand that an infinite group will have infinite number of subgroups.

[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.27" hover_enabled="0" _i="2" _address="0.1.0.2.2"]

We are considering here set of all subsets of a countable group G, which is a subset of the Power Set of G.
Given G is countable the Power Set of G is uncountable.
Now we know that \(2^d\), where d is the set of natural numbers denotes the cardinality of the Power Set of G which is an uncountable set as it is bijective to the Real Numbers [Here d is not finite by the way, d is countably infinite].
So it is kinda intuitive that it may be uncountable.
First I took the group (Z,+), but we all know that the subgroups of Z are nZ only. This doesn’t solve our purpose.
So naturally the next choice was the group (Q.+) whose subgroup is (Z,+).
While understanding the subgroups of (Q.+), the question is solved.

[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.27" hover_enabled="0" _i="3" _address="0.1.0.2.3"]

We need to understand the subgroups of (Q.+).
Consider any rational number q and the subgroup qZ of (Q.+) generated by q.
What if we take two rational numbers?
For simplicity check the subgroup generated by {1/2 , 1/3}.
Prove that the subgroup generated by {1/2 , 1/3} is (1/2.3)Z.{Observe that co-prime property of 2 and 3 is playing an important role}.

[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.27" hover_enabled="0" _i="4" _address="0.1.0.2.4"]

Now what if we take three mutually coprime natural numbers say a,b,c and see the subgroup generated by { 1/a , 1/b ,1/c}. In fact for simplicity take a,b,c to be primes.
Observe that it is of the form (1/a.b.c)Z.
Hence for every finite subset of primes, we generate a distinct subgroup of (Q,+).
Hence the total number of subgroups contained all the subsets of primes, which has a bijection with the Real Numbers as mentioned above.
So the answer is False.

Exercise

Find all the subgroups of (Q,+) with proof.

Watch the video

[/et_pb_text][et_pb_code _builder_version="3.26.4" _i="4" _address="0.1.0.4"]

Connected Program at Cheenta

Group with Quotient : TIFR GS 2018 Part A Problem 16

Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.27" text_font="Raleway||||||||" background_color="#f4f4f4" box_shadow_style="preset2" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" hover_enabled="0" _i="1" _address="0.0.0.1"]Let G be a finite group with a normal subgroup H such that G/H has
order 7. Then \(G \cong\) H × G/H. [/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25" _i="1" _address="0.1"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||" _i="0" _address="0.1.0"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.27" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px" hover_enabled="0" _i="0" _address="0.1.0.0"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="3.27" _i="0" _address="0.1.0.0.0" hover_enabled="0"]TIFR GS 2018 Part A Problem 23 [/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="3.27" _i="1" _address="0.1.0.0.1" open="off" hover_enabled="0"]Group Theory [/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.27" _i="2" _address="0.1.0.0.2" open="off" hover_enabled="0"]Medium [/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="3.27" _i="3" _address="0.1.0.0.3" open="off" hover_enabled="0"]Dummit and Foote [/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.23.3" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" _i="1" _address="0.1.0.1"]

Start with hints

[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="3.27" tab_text_color="#ffffff" tab_font="||||||||" background_color="#ffffff" hover_enabled="0" _i="2" _address="0.1.0.2"][et_pb_tab title="Hint 0" _builder_version="3.22.4" _i="0" _address="0.1.0.2.0"]Do you really need a hint? Try it first!

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.27" _i="1" _address="0.1.0.2.1" hover_enabled="0"]

This is also an interesting question.First of all we need to understand something in general.

If G is a finite group and H Δ G. So Consider the quotient group G/H.

Observe the following!

Lemma: If G ≈ H x G/H , then G/H is isomorphic to a normal subgroup of G. [Consider the projection homomorphism of G to (H,1) which contains G/H as the kernel.]
But in general G/H is not even a subgroup of G.

We will illustrate this by giving a simple example.

[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.27" _i="2" _address="0.1.0.2.2" hover_enabled="0"]

Naturally we took the group (Z,+) and we know all the subgroups of Z are nZ ,which are normal subgroups as Z is an abelian group.
Consider the quotient group Z/nZ.We know that this is not even isomorphic to a subgroup of Z.
Hence comes our counter-example.
G=Z, H=7Z. G/H =Z/7Z. But G is not isomorphic to 7Z x Z/7Z as Z/7Z is not isomorphic to a subgroup of Z.
Hence the answer is False.

[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.27" _i="3" _address="0.1.0.2.3" hover_enabled="0"]

But we will give an example where the given statement is also False.

