Decomposition of complex Hilbert space using infinite family of orthogonal subspaces Let $H$ be a Hilbert space over the complex numbers. Let $(X_n)_{n=1}^{\infty}$ be a family of closed subspaces such that: (i) $X_n \neq \{0\}$ and (ii) $X_n \perp X_m$ if $n \neq m$.
Set $X_0 := \cap_{n=1}^{\infty} X_n^\bot$ and let $P_n$ denote the orthogonal projection onto $X_n$.

a) Prove that each $x \in H$ admits a unique decomposition: $x = x_0 + \sum_{n=1}^{\infty}x_n$ where $x_i \in X_i$ for all $i\geq0$ and $x_n = P_nx$ for each $n\geq1$.

I am familiar with the fact that if $A$ is a closed subspace of a Hilbert space $H$ then there is a unique decomposition of each element $x \in H$ such that $x = a + \hat{a}$ where $a \in A$ and $\hat{a} \in A^\bot$.
I am also familiar with the fact that for two subspaces $L_1, L_2$ of $H$, that $(L_1+L_2)^\bot = L_1^\bot \cap L_2^\bot$.
It's not obvious to me, however, how this can all be put together and extended to the case where we're taking an infinite number of subspaces of a Hilbert space.

b) Given a bounded sequence $(\lambda)_{n=1}^{\infty}$ of non-zero elements in $\mathbb{C}$, show that one defines an element $A \in B(H)$ by setting: $Ax = \sum_{n=1}^{\infty} \lambda_n P_nx$ for each $x \in H$. Compute $|| A ||, \text{Ker}(A), A^*.$ Show that $A$ is normal. When is $A$ invertible?

For this part, I'm really not even sure where to start. Intuitively, it makes sense that one could define a bounded linear operator $Ax = \sum_{n=1}^\infty \lambda_n x_n$, but I'm not so sure about the rest.
Thanks in advance.
 A: First (a). Define the subspace $$K=\bigoplus_{n}X_n=\{\sum_{n=1}^\infty h_n: h_n\in X_n \text{ for all }n\geq1, \sum_{n=1}^\infty\|h_n\|^2<\infty\} $$
where the $\bigoplus$ notation refers to the Hilbert space direct sum.
Verify that $K$ is a closed subspace of $H$. Compute now $K^\bot$. If $x\in \bigcap_nX_n^\bot$ and $\sum_nh_n\in K$, then $\langle x,\sum_nh_n\rangle=\sum_n\langle x,h_n\rangle=0$, so $x\in K^\bot$. On the other hand, it is easy to see that $X_n\subset K$ for all $n\geq1$, so $K^\bot\subset X_n^\bot$ and this is true for all $n$, so $K^\bot\subset\bigcap_nX_n^\bot$. This proves that $K^\bot=\bigcap_nX_n^\bot:=X_0$.
By the theorem you are familiar with, if $x\in H$ then we have that $x$ is written uniquely as $x=x_0+y$, where $x_0\in X_0$ and $y\in K$ and $y=P(x)$, where $P$ is the orthogonal projection onto $K$.
Claim: $P(x)=\sum_{n=1}^\infty P_n(x)$. First note that, since the $P_n$ are orthogonal projections we have that the partial sums $\sum_{n=1}^NP_n$ are projections for all $N\geq1$ (these are self-adjoint idempotent operators; if this confuses you, take the space $K_N=\{\sum_{n=1}^Nh_n: h_n\in X_n\text{ for all }n=1,\dots,N\}$ and verify that the projection onto $K_N$ is equal to $\sum_{n=1}^NP_n$).
So, since projections are contractions, we have that $\|\sum_{n=1}^NP_n(x)\|^2\leq\|x\|^2$, i.e. (by Parseval's identity) $\sum_{n=1}^N\|P_n(x)\|^2\leq\|x\|^2$ and this shows that the series $\sum_{n=1}^\infty\|P_n(x)\|^2$ converges. Thus we have that $\sum_{n=1}^\infty P_n(x)\in K$.
If we show that $x-\sum_{n=1}^\infty P_n(x)\in X_0$ we are done, by the uniqueness of the decomposition. So take any element of $K$, say $\sum_{n=1}^\infty h_n$, where $h_n\in X_n$ and $\sum_{n}\|h_n\|^2<\infty$. Then
$$\langle x-\sum_nP_n(x),\sum_mh_m\rangle=\langle x,\sum_mh_m\rangle-\langle \sum_nP_n(x),\sum_mh_m\rangle=\sum_m\langle x,h_m\rangle-\sum_{n,m}\langle P_n(x),h_m\rangle=$$
$$=\sum_m\langle P_m(x),h_m\rangle-\sum_{n,m}\langle P_n(x),h_m\rangle=\sum_m\langle P_m(x),h_m\rangle-\sum_n\langle P_n(x),h_n\rangle=0$$
and we are done.
I am only hinting (b):
Show that $A$ is well-defined by (a). Observe that $\|Ax\|^2=\sum_n|\lambda_n|^2\|P_n(x)\|^2\leq\sup_n|\lambda_n|^2\cdot\|x\|^2$. Find the proper unit vectors $x$ to obtain $\|Ax\|^2\geq|\lambda_n|$ and conclude that $\|A\|=\sup_n|\lambda_n|$. Verify that $A^*x=\sum_n\bar{\lambda_n}P_n(x)$. The kernel should be easily computed at that point.
