1. Databases
2. Proof database
3. Any vector can be decomposed into one vector and another in its orthogonal complement

Any vector can be decomposed into one vector and another in its orthogonal complement

Theorem : Given any vector $v$ and another vector $u$, $\| u \| \neq 0$, $u$ generating the one-dimensional subspace $W = \textrm{Span}(u)$, then $v$ can be written as

\begin{equation} v = \alpha u + w \end{equation}

with a unique $\alpha \in \mathbb{R}$ and a unique $w \in W^\perp$.

Proof : First project $v$ onto $u$, via

\begin{equation} \mathrm{Proj_u(v)} = \frac{\langle v, u \rangle}{\langle u, u \rangle} u \end{equation}

If we decompose our vector $v$ via

\begin{equation} v = \mathrm{Proj_u(v)} + (v - \mathrm{Proj_u(v)}) \end{equation}

Then we need to show that the second part is orthogonal to $u$.

\begin{eqnarray} \langle v - \mathrm{Proj_u(v)}, u \rangle &=& \langle v, u \rangle - \langle \mathrm{Proj_u(v)}, u \rangle\\ &=& \langle v, u \rangle - \frac{\langle v, u \rangle}{\langle u, u \rangle} \langle u, u \rangle\\ &=& \langle v, u \rangle - \langle v, u \rangle\\ &=& 0 \end{eqnarray}

Now let's show that this decomposition is unique. Consider a second decomposition :

\begin{equation} v = \alpha' u + w' \end{equation}

As $w'$ is orthogonal to $u$, we can just substract the two :

\begin{equation} v - v= (\alpha - \alpha') u + (w - w') \end{equation}

Being in different subspaces, $\alpha - \alpha'$ remains in $W$ and $w - w'$ in $W^\perp$, and therefore each must be equal to zero independently. Our decomposition is therefore unique.