Consider the Dihedral group on n elements\(D_n\) as a subgroup of O(2) [The orthogonal group in R^2.] There is a homomorphism (Determinant) from O(2) → {-1,1},whose kernel is SO(2).
Hence consider the homomorphism from \(D_n\) → {-1,1} formed by the composition of inclusion homomorphism and the determinant homomorphism.
Observe that the Kernel of the above defined homomorphism is the Rotation Group of angle 2 π /n and the Quotient Group is the Reflection Group around a specific line(?)[Which is essentially Z/2Z.]
But observe that Dn is not isomorphic to Rotation Group of angle 2 π /n x Z/2Z .[As there is an interaction between rotation and reflection. \(ref.rot.ref=rot^{-1}\) .]

[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.27" _i="4" _address="0.1.0.2.4" hover_enabled="0"]

#Bonus

Prove that the finite subgroups of the group of rigid body motion are only

Rotation Group of Angle 2 π /n for all n in N.
Dihedral Group \(D_n\)

Watch the video

[/et_pb_text][et_pb_code _builder_version="3.26.4" _i="4" _address="0.1.0.4"]

Connected Program at Cheenta

Symmetric groups of order 30: TIFR GS 2018 Part A Problem 23

Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.27" text_font="Raleway||||||||" background_color="#f4f4f4" box_shadow_style="preset2" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" _i="1" _address="0.0.0.1"]The permutation group \(S_{10}\) has an element of order 30.[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25" _i="1" _address="0.1"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||" _i="0" _address="0.1.0"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.27" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px" _i="0" _address="0.1.0.0"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="3.27" hover_enabled="0" _i="0" _address="0.1.0.0.0"]TIFR 2018 Part A Problem 23[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="3.27" hover_enabled="0" _i="1" _address="0.1.0.0.1" open="off"]Group Theory[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.27" hover_enabled="0" _i="2" _address="0.1.0.0.2" open="off"]Easy[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="3.27" hover_enabled="0" _i="3" _address="0.1.0.0.3" open="off"]Dummit and Foote[/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.23.3" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" _i="1" _address="0.1.0.1"]

Start with hints

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.27" hover_enabled="0" _i="1" _address="0.1.0.2.1"]

Consider S={1,2,...,10}.\(S_{10}\) be the permutation group on S.

What will you do if one asked for a subgroup of order 3!=6?

We all would have taken the subgroup of \(S_3\) embedded in \(S_{10}\) right? Call the subgroup H taking identity transformation on {4,5,6,..10} and embedding in {1,2,3}.

[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.27" hover_enabled="0" _i="2" _address="0.1.0.2.2"]

What do you do if one asked for a subgroup of order 5?

We will try to form a cyclic subgroup of order 5. We need to find a generator. Can you see it?
Keeping {1,2,.,5} same and taking 5 elements {6,7,..,10} and observe the permutation \(i\mapsto i+1\) for i=6,7,8,9 and \(10\mapsto 7\).Take the subgroup generated by this element.Observe this is a cyclic subgroup of order 5.Call this subgroup K.
Now any idea how to combine this?

[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.27" hover_enabled="0" _i="3" _address="0.1.0.2.3"]

Observe that HK is a set of 30 elements. Does it seem HK is a subgroup?
Lemma: H and K are two subgroups of G. HK is a subgroup of G iff KH=HK.
Using the lemma prove that HK is really a subgroup of \(S_{10}\) of order 30.
Observe that the selection of disjoint elements of H and K is the main reason behind this!
Hence the answer is True.

Watch the video

[/et_pb_text][et_pb_code _builder_version="3.26.4" _i="4" _address="0.1.0.4"]

Connected Program at Cheenta

Matrix additive-multiplicative :TIFR 2018 Part A Problem 19

Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.27" text_font="Raleway||||||||" background_color="#f4f4f4" box_shadow_style="preset2" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" hover_enabled="0" _i="1" _address="0.0.0.1"]Let A, B ∈ \(M_n(\Bbb R)\) be such that A + B = AB. Then AB = BA. [/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25" _i="1" _address="0.1"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||" _i="0" _address="0.1.0"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.27" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px" _i="0" _address="0.1.0.0"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="3.27" hover_enabled="0" _i="0" _address="0.1.0.0.0"]TIFR 2018 Part A Problem 19[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="3.27" hover_enabled="0" _i="1" _address="0.1.0.0.1" open="off"]Linear Algebra[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.27" hover_enabled="0" _i="2" _address="0.1.0.0.2" open="off"]Medium[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="3.27" hover_enabled="0" _i="3" _address="0.1.0.0.3" open="off"]Linear Algebra, Hoffman and Kunze[/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.23.3" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" _i="1" _address="0.1.0.1"]

Start with hints

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.27" hover_enabled="0" _i="1" _address="0.1.0.2.1"]

We need to play around with symbols.

Try to make (AB-BA) containing equation so that we can get the idea how to show it 0.This is a symmetric operation so remember if you use only B to get something you can also use A to get that so we have to take help of the symmetry.

