# 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.

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