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I have that $E=\{e_1,e_2\}^\perp$, and that $(e_n)_{n=1}^\infty$ is an orthonormal basis for the Hilbert space. Furthermore, I have that $P_Eh$ is called orthogonal projection of $h$ onto $E$.

I have to show that: $$P_Eh=h-\langle h,e_1\rangle e_1-\langle h,e_2\rangle e_2\quad\forall h\in H,$$ where $H$ is the Hilbert space.

I think I can use that $P_Eg=\sum_{n=1}^N\langle g,e_n\rangle e_n$,but I'm quite stuck, so any help would be greatly appreciated.

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  • $\begingroup$ What does $P_Eh$ mean? $\endgroup$ – André Porto Nov 4 '18 at 18:15
  • $\begingroup$ What do you mean by "the basis"? An orthonormal basis, perhaps? There is not a unique basis for any Hilbert space, except the trivial one. $\endgroup$ – user593746 Nov 4 '18 at 19:18
  • $\begingroup$ Thank you, André and Zvi - I have improved my question. $\endgroup$ – Frederik Nov 4 '18 at 19:23
  • $\begingroup$ I would start by defining $Q(h)$ to be $h-\langle h,e_1\rangle e_1-\langle h,e_2\rangle e_2$. Then, I would prove that (1) $Q$ is a Hermitian operator, (2) $Q^2=Q$, and (3) the range of $Q$ is precisely $E$. This would show that $Q$ is the orthogonal projection onto $E$, making $Q=P_E$. $\endgroup$ – user593746 Nov 4 '18 at 19:55
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    $\begingroup$ @DionelJaime I think the question has sufficient information. He did say "I have that $P_Eh$ is called orthogonal projection of $h$ onto $E$". In a Hilbert space, there exists exactly one orthogonal projection onto a closed subspace (if my memory doesn't fail me), so $P_E$ is well-defined here. $\endgroup$ – user593746 Nov 4 '18 at 20:45
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In my notation, $\langle\bullet,\bullet\rangle$ is linear in the first variable and anti-linear in the second variable. Let $Q:H\to H$ be defined by $$ Qh=h-\langle h,e_1\rangle e_1-\langle h,e_2\rangle e_2$$ for all $h\in H$. Note that $Q$ is hermitian because

\begin{align}\langle Qu,v\rangle &=\big\langle u-\langle u,e_1\rangle e_1-\langle u,e_2\rangle e_2,v\big\rangle\\&=\langle u,v\rangle -\langle u,e_1\rangle \langle e_1,v\rangle -\langle u,e_2\rangle \langle e_2,v\rangle\\&=\langle u,v\rangle -\langle u,e_1\rangle \overline{\langle v,e_1\rangle} -\langle u,e_2\rangle \overline{\langle v,e_2\rangle}\\&=\big\langle u,v-\langle v,e_1\rangle e_1-\langle v,e_2\rangle e_2=\langle u,Qv\rangle.\end{align}

Next, we prove that $Q$ is a projection. That is, $Q^2=Q$. To show this, let $h\in H$ be arbitrary. We have

\begin{align} Q^2h&=Q(Qh)=Q\big(h-\langle h,e_1\rangle e_1-\langle h,e_2\rangle e_2\big)\\&=Qh-\langle h,e_1\rangle Qe_1-\langle h,e_2\rangle Qe_2\\&=Qh-\langle h,e_1\rangle \big(e_1-\langle e_1,e_1\rangle e_1-\langle e_1,e_2\rangle e_2\big) -\langle h,e_2\rangle \big(e_2-\langle e_2,e_1\rangle e_1-\langle e_2,e_2\rangle e_2\big) \\&=Qh-\langle h,e_1\rangle (e_1-e_1-0)-\langle h,e_2\rangle (e_2-0-e_2\rangle\\&=Qh-\langle h,e_1\rangle \cdot 0-\langle h,e_2\rangle\cdot 0= Qh.\end{align}

Now, observe that $Qe_k=e_k$ for $k=3,4,5,\ldots$ but $Qe_1=Qe_2=0$. Therefore, for any $h\in H$, $Qh\perp e_1$ and $Qh\perp e_2$. This is because

$$\langle Qh,e_k\rangle =\langle h,Qe_k\rangle =\langle h,0\rangle =0$$

for $k=1,2$, so $Qh\in \{e_1,e_2\}^\perp =E$. This proves that $\operatorname{im}Q\subseteq E$. The final task to show that for any $h\in E$, $Qh=h$, and this establishes the claim that $\operatorname{im} Q=E$. That is, $Q=P_E$. To see this, we suppose that $h\in E$. Thus, $h\perp e_1$ and $h\perp e_2$, so $\langle h,e_1\rangle=\langle h,e_2\rangle=0$. That is,

$$Qh=h-\langle h,e_1\rangle e_1-\langle h,e_2\rangle e_2=h-0e_1-0e_2=h.$$

I think it is generally true that if $\{e_1,e_2,e_3,\ldots\}$ is an orthonormal basis of a separable Hilbert space $H$ and $P$ is the orthogonal projection onto a closure of the subspace spanned by $\{e_k:k\in A\}$, where $A$ is a subset of $\Bbb N_1$, then $$Ph=\sum_{k\in A}\langle h,e_k\rangle e_k=h-\sum_{k\in \Bbb N_1\setminus A} \langle h,e_k\rangle e_k$$ for all $h\in H$. In other words, $P$ is the projection onto the orthogonal complement of $\{e_k:k\in\Bbb{N}_1\setminus A\}$.

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