# Are projection and norm enough to define an inner product?

Given an inner product, one can define a projection and a norm. Can we do the opposite?

That is, suppose we have:

• a complex vector space V
• a norm $|V|^2 : V \rightarrow \mathbb{R}$ such that:
• is posite definite
• $|\alpha V|^2 = |\alpha|^2 |V|$ with $\alpha \in \mathbb{C}$
• a family of projections $\forall v \in V \; \exists \mathbf{P}_v : V \rightarrow V$ such that:
• $\mathbf{P}_v ^2 = \mathbf{P}_v$
• $\mathbf{P}_v (\alpha w_1 + \beta w_2) = \alpha \mathbf{P}_v (w_1) + \beta \mathbf{P}_v (w_2)$
• $\forall w \in V$ $\mathbf{P}_v(w)=\alpha v$ with $\alpha \in \mathbb{C}$
• $\mathbf{P}_v(v)=v$

Is the map $\langle \cdot , \cdot \rangle : V \times V \rightarrow \mathbb{C}$ defined such that $\mathbf{P}_v(w)=\frac{\langle v, w \rangle}{|v|^2} v$ an inner product?

$\langle \cdot , \cdot \rangle$ is positive definite. $\mathbf{P}_v(v) = v$. $\langle v, v \rangle = |v|^2$. The norm induced by the map is the original norm which is positive definite.

$\langle \cdot , \cdot \rangle$ is linear in the second argument because the projection is linear.

Conjugate symmetry is the only thing that I am missing. Is it an extra requirement, or can it be derived by the previous?

• You also need some axiom on the family of projections relating the indexes, i.e., the indexes $v$ in $P_v$ should satisfy some property. Otherwise $P_v=0$ satisfies all of this but is clearly useless. Something like $|v_1|^2P_{v_1}+|v_2|^2P_{v_2}=|v_1+v_2|^2P_{v_1+v_2}$, maybe? (But not this, because again $P_v=0$ is also a counter-example). Ok, I think you also need $P_v(v)=v/|v|$, so these counter-examples don't work anymore Sep 15, 2016 at 20:01
• The third condition was supposed to do that... Maybe I should just require $\mathbf{P}_v (v) = v$ more explicitly? Sep 15, 2016 at 20:09
• Yes, this is necessary (I made a mistake: it should've been $P_v(v)=v$, as you noted), otherwise, we have some counter-examples as above. This should also be neccessary for the inner-product to be non-degenerate. Sep 15, 2016 at 20:11

Since you require nothing on the family of functionals/projections with respect to $$v$$, there are plenty of functionals/projections violating the conjugate symmetry.