Question

Question #3d7c3

Answer 1

Using the first fact:

`bb(A)bb(B) = bb(A)`

Premultiplying by `bb(A)^(-1)` we get:

`bb(A)^(-1)bb(A)bb(B) => bb(A)^(-1)bb(A) => bb(B)= bb(I)`

And so

`bb(B)^2 = bb(B)bb(B) = bb(B)bb(I) = bb(B)`
`=>bb(B)` is an idempotent matrix QED

And using the second fact:

`bb(B)bb(A) = bb(A)`

Premultiplying by `bb(B)^(-1)` we get:

`bb(B)^(-1)bb(B)bb(A)= bb(A)^(-1)bb(A) => bb(A)= bb(I)`

And so

`bb(A)^2 = bb(A)bb(A) = bb(A)bb(I) = bb(A)`
`=>bb(A)` is an idempotent matrix QED