C (AB-BA) ^2= (AB-BA) ^2*C

[TheWrathOfMath]

A, B and C are three matrices of order 2x2. Prove that C (AB-BA) ^2= (AB-BA) ^2*C.

I let A=(a b and B=(q r)
c d s t

I found that the sum of the diagonals' entries of AB-BA is zero. In other words, tr(AB-BA)=0.

AB-BA=(br-cq________aq+bs-bp-dq)
_________(cp+dr-ar-cs__cq-br)

I do not know how to proceed.
I do not think that I will need to raise AB-BA to the power of 2.
There must be a quicker and cleaner solution.
Or should I just raise AB-BA to the power of two, let C=(e f) and substitute it in the equation?
________________ ___________________________________________g h

 

[blamocur]

Hint: for all 2x2 matrices M if tr(M) = 0 then M^2 = a*I where 'a' is a scalar and 'I' is 2x2 identity matrix

 

[TheWrathOfMath]

Hint: for all 2x2 matrices M if tr(M) = 0 then M^2 = a*I where 'a' is a scalar and 'I' is 2x2 identity matrix

We did not learn the Cayley-Hamilton Theorem yet

 

[topsquark]

See here for the full problem statement.

-Dan

 

[TheWrathOfMath]

Hint: for all 2x2 matrices M if tr(M) = 0 then M^2 = a*I where 'a' is a scalar and 'I' is 2x2 identity matrix

Never mind. I solved it.
Thank you for the assistance

 

[TheWrathOfMath]

See here for the full problem statement.

-Dan

I solved it.
Thank you very much for the assistance

 

[Steven G]

I do not think that I will need to raise AB-BA to the power of 2.
There must be a quicker and cleaner solution.
Or should I just raise AB-BA to the power of two, let C=(e f) and substitute it in the equation?
________________ ___________________________________________g h

Yes, that method will work fine.