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

How is this connected with Positron Emission Tomography?

3 Answers
Mar 11, 2018

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

Mar 11, 2018

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.

Mar 11, 2018

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