Doubts about the proof of $q^*q = qq^*$ property for quaternions.

Question

I have a problem with the proof of this quaternions property, included below. $(q^∗)^∗ = [s, −(−v)] = [s, v] = q$，where $q = [s, v]$ and $v = [x, y, z]$. $$q^∗q = (q^∗)(q^∗)^∗= ∥q^∗∥^2= s^2 + x^2 + y^2 + z^2= ∥q∥^2= qq^∗$$ so we got $q^*q = qq^*$.

Sorry my question is not clearly , my problem is why $$q^∗(q^∗)^∗=||q^∗||^2$$ not a $$q^∗(q^∗)^∗=||q^∗||^{{2}∗}$$ Why in the last equation we lost one of the conjugate marks?

Answer 1

I'm not sure about which conjugate are you asking, so let's go step by step. $$q^*q = q^*(q^*)^* $$ Here we used the property $(q^*)^* = q$ proved earlier.

$$q^*(q^*)^* = ||q^*||^2 $$ Here, we used that $rr^* = ||r||^2$ with $r = q^*$.

$$||q^*||^2 = s^2+x^2+y^2+z^2 $$ Here is the definition, $||q^*||^2 = s^2+(-x)^2+(-y)^2+(-z)^2 = s^2+x^2+y^2+z^2$, the author didn't bother with writing out the step in between.

$$s^2+x^2+y^2+z^2 = ||q||^2 $$ This is also the definition. $$||q||^2 = qq^* $$ Here the property $rr^* = ||r||^2$ is used again, this time with $r = q$.

Answer 2

Conjugate operation is involutive, so $(q^*)^* = q$.

Thus, since $qq^*$ is real, $qq^* = (qq^*)^*=q^*(q^*)^*=q^*q$.