Understand the problem
Source of the problem
Start with hints
- 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.
- 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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