Are there any exceptions?

I was thinking proof by contradiction i.e. define $\langle f,g\rangle\ \neq0$ for two orthogonal elements of the product space, but positive definiteness would require one of the two elements to be zero.

EDIT: I was told that additivity in the second coordinate of any inner product space $\langle f,g\rangle$ logically follows from and can be proven using only these four properties of inner product spaces:

  1. $\langle f,g\rangle=\langle g,f\rangle$ (symmetry)
  2. $\langle f+h,g\rangle=\langle f,g\rangle+\langle h,g\rangle$ (additivity)
  3. $\langle \text{c}f,g\rangle=\text{c}\langle f,g\rangle$
  4. $\langle f,f\rangle>0\,\,\forall\,\left\{f\neq0\right\}\in V$, where $V$ is some linear space (posititive definiteness).

I'm thinking that 1. and 2. imply $\langle f,g+h\rangle=\langle f,g\rangle+\langle f,h\rangle$; how do I use these axioms to explicitly verify this?

  • 3
    $\begingroup$ It has to be additive bilinear in both coordinates otherwise it is not an inner product. $\endgroup$
    – Timbuc
    Mar 25 '15 at 17:39
  • 1
    $\begingroup$ I get that - but how do I show that being additive bilinear in one coordinate implies the same for the other coordinate, for any inner product? $\endgroup$ May 2 '15 at 23:33

Yes. This is the additivity property of the inner product space.

  • $\begingroup$ I need to prove this using only the four axioms listed above. The wiki article, while conceptually useful, gestures to concepts not addressed in my linear algebra class. What would a proof using only these four look like? $\endgroup$ May 2 '15 at 23:34

$\langle f,g+h\rangle = \langle g+h, f\rangle$ by symmetry. Then, $\langle g+h,f\rangle = \langle g,f\rangle + \langle h,f\rangle = \langle f,g\rangle + \langle f,h\rangle$ by additive in first coordinate and then symmetry.


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