It seems to me that if the base field is the real numbers, then we have linearity in both arguments i.e. $\langle u + v, w + z\rangle = \langle u,w\rangle + \langle u,z\rangle + \langle v,w\rangle + \langle v,z\rangle$ because we know $\langle x,y\rangle = \langle y,x\rangle$ for any $y,x$
Do we only define inner products to be linear in their first argument in case the base field is the complex numbers?
Could we have just defined inner products over the real numbers to say that inner products are linear in both arguments?
Best Answer
The issue is that if you do so, you'll get $$\Vert \lambda u \Vert^2 = \langle \lambda u, \lambda u \rangle = \lambda^2 \langle u, u \rangle = \lambda^2 \Vert u \Vert^2$$ and you can't ensure that all those numbers are real numbers. While if $$\Vert \lambda u \Vert^2 = \langle \lambda u, \lambda u \rangle = \lambda \overline{\lambda} \langle u, u \rangle = \vert \lambda \vert^2 \Vert u \Vert^2$$ you get compatibility with the definition of a norm.