Let $V$ be a vector space over field $\mathbb{F}$, and let $W_1,W_2 \subseteq V$ be sub-spaces such that $V=W_1 \oplus W_2$.
Let $\langle,\rangle_i$ be an inner product on $W_i$ for $i=1,2$.
Prove that there exists ONE (and only) inner product on $V$ satisfying:
a. $W_2=W_1^{\perp}$
$\langle,\rangle=\langle,\rangle_i$ for $i=1,2$ and foe all $u,v \in W$

What I did:
I proved that for any $W \subseteq V$: $W \oplus W^{\perp}=V$ so maybe I can say that $W_2$ (from the question) is equal to $W_1^{\perp}$ as required in a.
I didn't quite understand how to prove that there is only one inner product that satisfy these requirements.
Any hints/ideas?


Your thought process shows a way to decompose $V$ into a subspace and its orthogonal complement, but it doesn't show that if $V=W_1\oplus W_2$, then $W_2=W_1^\perp$ for any inner product. In fact, this is false; take $V=\mathbb{R}^2$, $W_1=span((1,0))$, $W_2=span((1,1))$ so that $W_1^\perp = span((0,1))\neq W_2$ with the standard Euclidean inner product.

Hint: If you take $(u_1,\dots,u_n)$ to be a basis of $W_1$ and $(v_1,\dots,v_m)$ to be a basis of $W_2$, can you show that adjoining these two bases yields a basis of $V$ given $V=W_1\oplus W_2$? Can you define an inner product using this basis of $V$ to ensure that $\langle u_i,v_k\rangle =0$ for each $u_i$ and $v_k$?

  • $\begingroup$ Hmmm can I say that if the norm satisfy parallelogram law than it's uniquely defiened? $\endgroup$ – user114138 Jan 3 '15 at 11:24
  • $\begingroup$ If you can prove that the norm is unique, then yes. $\endgroup$ – Michael M Jan 3 '15 at 15:27
  • $\begingroup$ I succeeded thank you! But can you please explain why proving the norm is unique proves the uniqueness of the inner product? $\endgroup$ – user114138 Jan 4 '15 at 10:55
  • $\begingroup$ If you know that the norm comes from an inner product, you can uniquely define the inner product as $\langle u,v\rangle=\frac{\|u+v\|^2-\|u-v\|^2+\|u+iv\|^2i-\|u-iv\|^2i}{4}$ (on a complex space $V$). Of course not every norm comes from an inner product, but it is possible to prove that every norm satisfying the parallelogram law comes from an inner product. See here: math.stackexchange.com/questions/21792/… $\endgroup$ – Michael M Jan 4 '15 at 14:48

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