Question 904029
We want to show that C is an idempotent matrix


<a href="http://mathworld.wolfram.com/IdempotentMatrix.html">http://mathworld.wolfram.com/IdempotentMatrix.html</a>


That means we want to show that C*C = C is true.


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{{{C = X*A*X^(-1)}}} Start with the given equation.


{{{C*C = (X*A*X^(-1))*C}}} Right multiply both sides by C (see note below)


{{{C*C = (X*A*X^(-1))*(X*A*X^(-1))}}} Plug in {{{C = X*A*X^(-1)}}} on the right side.


{{{C*C = X*A*(X^(-1)*X)*A*X^(-1)}}} Use the associative property of matrix multiplication.


{{{C*C = X*A*(I)*A*X^(-1)}}} The expression {{{X^(-1)*X}}} is equal to {{{I}}} since X is nonsingular (ie X is invertible)


{{{C*C = X*(A*I)*A*X^(-1)}}} Use the associative property of matrix multiplication.


{{{C*C = X*A*A*X^(-1)}}} Matrix Multiplicative Identity Property 


{{{C*C = X*(A*A)*X^(-1)}}} Use the associative property of matrix multiplication.


{{{C*C = X*A*X^(-1)}}} Matrix A is idempotent, so A*A = A


{{{C*C = C}}} Replace the right side with what it really is


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Because we've shown that {{{C*C = C}}} is true, this proves that C is an idempotent matrix.



Note: You can also use left multiplication to get to the same result (in a very slightly different way, nothing too major though). Left multiplication is a bit different than right multiplication because matrix multiplication is NOT commutative. The proof is effectively the same as shown above which is why I don't need to show it, but it's important to keep this in mind.