[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.27" hover_enabled="0" _i="2" _address="0.1.0.2.2"]

Playing with the equation and multiplying the equation once left and once right by B we get A.B+B.B=A.B.B and B.A + B.B = B.A.B
Now subtract it to get AB-BA term,which gives (AB-BA)=(AB-BA)B => (AB-BA)(I-B)=0
Similarly use A in place of B to get (I-A)(AB-BA)=0

[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.27" hover_enabled="0" _i="3" _address="0.1.0.2.3"]

Now if somehow we can show (I-A) or (I-B) is invertible then we are done.
So we go back to the actual equation. Remember the identity (1-a)(1-b)=1-a-b+ab.
Observe that (I-A)(I-B)=I follows from the actual equation.Hence the result follows as (I-A) and (I-B) are invertible and are inverses of each other.
The answer is therefore True.

[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.27" hover_enabled="0" _i="4" _address="0.1.0.2.4"]

U+T are linear operators on R^n and U+T=UT.Then Img(T)=Img(U) and Ker(T)=Ker(U)

Prove that Ker(T) is a subspace of Ker(U).
Take adjoint of the equation and observe that Ker(U*) is a subspace of Ker(T*).
Now we know that Ker(U*) is orthogonal to Img(U) and same goes for T.
Hence conclude by that Img(T) is a subspace of Img(T).
Therefore prove that Img(T)=Img(U) and Ker(T)=Ker(U) by Rank Nullity Theorem.

Watch the video

[/et_pb_text][et_pb_code _builder_version="3.26.4" _i="4" _address="0.1.0.4"]

Connected Program at Cheenta

Commutative does not commute in matrices: TIFR 2018 Part A, Problem 11

Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.27" text_font="Raleway||||||||" background_color="#f4f4f4" box_shadow_style="preset2" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" hover_enabled="0" _i="1" _address="0.0.0.1"]True or False: Let A, B, C ∈ M3(\(\mathbb{R}\)) be such that A commutes with B, B commutes with C and B is not a scalar matrix. Then A commutes with C. [/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25" _i="1" _address="0.1"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||" _i="0" _address="0.1.0"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.27" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px" hover_enabled="0" _i="0" _address="0.1.0.0"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="3.27" _i="0" _address="0.1.0.0.0" hover_enabled="0"]TATA INSTITUTE OF FUNDAMENTAL RESEARCH GS-2018 (Mathematics) [/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="3.27" _i="1" _address="0.1.0.0.1" open="off" hover_enabled="0"]Linear Algebra [/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.27" _i="2" _address="0.1.0.0.2" open="off" hover_enabled="0"]Medium [/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="3.27" _i="3" _address="0.1.0.0.3" open="off" hover_enabled="0"]Linear Algebra Hoffman and Kunze [/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.23.3" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" _i="1" _address="0.1.0.1"]

Start with hints

[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="3.27" tab_text_color="#ffffff" tab_font="||||||||" background_color="#ffffff" hover_enabled="0" _i="2" _address="0.1.0.2"][et_pb_tab title="Hint 0" _builder_version="3.22.4" _i="0" _address="0.1.0.2.0"]Do you really need a hint? Try it first!

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.27" _i="1" _address="0.1.0.2.1" hover_enabled="0"]First we need to get some idea whether or not this may be true or false . As a result we need to make some calculations. The first step to approach is always to build the expression of AC - CA & then see whether it is zero or not [/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.27" _i="2" _address="0.1.0.2.2" hover_enabled="0"]Given AB = BA & BC = CB . Prove the following ! B(AC) = AC(B) & B(CA) = (CA)B and then subtract to get ( AC - CA )B = B(AC -CA) Now this is not obvious if DB = BD & B being non- scalar matrix then D = 0 is too strong statement to be true So this gives us idea that it maybe false . Now to prove it false we need to construct a counter example [/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.27" _i="3" _address="0.1.0.2.3" hover_enabled="0"]We can take approach using beautiful fact of matrices that they are transformation of spaces . Now given they are transformation of spaces and this is sort of abelian character showing up , we seek help from Groups [/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.27" _i="4" _address="0.1.0.2.4" hover_enabled="0"]You know why B is restricted to be non scalar because they form the centre of the \(GL_n\)(F) So we approach it in the following way If we can find a group of 3 x 3 matrices with non trivial centre . If we search for centre of groups then the only example available to us is Heisenberg Group In mathematics, the Heisenberg group

H

, named after Werner Heisenberg, is the group of 3×3 upper triangular matrices of the form

\begin{pmatrix} 1 & a & c\\ 0 & 1 & b\\ 0 & 0 & 1\\ \end{pmatrix}

under the operation of matrix multiplication. Elements a, b and c can be taken from any commutative ring with identity, often taken to be the ring of real numbers(resulting in the "continuous Heisenberg group") or the ring of integers (resulting in the "discrete Heisenberg group"). The continuous Heisenberg group arises in the description of one-dimensional quantum mechanical systems, especially in the context of the Stone–von Neumann theorem. More generally, one can consider Heisenberg groups associated to n-dimensional systems, and most generally, to any symplectic vector space. [/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.23.3" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" _i="3" _address="0.1.0.3"]

