# Unique decomposition of a vector in finite-dim Hilbert space

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!

• This is false. Let $\mathcal{H}=\Bbb{C}^2$ with its usual inner product. Let $L$ and $L^\perp$ be the $x$ and $y$ axes. Then $v=(1,1)=(1,0)+(0,1)=2(1/2,0)+2(0,1/2)$. – jgon May 12 at 20:01
• Right, sorry, forgot to mention the normalization conditions on the vectors. All three vectors should have unit norm. – Marsl May 12 at 20:06
• That mostly fixes things, but you'll also want $a,b \ge 0$ rather than $a,b > 0$, otherwise you can't express vectors lying completely in $L$ or $L^\perp$. – jgon May 12 at 20:14

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).
• @Marsl, yes there was a typo. That should have been $\|w'\|$ in the denominator. – jgon May 12 at 23:03