# Show that $P_Eh=h-\langle h,e_1\rangle e_1-\langle h,e_2\rangle e_2$

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.

• What does $P_Eh$ mean? – André Porto Nov 4 '18 at 18:15
• 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. – user593746 Nov 4 '18 at 19:18
• Thank you, André and Zvi - I have improved my question. – Frederik Nov 4 '18 at 19:23
• 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 $Q=P_E$. – user593746 Nov 4 '18 at 19:55
• @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. – user593746 Nov 4 '18 at 20:45

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 $$\{e_k:k\in\Bbb{N}_1\setminus A\}$$.