Watch the video

[/et_pb_text][et_pb_code _builder_version="3.26.4" _i="4" _address="0.1.0.4"]

Connected Program at Cheenta

Spanning matrix space by niltopent matrices: TIFR 2018 Part A, Problem 15

Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.27" text_font="Raleway||||||||" background_color="#f4f4f4" box_shadow_style="preset2" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" hover_enabled="0" _i="1" _address="0.0.0.1"]The set of nilpotent matrices in \(M_3(R)\) spans \(M_3(R)\) considered as an R-vector space (a matrix A is said to be nilpotent if there exists n ∈ N such that \(A^n = 0)\). [/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25" _i="1" _address="0.1"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||" _i="0" _address="0.1.0"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.27" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px" hover_enabled="0" _i="0" _address="0.1.0.0"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="3.27" _i="0" _address="0.1.0.0.0" hover_enabled="0"]TIFR 2018 Part A, Problem 15 [/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="3.27" _i="1" _address="0.1.0.0.1" open="off" hover_enabled="0"]LINEAR ALGEBRA [/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.27" _i="2" _address="0.1.0.0.2" open="off" hover_enabled="0"]Medium [/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="3.27" _i="3" _address="0.1.0.0.3" open="off" hover_enabled="0"]Linear Algebra, Hoffman and Kunze [/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.23.3" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" _i="1" _address="0.1.0.1"]

Start with hints

[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="3.27" tab_text_color="#ffffff" tab_font="||||||||" background_color="#ffffff" hover_enabled="0" _i="2" _address="0.1.0.2"][et_pb_tab title="Hint 0" _builder_version="3.22.4" _i="0" _address="0.1.0.2.0"]Do you really need a hint? Try it first!

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.27" _i="1" _address="0.1.0.2.1" hover_enabled="0"]

Let’s first try an easier problem first :

Whether there exists a and b such that I = a N + b M,where N and M are nilpotent matrices and a,b are real numbers ?
To use the given definition to prove or disprove this we need to take powers and things will become too messy for larger n though here it is 3.[We have to take only \(A^3=0\) it is a small exercise to verify]

[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.27" _i="2" _address="0.1.0.2.2" hover_enabled="0"]

So let us recapitulate some properties of nilpotent matrices: They are the matrices which has only eigenvalue 0. How to use this?
We wish to have some linear operator acting on the right side for the RHS is a linear combination and the natural choice is Trace.
Trace of a nilpotent matrix is 0.

[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.27" _i="3" _address="0.1.0.2.3" hover_enabled="0"]

Now assume that I is a finite linear combination of Nilpotent matrices and then take Trace Operator on both sides.
Thus LHS has trace n and RHS has trace 0.(Contradiction)
So the answer is False.

[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.27" _i="4" _address="0.1.0.2.4" hover_enabled="0"]

Food for Thought:

Observe that the order of the matrices is not at all a big issue here!
Does span of the nilpotent matrices contain all the matrices with trace 0?
Hint:Try to find all the 2x2 nilpotent matrices and check the above statement out!

Watch the video

[/et_pb_text][et_pb_code _builder_version="3.26.4" _i="4" _address="0.1.0.4"]

Connected Program at Cheenta

About TIFR

TIFR Entrance for Integrated M.Sc-Ph.D. Programme in Mathematics

Eligibility

Duration of the course

Application Procedure

Selection Procedure

TIFR GS Entrance Exam Pattern

Marking System

Syllabus

How to prepare for TIFR Entrance

Books for TIFR Entrance

Books for Linear Algebra

Books for Abstract Algebra

Books for Real Analysis

Books for Topology

Books for Vector Calculus and Differential Equation

Preparation Tips

Understand the problem

Start with hints

Watch the video

Connected Program at Cheenta

Similar Problems

Understand the problem

Start with hints

Watch the video

Connected Program at Cheenta

Similar Problems

Understand the problem

Start with hints

Watch the video

Connected Program at Cheenta

Similar Problems

Understand the problem

Start with hints

Watch the video

Connected Program at Cheenta

Similar Problems

Understand the problem

Start with hints

Watch the video

Connected Program at Cheenta

Similar Problems

Understand the problem

Start with hints

Watch the video

Connected Program at Cheenta

Similar Problems

Understand the problem

Start with hints

Watch the video

Connected Program at Cheenta

Similar Problems

Understand the problem

Start with hints

Watch the video

Connected Program at Cheenta

Similar Problems

Understand the problem

Start with hints

Watch the video

Connected Program at Cheenta

Similar Problems