0
$\begingroup$

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?

$\endgroup$
0
$\begingroup$

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 $v_k$?

$\endgroup$
4
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.