Unique decomposition of a vector

Question

Show that every vector $\vec{u}$$\in$$U$ can be uniquely decomposed into $\vec{u}$$=$$\vec{u}_{1}$$+$$\vec{u}_{2}$ where $\vec{u}_{1}\in{W}\subset{U}$ and $\vec{u}_{2}\in{W^⊥}$.

Answer 1

Suppose we have $W, W^{\perp}$ subspaces of $U$. Let $u\in U$ be given. $u$ has a decomposition into $P\rvert_W(u)+P\rvert_{W^{\perp}}(u)=u.$ Here, $P\rvert_W$ denotes the projection map onto the subspace $W.$ It is sufficient to show that $W\oplus W^{\perp}=U$.

The proof is simple. We know that $W\cap W^{\perp}=\{0\}$, by definition of the orthogonal complement. Also from the standard definition of the orthogonal complement, we know that $$dim(W)+dim(W^{\perp})=dim(U).$$ So, $W+W^{\perp}=U$, and we know that the sum is direct from the intersection of the spaces being trivial. So, by definition of a direct sum of vector subspaces, each $u\in U$ has a unique representation as a sum of $x_1\in W, x_2\in W^{\perp}$. In this case, $x_1=P\rvert_W(u), x_2=P\rvert_{W^{\perp}}(u)$.