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

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The discussion clarifies that while A being proportional to the identity matrix implies it is always diagonal, it does not inherently prove that B is diagonalizable. Instead, B is assumed to be diagonalizable, and a similarity transformation exists that can diagonalize both A and B if they commute. The identity matrix's properties ensure that any transformation applied to it will result in the identity matrix itself. Thus, any transformation that diagonalizes B will not alter A, maintaining its diagonal form. The key takeaway is the relationship between A and B's diagonalizability is contingent on their commutation.
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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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