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College Mathematics

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
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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