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?

  • $\begingroup$ 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 $\endgroup$ Sep 15, 2016 at 20:01
  • $\begingroup$ The third condition was supposed to do that... Maybe I should just require $\mathbf{P}_v (v) = v$ more explicitly? $\endgroup$
    – Carcassi
    Sep 15, 2016 at 20:09
  • $\begingroup$ 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. $\endgroup$ Sep 15, 2016 at 20:11

1 Answer 1


The family of projections with one-dimensional range is a red herring: It is equivalent to require a family of linear functionals (if you assume the axiom of choice). In the case of a normed space, it is equivalent to require a family of bounded projections/bounded linear functionals by Hahn-Banach (again, if you assume the axiom of choice).

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

A prominent family of examples are semi-inner products, see https://en.wikipedia.org/wiki/L-semi-inner_product (The examples given there are linear in the first argument instead of the second, but this is of course just a notational difference.)

  • $\begingroup$ Thanks! Counter-examples makes it a clear no. $\endgroup$
    – Carcassi
    Feb 18, 2021 at 16:44

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