Why Might the Square of a Sum Not Equal the Sum of Squares for Matrices?

  • Thread starter Thread starter cmab
  • Start date Start date
cmab
Messages
32
Reaction score
0
(e+f)^2 = e^2+2ef+f^2
Why is this not neccesary true for squares matrices for all size?
 
Physics news on Phys.org
(e+ f)^2= (e+f)(e+f)= e^2+ fe+ef+ fe. Remember that multiplication of matrices is not, in general, commutative!
 
HallsofIvy said:
(e+ f)^2= (e+f)(e+f)= e^2+ fe+ef+ fe. Remember that multiplication of matrices is not, in general, commutative!
Do you mean (e+f)2 = e2+fe+ef+f2? I might be misinterpreting it but I can't see where that extra "fe" came from and where the "f2" went.
 
Yeah, that's what he meant: (e+f)² = e² + ef + fe + f².
 
One of these days, I'm going to learn to type!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top