SRL Archive - Mathematics

The statement is:
Show that if a matrix, A, is idempotent, then 2A-I is invertible and it is its own inverse.

I did
(2A-I)*I
=(2A-I)*A*A^-1
=(2A^2-A)*A^-1
=(2A-A)*A^-1
=A*A^-1 = I
and I is invertible, thus 2A-I is invertible.

Have I done something that is not always true here? or is 2A-I really equal to I? meaning A = I
(Idempotent means that A^n = A for any matrix A and n being an element of Z+

Just seems too neat

I - A is idempotent

( 2A - I )( 2A - I ) = 4A² - 2AI - 2IA + I² = 4A - 2A - 2A + I = I, which is the identity :)

hmm thats pretty neat, almost TOO neat. I think I have done something wrong, or A=I

Now moving on to getting bloody vandermonde determinantes through row operations… :P