Question 39572
    I need to show (in general) that either matrix A is "singular" or "A^2 = A^(-1)". A is an n*n square matrix such that A^4 = A. I want to know how to do this problem and what happens if the exponent values are different.(for example, odd number exponents?)
PROCEDURE IS TO,TRANSPOSITION TO LEFT OR RIGHT SIDE OF THE EQUATION, RIGHT OR LEFT MULTIPLY BOTH SIDES OF THE EQUATION WITH A OR A^(-1),TAKING COMMON FACTORS ACCORDINGLY....NOTING THAT A^N=A*A*A*...N TIMES AND 
A*A^(-1)=A^(-1)*A=I...AND A*B=0 IMPLIES A=0 OR B=0.....
IN THIS CASE WE HAVE 
A^4=A
(A^4)-A=0
A(A^3-I)=0...HENCE A=0...OR.....A^3-I=0
A^3=I...MULTIPLY WITH A^-1 BOTH SIDES
A^2*A*(A^-1)=I*(A^-1)=A^-1
A^2=A^(-1)