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#