Group theory? This solution doesn't make sense....

applestrudle
Messages
64
Reaction score
0
Case 2:
I get that D = c I means A must also be proportional to I but how does that mean B must be diagonal?

Question:
Screen Shot 2015-07-22 at 17.14.32.png

Answers:
Screen Shot 2015-07-22 at 17.17.04.png


Screen Shot 2015-07-22 at 17.18.17.png


Screen Shot 2015-07-22 at 17.18.59.png

Screen Shot 2015-07-22 at 17.20.00.png
 
Physics news on Phys.org
It does not show that B is diagonalizable. B is presumed to be such. The point is that there is a transform that makes B diagonal and also makes A diagonal IFF A and B commute. The particular case of A proportional to the identity matrix means that A is always diagonal. Because every similarity transform on the identity matrix simply gives back the identity matrix. So, in this case, any similarity transform that makes B diagonal will leave A unchanged and still diagonal.
 
  • Like
Likes WWGD
DEvens said:
It does not show that B is diagonalizable. B is presumed to be such. The point is that there is a transform that makes B diagonal and also makes A diagonal IFF A and B commute. The particular case of A proportional to the identity matrix means that A is always diagonal. Because every similarity transform on the identity matrix simply gives back the identity matrix. So, in this case, any similarity transform that makes B diagonal will leave A unchanged and still diagonal.
THANKS! :D
 
A sphere as topological manifold can be defined by gluing together the boundary of two disk. Basically one starts assigning each disk the subspace topology from ##\mathbb R^2## and then taking the quotient topology obtained by gluing their boundaries. Starting from the above definition of 2-sphere as topological manifold, shows that it is homeomorphic to the "embedded" sphere understood as subset of ##\mathbb R^3## in the subspace topology.
Back
Top