0
$\begingroup$

Show that $P^{-1}AP=B$ where $P=(I-(A+B))$.Given that $A^2=A$ and $B^2=B$. .Please explain this step by step since I am unable to solve this problem .I only solve $A(I-(A+B))=A-A^2-AB=-AB$ since $A^2=A$.

  • 0
    Note that $P$ might not be invertible, so, to be precise, you can only show that $P^{-1}AP=B$ if $P$ is invertible (whereas $AP=PB$ is independent of that).2012-08-02

2 Answers 2

1

We have $ A(I-(A+B)) = A - A^2 - AB = -AB $ Also we have $ (I-(A+B))B = B - AB - B^2 = -AB $ So we can conclude the the LHS on both equations are equal, i.e. $ A(I-(A+B)) = (I-(A+B))B $ which is $AP = PB,$ and if $P$ is invertible, we have $P^{-1} AP = P^{-1}PB = B.$

  • 0
    I got it from Karolis comment but thank you for complete solution2012-08-02
1

$PB=(I-(A+B))B=B-AB-B^2=-AB\\AP=A(I-(A+B))=A-A^2-AB=-AB$