Understand the problem

Let A, B ∈ \(M_n(\Bbb R)\) be such that A + B = AB. Then AB = BA.
Source of the problem
TIFR 2018 Part A Problem 19
Topic
Linear Algebra
Difficulty Level
Medium
Suggested Book
Linear Algebra, Hoffman and Kunze

Start with hints

Do you really need a hint? Try it first!

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.
  • 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
  • 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.
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

Connected Program at Cheenta

College Mathematics Program

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.

Similar Problems

Application of eigenvalue in degree 3 polynomial: ISI MMA 2018 Question 14

This is a cute and interesting problem based on application of eigen values in 3 degree polynomial .Here we are finding the determinant value .

Order of rings: TIFR GS 2018 Part B Problem 12

This problem is a cute and simple application on the ring theory in the abstract algebra section. It appeared in TIFR GS 2018.

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

This problem is a cute and simple application on the number theory in classical algebra portion. It appeared in TIFR GS 2018.

Group in graphs or graphs in groups ;): TIFR GS 2018 Part A Problem 24

This problem is a cute and simple application on the graphs in groups in the abstract algebra section. It appeared in TIFR GS 2018.

Problems on quadratic roots: ISI MMA 2018 Question 9

This problem is a cute and simple application on the problems on quadratic roots in classical algebra,. It appeared in TIFR GS 2018.

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

This problem is a cute and simple application on the order of a countable groups in the abstract algebra section. It appeared in TIFR GS 2018.

Coloring problems: ISI MMA 2018 Question 10

This problem is a cute and simple application of the rule of product or multiplication principle in combinatorics,. It appeared in TIFR GS 2018.

Diagonilazibility in triangular matrix: TIFR GS 2018 Part A Problem 20

This problem is a cute and simple application on the diagonilazibility in triangular matrix in the abstract algebra section. It appeared in TIFR GS 2018.

Multiplicative group from fields: TIFR GS 2018 Part A Problem 17

This problem is a cute and simple application on the Multiplicative group from fields in the abstract algebra section. It appeared in TIFR GS 2018.

Group with Quotient : TIFR GS 2018 Part A Problem 16

This problem is a cute and simple application on Group theory in the abstract algebra section. It appeared in TIFR GS 2018.