Proving B is Involutory if A is Idempotent

[radou]

Let A and B be square matrices of order n, such that B = 2A - I. (1) One has to proove that B is involutory (B^2 = I) <=> A is idempotent (A^2 = A).

Starting off with direction <= , one multiplies (1) with A from the right and from the left, separately, and obtains the results BA = A, and AB = A, respectively. This implies B = I, and I is an involutory matrix, so that direction is prooved. (I hope, that's why I'm asking.)

The second direction, => , causes trouble and I'd appreciate any help. I tried multiplying (1) with B, but it didn't seem to help. Thanks in advance.

 

[StatusX]

BA = A and AB = A does not give B = I. For example, take A=0 (there are other examples too). It is true that B=I if BA = A and AB = A holds for all A, but that isn't the case here.

Just square both sides of B=2A-I. This should give both directions pretty easily.

 

[radou]

BA = A and AB = A does not give B = I. For example, take A=0 (there are other examples too). It is true that B=I if BA = A and AB = A holds for all A, but that isn't the case here.

Just realized that, thanks.

Just square both sides of B=2A-I. This should give both directions pretty easily.

Did that, and prooved both directions. Thanks again!