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

applestrudle
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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?

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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.
 
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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.
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I posted this question on math-stackexchange but apparently I asked something stupid and I was downvoted. I still don't have an answer to my question so I hope someone in here can help me or at least explain me why I am asking something stupid. I started studying Complex Analysis and came upon the following theorem which is a direct consequence of the Cauchy-Goursat theorem: Let ##f:D\to\mathbb{C}## be an anlytic function over a simply connected region ##D##. If ##a## and ##z## are part of...
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