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# Matrix additive-multiplicative :TIFR 2018 Part A Problem 19

This problem is a cute and simple application of additive and multiplicative properties of matrices in the linear algebra section. It appeared in TIFR GS 2018.

# 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
Linear Algebra
Medium
##### Suggested Book
Linear Algebra, Hoffman and Kunze

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.

# 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.

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