Prove that there is a unique inner product on $V$ [duplicate]

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 $v,u\in W_i$: $\langle v, u \rangle = \langle v, u \rangle _i$

Uniqueness:
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 StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); May 8 at 3:38

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.
• Cool. Should I explain further why is it an inner-product? (i.e. $\langle x,x \rangle > 0$) – AlonAlon Jan 5 '15 at 19:31
• 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 $W_2 \subset W_1^\perp$.) – John Hughes Jan 5 '15 at 19:37
• @AlonAlon: Because the decomposition of any vector into components in $W_1, W_2$ is unique. Then the orthogonality follows by definition. – copper.hat Jan 5 '15 at 20:39