# Inner product alternative definition

I'm trying to make an alternative (but equivalent) definition of an inner product. I prefer to use arbitrary sums with $\sum_i v_i$ instead of sum of just two vectors $v+w$, and few but clear axioms.

An inner product is a function $\langle \cdot , \cdot \rangle : V \times V \to \mathbb{K}$ with $\mathbb{K} = \mathbb{R}$ or $\mathbb{C}$, that satisfies

This is my list of axioms, I'll call it a $B$-inner product, I need help to know if this is correct:

B1. $\langle \sum_i \lambda_i v_i , \sum_j \mu_j w_j \rangle = \sum_i \sum_j \lambda_i \overline{\mu_j} \langle v_i, w_j \rangle$

for arbitrary $v_i, w_j \in V$, $\lambda_i, \mu_j \in \mathbb{K}$

B2. $\langle v,v \rangle = 0 \Rightarrow v=0$.

I know the actual definition of inner product (I call it a $A$-inner product) is

A1. $\langle v_1 + v_2, w \rangle = \langle v_1,w \rangle + \langle v_2, w \rangle$

A2. $\langle \lambda v, w \rangle = \lambda \langle v,w \rangle$

A3. $\langle v,w \rangle = \overline{ \langle w, v \rangle }$

A4.1 $\langle v,v \rangle \geq 0$

A4.2 $\langle v,v \rangle = 0 \Leftrightarrow v=0$

Clearly an $A$-inner product is a $B$-inner product.

And clearly $B1$ implies $A1$ and $A2$. At least in finite dimension I think it also implies $A3$ (but not sure): I write $v$ and $w$ in an orthonormal basis, apply $B1$ on $\langle w,v \rangle$, and conjugate, it gives the same than the product $\langle v,w \rangle$.

I think it also satisfy $A4.1$ as the product of a complex number by its conjugate always gives a non-negative real number.

Is my definition equivalent? Do I also need $v=0 \Rightarrow \langle v,v \rangle = 0$? What other axioms should I add to make it equivalent?

Thanks.

• Are the sums in your definition finite? – Omnomnomnom Dec 3 '14 at 17:37
• Yes are sums are finite. How can I prove $A3$ in the infinite-dimensional case? I used that a orthonormal basis exists to prove that $\langle v,w \rangle = \sum_i \lambda_i \overline{\mu_i} = \overline{\sum_i \overline{\lambda_i} \mu_i} = \overline{ \langle w, v \rangle }$ – dami Dec 3 '14 at 18:17
• My proof of $A4.1$ is $\langle v,v \rangle = \langle \sum_i \lambda_i v_i, \sum_i \lambda_i v_i \rangle = \sum_i \lambda_i \overline{\lambda_i} = \sum_i |\lambda_i|^2 \geq 0$ – dami Dec 3 '14 at 18:20
• $\langle a,b \rangle = -ab, \mathbb{R}\times\mathbb{R}\rightarrow \mathbb{R}$ satisfies $B1$ and $B2$, but not $A4.1$ – Gregor de Cillia Jan 29 '15 at 3:45
• Gregor: you are right. I now think B1 and B2 are just substitutes to A1 and A2. And the other axioms A3, A4.1 A4.2 are necessary. – dami Jan 30 '15 at 4:24

Assuming the sums $\sum_i,\sum_j$ are over finitely many indices, your definition is equivalent to the standard one.
Your axiom $$B1$$ states that $$\langle \cdot, \cdot \rangle$$ is bilinear and the axiom $$B2$$ states that $$\langle \cdot, \cdot \rangle$$ is non-degenerate. On the other hand, axioms $$A1-A4$$ states that the inner product is (conjugate) symmetric and positive definite. No doubt, that non-symmetric, non-positive definite forms exists. So, the axioms $$(A)$$ are strictly stronger.
In your proof of $$A3$$ and $$A4.1$$ you implicitly use the fact that $$\langle \cdot, \cdot \rangle$$ has an othonormal basis, that is a system of vectors $$(e_i)_{i \in I}$$ such that $$\langle e_i, e_j \rangle = 1 \iff i = j$$ and $$\langle e_i, e_j \rangle = 0 \iff i \neq j$$. And you wrote $$v = \lambda_i e_i,w = \mu_i e_i$$. In fact, in any finite dimensional vector space, any inner product admits an orthonormal basis. So, if you add axiom $$B'3:\text{orthonormal basis exists},$$ or, equivalently, rewrite axiom $$B1$$ as $$B'1: \left\langle \sum_i\lambda_i e_i, \sum_i\mu_i e_i \right\rangle = \sum_i \lambda_i\overline{\mu_i},$$ which resembles standard definition of the dot product, your axioms would be equivalent. To sum all up, you successfully proved that any non-degenerate form, which admits an orthonormal basis, is an inner product.
However, it turns out that in infinite dimension not every inner product admits an orthonormal basis. Whence, the updated axioms $$(B')$$ are stronger than axioms $$(A)$$.