Why Do the Positions of C and C^-1 Switch in Matrix Diagonalization Definitions?

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The discussion centers on the definitions of similar matrices and their diagonalization, specifically the order of the matrices C and C^-1. The confusion arises from the textbook's notation, where A is expressed as A = C^-1BC for similar matrices and A = CDC^-1 for diagonalized matrices. Participants clarify that the order of multiplication is crucial, and the definitions provided are indeed correct. When calculating A^n, the switching of C and C^-1 is intentional and follows from the properties of matrix multiplication. The conversation concludes that the definitions are consistent and valid within the context of linear algebra.
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Homework Statement


My linear algebra textbook defines...

similar matrices: A = C^-1BC
diagonalized similar matrices: A = CDC^-1
A^n = C^-1*D^n*C

Why do the C^-1 and C's get switched around between the definitions? Doesn't order of multiplication matter? Are these the correct definitions? Is A^n really the opposite of the definition for diagonalized similar matrices, or is this an error?


Homework Equations





The Attempt at a Solution

 
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cookiesyum said:
My linear algebra textbook defines...

similar matrices: A = C^-1BC
That couldn't possibly have been the definition your book gave. What did you omit?
 
Hurkyl said:
That couldn't possibly have been the definition your book gave. What did you omit?

Nope, that's it! He has us using his own lecture notes, instead of a published textbook. This is why I thought it could potentially be an error.

So if I'm diagonalizing a matrix, there exists C such that A = CDC^-1 but when I want to find A^n, I solve C^-1*D^n*C (the C inverse and C switch spots). Is that on purpose?
 
cookiesyum said:
Nope, that's it! He has us using his own lecture notes, instead of a published textbook. This is why I thought it could potentially be an error.
Nothing else, really? Nothing like
A and B are similar matrices if ...​
or
... if there exists an invertible matrix C such that ...​
?




P.S. if A = C D C^{-1}, then we do indeed have A^n = C D^n C^{-1}. (For nonnegative integers n. Negative is allowed if D is invertible) And, of course, if we have the equation A = C^{-1} D C then we can infer A^n = C^{-1} D^n C
 
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