Question

Under what non-trivial circumstances does `(A+B)^2=A^2+B^2`?

How is this connected with Positron Emission Tomography?

Answer 1

Under the circumstance that `AB=0`

Explanation:

We want to find when `(A+B)^2=A^2+B^2`.

We start by expanding the left hand side using the perfect square formula

`(A+B)^2=A^2+2AB+B^2`

So we see that `(A+B)^2=A^2+B^2` iff `2AB=0`

Answer 2

See below.

Explanation:

If `A, B` are vectors then

`(A+B)cdot(A+B) = norm(A)^2+2 A cdot B+norm(B)^2 = norm(A)^2+norm(B)^2`

then necessarily `A cdot B = 0 rArr A bot B` so `A,B` are orthogonal.

Answer 3

Some possibilities...

Explanation:

Given:

`(A+B)^2 = A^2+B^2`

A couple of possibilities...

Matrices

If `A` and `B` are matrices with `AB = -BA`

For example:

`A = ((0, -1, 0, 0), (1, 0, 0, 0), (0, 0, 0, -1), (0, 0, 1, 0))`

`B = ((0, 0, -1, 0), (0, 0, 0, 1), (1, 0, 0, 0), (0, -1, 0, 0))`

Then:

`AB = ((0, 0, 0, -1), (0, 0, -1, 0), (0, 1, 0, 0), (1, 0, 0, 0))`

`BA = ((0, 0, 0, 1), (0, 0, 1, 0), (0, -1, 0, 0), (-1, 0, 0, 0))`

Field of characteristic `2`

In a field of characteristic `2`, any multiple of `2` is `0`

So:

`(A+B)^2 = A^2+color(red)(cancel(color(black)(2AB)))+B^2 = A^2+B^2`