MHB Commutativity in the linear transformation space of a 2 dimensional Vector Space

quarkine
Messages
2
Reaction score
0
A variant of a problem from Halmos :
If AB=C and BA=D then explain why (C-D)^2 is commutative with all 2x2 matrices if A and B are 2x2 matrices.
This result does not hold for any other nxn matrices where n > 2. Explain why.

Edit: I tried to show that ((C-D)^2) E - E((C-D)^2) is identically zero. But that didn't work.
A guess is that since the above matrix commute with any 2x2, it has to be of the form bI (where b is a scalar anad I the indentity), which can be confirmed by brute calculation but I am searching for a better way.
 
Last edited:
Physics news on Phys.org
Can you give your thoughts and/or show what you have tried so our helpers know exactly where you are stuck and how best to help?
 
quarkine said:
If AB=C and BA=D then explain why (C-D)^2 is commutative with all 2x2 matrices if A and B are 2x2 matrices.
This is known as Hall's identity. The proof, in very brief outline, goes like this. The commutator $[A,B] = AB-BA$ has trace zero. Its characteristic equation is therefore of the form $\lambda^2 = \mathrm{const.}$ It then follows from the Cayley–Hamilton theorem that $[A,B]^2$ is a multiple of the identity.
 
Thread 'Derivation of equations of stress tensor transformation'
Hello ! I derived equations of stress tensor 2D transformation. Some details: I have plane ABCD in two cases (see top on the pic) and I know tensor components for case 1 only. Only plane ABCD rotate in two cases (top of the picture) but not coordinate system. Coordinate system rotates only on the bottom of picture. I want to obtain expression that connects tensor for case 1 and tensor for case 2. My attempt: Are these equations correct? Is there more easier expression for stress tensor...

Similar threads

Back
Top