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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?

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  • $\begingroup$ When choosing definitions, mathematicians usually like to make them as simple as possible, to make it easier to prove that a particular object satisfies that definition. The linearity in the second coordinate (over $\mathbb{R}$) follows as a lemma. $\endgroup$
    – vadim123
    Sep 11 '15 at 14:28
  • $\begingroup$ Perhaps relevant: math.stackexchange.com/questions/1429174. $\endgroup$
    – joriki
    Sep 11 '15 at 14:31
  • $\begingroup$ I changed $<u,v>$ to $\langle u,v\rangle$. That is standard usage. ${}\qquad{}$ $\endgroup$ Sep 11 '15 at 14:34
  • $\begingroup$ It depends what standard means. Even $\langle u | v\rangle$ has been standard for long time now (if you are a physicist). $\endgroup$ Mar 29 at 11:56
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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.

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We might take an arbitrary bilinear from instead, but: We want to be able to define a norm (and ultimately a topology) from our inner product. First of all, this prevents us from talking about inner products in characteristic $\ne 0$. We also get difficculties if the groud field is larger than $\mathbb C$. Remains to cover the fields from $\mathbb Q$ up to $\mathbb C$, of which $\mathbb R$ is just a special case. To cover all cases at once, it suffices to adjust the definition to the case of $\mathbb C$, where we need the conjugate symmetry (and hence sesquilinearity) instead of bilinerity in order to have $\langle x,x\rangle\in\mathbb R$ (for our intended norm).

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They are sesquilinear (which means one and a half times linear), not merely linear in the first variable, i.e. it is additive also in the second variable, and $$f(u,\lambda v)=\bar\lambda f(u,v)$$ This is to achieve the axioms of a norm in the context of $\mathbf C$-vector spaces: $$f(\lambda ua,\lambda u)=\lambda\bar\lambda f(u,u)=\lvert\lambda\rvert^2f(u,u). $$

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  • $\begingroup$ I'm sorry, but I am awful at remembering mathematical notation...does this bar represent the conjugate of lambda and if lambda is a real number does this just become $f(u,\lambda v) = \lambda f(u,v)$ $\endgroup$
    – user111707
    Sep 12 '15 at 6:01
  • $\begingroup$ You're absolutely right. $\endgroup$
    – Bernard
    Sep 12 '15 at 8:40
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If it's a real inner product, then it's symmetric, i.e. $\langle x,y\rangle = \langle y, x\rangle$, so linearity in one argument implies linearity in the other; hence keeping the list of axioms non-redundant requires mentioning only one explicitly.

If it's a complex inner product, then $\langle x,y\rangle$ and $\langle y,x\rangle$ are not equal to each other but conjugates of each other. In that case it's conjugate-linear in the second argument, i.e. $\langle x,\lambda y\rangle$ is not $\lambda\langle x,y\rangle$, but $\bar\lambda\langle x,y\rangle$.

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  • $\begingroup$ Thank you...this makes it clear. I figured that if we ONLY cared about real inner product spaces we could say that it's linear in both arguments but all of the definitions (I read) for an inner product space explicitly mentioned linearity only in the first argument and I wanted to make sure I wasn't missing anything because it seemed to me like if we are ONLY talking about real inner product spaces, we have linearity in both arguments $\endgroup$
    – user111707
    Sep 12 '15 at 6:03
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It seems strange that nobody has given the following proof. The most easy way to understand this is working with complex numbers, and for simplicity I will work in $\mathbb{C}^n$. A function $(\cdot, \cdot)$ from $\mathbb{C}^n \times \mathbb{C}^n$ to $\mathbb{C}$ is an inner product if it satisfies (note that I switched a little bit the inner product definition!)

  1. $(\cdot, \cdot)$ is linear in the second argument, $$\left(\vec{v},\sum_{i} c_i \vec{w}_i \right)=\sum_{i} c_i (\vec{v},\vec{w}_i),$$ with $\vec{v},\vec{w}_i \in \mathbb{C}^n$.

  2. $(\vec{v},\vec{w})=(\vec{w},\vec{v})^*$, where $*$ means complex conjugation.

  3. $(\vec{v},\vec{v})\geq 0$ with equality if and only if $\vec{v}=0$.

By definition, the inner product is linear in the second entry always. It is the first entry that one has to prove that it is conjugate-linear. But then take property 2, i.e.

\begin{equation} \begin{split} \left(\sum_i c_i\vec{v}_i,\vec{w} \right) &= \left(\vec{w},\sum_i c_i\vec{w} \right)^*\\ &= \sum_i c_i^*\left(\vec{w},\vec{v}_i \right)^*\\ &= \sum_i c_i^*\left(v_i,w_i \right), \end{split} \end{equation}

where I have used properties 1 and 2 of the inner product and the fact that, for two complex numbers $z_1,z_2$ ($\left(\vec{a},\vec{b} \right)\in\mathbb{C}$) $(z_1,z_2)^*=z_1^* z_2^*$. If you instead work with real numbers, then the complex conjugation does not affect the first argument, obviously...

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