1. Having any matrices $A,B$ and vectors $x,y$ (real/complex, singular/regular, rectangular, infinite size, etc.) with appropriate size (so the difference exist but other than scalar) does this always hold (distributive law)? $$(Ax-By)^H(Ax-By)=x^HA^HAx-y^HB^HAx-x^HA^HBy+y^HB^HBy$$ ($()^H$ is hermitian transpose). I know that it is trivial and it holds for ''usual'' cases, I am just curious if it holds for any (imaginable) case.

2. Is it true that having complex entries $y^HB^HAx\ne x^HA^HBy$ (after transposing, it becomes complex conjugate), but for any real case $y^HB^HAx = x^HA^HBy$?

Thanks.

• I think yes and yes. – Peter Franek Dec 15 '14 at 17:33

1. Yes, of course. Assuming you are comfortable with the following facts:

a) Matrix multiplication is associative

b) Matrix multiplication is distributive

c) $(AB)^H = B^HA^H$

d) $(A+B)^H = A^H+B^H$

Then what you have written is an immediate consequence of these properties.

1. Yes. If $a$ is a real scalar, then $a^H = a$. Nothing changes if you write $a$ as a product of complex matrices.