Question

If B is a square matrix and B²=B, then prove that A = I -B satisfies A²=A and AB=BA=O.

Answer 1

Answer:

If B is a square matrix and B²=B, then prove that A = I -B satisfies A²=A and AB=BA=O.

Given:

$\text{B is a square matrix and }B^2=B$

$A=I-B$

1.

$A^2$

$=A.A$

$=(I-B)(I-B)$

$=I^2-IB-IB+B^2$

$=I-B-B+B$        $(using\:B^2=B)$

$=I-B$

$=A$

$\implies\:\boxed{A^2=A}$

2.

$AB$

$=(I-B)B$

$=IB-B^2$

$=B-B$      $(using\:B^2=B)$

$=\bf{0}$

$BA$

$=B(I-B)$

$=BI-B^2$

$=B-B$       $(using\:B^2=B)$

$=\bf{0}$

$\implies\:\boxed{AB=BA=0}$

Hence proved