How can I show that every vector in $U\oplus W$ can be uniquely decomposed into a sum of vectors in $U$ and $W$?
Let $V$be an inner product space over $\mathbb{R}$ (here, the inner product is denoted by $\langle,\rangle$).
a) Let $u$, $v$ be two vectors in $V$. Show that $u=v$ if $\langle u,w\rangle = \langle u,w \rangle$ for any $w$ in $V$.
b) Let $U$, $W$ be two subspaces of $V$. Prove that the sum $U+W$ is a direct sum $U\oplus W$ if $U$ and $W$ are orthogonal.
c) Show that every vector in $U\oplus W$ can be uniquely decomposed into a sum of vectors in $U$ and $W$.
I can solve a, b but can't c.
c): Let $v \in U \oplus W$.
Note that by definition, we already have that $v = u + w$ for some $u \in U,\; w \in W$.
We wish to show that this choice of $u$ and $w$ is unique. Suppose that we have $u' \in U, \; w' \in W$ such that $v = u' + w'$. We show that $u = u'$ and $w = w'$.
First, we note that $u + w = v = u' + w'$ and thus $u + w = u' + w'$. This can be rearranged to get $$u - u' = w' - w.$$
Now, note that the LHS is an element of $U$ and RHS of $W$. As $U \cap W = \{0\}$, we must have that both sides are $0$. This gives us that $u = u',\; w = w'$, as desired.