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Let $V$, a vector space over $\mathbb{F}$ and $W_1,W_2 \subseteq V$ such that $W_1 \oplus W_2 = V$. For $i=1,2$, let $\langle , \rangle $ on $W_i$. Prove that there is a unique inner product on $V$ such that:

  1. $W_2 = W^\perp_1$
  2. For $i=1,2$ and for all $v,u\in W_i$: $\langle v, u \rangle = \langle v, u \rangle _i$

Let $v = w_1 + w_2$ and $u = w_1' + w_2'$

$$\langle u, v \rangle_V = \langle w_1,w_1' \rangle + \langle w_1,w_2' \rangle + \langle w_2,w_1' \rangle + \langle w_2,w_2' \rangle = \langle w_1,w_1' \rangle + \langle w_2,w_2' \rangle $$

So the inner product is determined uniquely by $\langle, \rangle_i $ for $i=1,2$.

How do I show the existence of such inner product?


marked as duplicate by Lord Shark the Unknown, Martin Argerami linear-algebra May 8 at 3:38

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You can define $$ \langle p, q \rangle = \langle p_1, q_1 \rangle_1 + \langle p_2, q_2\rangle_2 $$ where $p = p_1 + p_2$ is the decomposition of $p$ into a sum of vectors in $W_1$ and $W_2$, and similarly for $q = q_1 + q_2$. Because it's a direct sum, this decomposition is unique, so this expression is well-defined.

And if $p \in W_1$ and $q \in W_2$, we get $\langle p, q \rangle = 0$, as needed for the orthogonality condition.

  • $\begingroup$ Cool. Should I explain further why is it an inner-product? (i.e. $\langle x,x \rangle > 0$) $\endgroup$ – AlonAlon Jan 5 '15 at 19:31
  • $\begingroup$ You should indeed prove that it's symmetric, positive definite, and that under this inner product, $W_2$ is in fact $W_1^\perp$. (All I showed was that $W_2 \subset W_1^\perp$.) $\endgroup$ – John Hughes Jan 5 '15 at 19:37
  • $\begingroup$ @AlonAlon: Because the decomposition of any vector into components in $W_1, W_2$ is unique. Then the orthogonality follows by definition. $\endgroup$ – copper.hat Jan 5 '15 at 20:39

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