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!

  • $\begingroup$ 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 $v=(1,1)=(1,0)+(0,1)=2(1/2,0)+2(0,1/2)$. $\endgroup$ – jgon May 12 at 20:01
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    $\begingroup$ Right, sorry, forgot to mention the normalization conditions on the vectors. All three vectors should have unit norm. $\endgroup$ – Marsl May 12 at 20:06
  • $\begingroup$ 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 $L^\perp$. $\endgroup$ – jgon May 12 at 20:14

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).

  • $\begingroup$ is there a typo in the displayed line? I do not see how this is in agreement with the unit normalization on all three vectors. $\endgroup$ – Marsl May 12 at 22:06
  • $\begingroup$ @Marsl, yes there was a typo. That should have been $\|w'\|$ in the denominator. $\endgroup$ – jgon May 12 at 23:03

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