Let $\mathcal{H}$ be a finite-dimensional Hilbert space and $L$ and $L^{\perp}$ be a subspace and its orthogonal complement such that $$L\oplus L^{\perp}=\mathcal{H}$$.
Show that any vector $\boldsymbol {v}\in \mathcal{H}$ of norm $|\boldsymbol{v}|=1$ has a unique decomposition $$ \boldsymbol{v} = a \boldsymbol{u}+b\boldsymbol{w} $$ for $\boldsymbol{u} \in L$ and $\boldsymbol{w} \in L^{\perp}$ and $|\boldsymbol{u}|=|\boldsymbol{w}|=1$ . In particular, show that $a,b\geq 0 $ and real!
The proof is as follows.
Existence: Since $\mathcal{H} = L\oplus L^\perp$, $v=u'+w'$ for unique $u'\in L$ and $w'\in L^\perp$. Then if $u'=0$ or $w'=0$, we have $v=w'$ or $v=u'$, so $v=v+0$ or $v=0+v$ is a decomposition. Otherwise, we can write $$v= \|u'\| \frac{u'}{\|u'\|} + \|w'\|\frac{w'}{\|w'\|},$$ and this is a valid decomposition.
Uniqueness: Observe that we must have $au = \pi_L(v)$, $bw=\pi_{L^\perp}(v)$, and $a=\|au\|$, $b=\|bw\|$, so $a,b,u,w$ are uniquely determined (as long as $a,b\ne 0$, otherwise the corresponding unit vector is undetermined).