Projection on the orthocomplement of a finite-dim Hilbert space

Question

Let $\mathcal{H}$ be an infinite-dimensional Hilbert space. Moreover, let $\mathcal{P}_2 \subset \mathcal{H}$ and $\mathcal{P}_3 \subset \mathcal{H}$ be a finite-dimensional and an infinite-dimensional closed sub-spaces of $\mathcal{H}$. Let's indicate the orthocomplement of $\mathcal{P}_2$ in $\mathcal{P}_2 + \mathcal{P}_3$ as: \begin{equation*} \mathcal{W} \triangleq (\mathcal{P}_2 + \mathcal{P}_3) \cap \mathcal{P}_2^\perp. \end{equation*}

Let's consider now $h \in \mathcal{H}$. How can I evaluate the orthogonal projection of $h \in \mathcal{H}$ onto $\mathcal{W}$ :

\begin{equation*} P_{\mathcal{W}}(h) = P_{(\mathcal{P}_2 + \mathcal{P}_3) \cap \mathcal{P}_2^\perp}(h) = ? \end{equation*}

Answer 1

Generally, if $A, B \subseteq \mathcal{H}$ are two closed subspaces, then $P_{A \cap B} = \lim_{n \rightarrow \infty} (P_AP_BP_A)^n$. In our case, let $A = \mathcal{P}_2^\perp$ and $B = \mathcal{P}_3^\perp$. Then $P_A = 1 - P_{\mathcal{P}_2}$ and $P_B = 1 - P_{\mathcal{P}_3}$, so,

$$P_{A \cap B} = \lim_{n \rightarrow \infty} [(1 - P_{\mathcal{P}_2})(1 - P_{\mathcal{P}_3})(1 - P_{\mathcal{P}_2})]^n = \lim_{n \rightarrow \infty} (1 - P_{\mathcal{P}_2} - P_{\mathcal{P}_3} + P_{\mathcal{P}_2}P_{\mathcal{P}_3} + P_{\mathcal{P}_3}P_{\mathcal{P}_2} - P_{\mathcal{P}_2}P_{\mathcal{P}_3}P_{\mathcal{P}_2})^n$$

Now, we also have,

$$\mathcal{W} = (\mathcal{P}_2 + \mathcal{P}_3) \cap \mathcal{P}_2^\perp = (\mathcal{P}_2^\perp \cap \mathcal{P}_3^\perp)^\perp \cap \mathcal{P}_2^\perp = (A \cap B)^\perp \cap A$$

Since $A \cap B \subseteq A$, we have,

$$P_\mathcal{W} = P_A - P_{A \cap B} = 1 - P_{\mathcal{P}_2} - \lim_{n \rightarrow \infty} (1 - P_{\mathcal{P}_2} - P_{\mathcal{P}_3} + P_{\mathcal{P}_2}P_{\mathcal{P}_3} + P_{\mathcal{P}_3}P_{\mathcal{P}_2} - P_{\mathcal{P}_2}P_{\mathcal{P}_3}P_{\mathcal{P}_2})^n$$

The limit, I should point out, is usually in the strong operator topology. Though in this case because $\mathcal{P}_2$ is finite-dimensional, the convergence happens in norm as well. Regardless, the following is always true: for any $h \in \mathcal{H}$,

$$P_\mathcal{W}(h) = h - P_{\mathcal{P}_2}(h) - \lim_{n \rightarrow \infty} [(1 - P_{\mathcal{P}_2} - P_{\mathcal{P}_3} + P_{\mathcal{P}_2}P_{\mathcal{P}_3} + P_{\mathcal{P}_3}P_{\mathcal{P}_2} - P_{\mathcal{P}_2}P_{\mathcal{P}_3}P_{\mathcal{P}_2})^n](h)$$

Answer 2

Thanks for your answer ! Building upon some results that I found in a book, I derived another expression that I would like to verify with you to check the consistency with your answer (since I'm not sure that my expression is correct).

We can note that: \begin{equation*} \mathcal{W}\triangleq (\mathcal{P}_2 + \mathcal{P}_3) \cap \mathcal{P}_2^\perp = P_{\mathcal{P}_2^\perp}(\mathcal{P}_3) = \mathcal{P}_3 - P_{\mathcal{P}_2}(\mathcal{P}_3), \end{equation*} that is the set of elements of $\mathcal{P}_2^\perp$ that can be written as the difference of an element of $\mathcal{P}_3$ and its projection onto $\mathcal{P}_2$. Let's express the finite dimensional space $P_{\mathcal{P}_2}(\mathcal{P}_3) \subset \mathcal{H}$ as the linear span of $k$ elements $\mathcal{H}$, i.e. $P_{\mathcal{P}_2}(\mathcal{P}_3) = [b_1,\ldots, b_k]$. Then, \begin{equation*} P_{\mathcal{W}}(h) = P_{\mathcal{P}_3}(h) - \sum_{j=1}^{d} \lambda_j(h)(b_j-P_{\mathcal{P}_3}(b_j)), \end{equation*} where \begin{equation*} \mathrm{dim}[b_1-P_{\mathcal{P}_3}(b_1),\ldots, b_k-P_{\mathcal{P}_3}(b_k)] = d \leq k, \end{equation*} \begin{equation*} \boldsymbol{\lambda}(h) = \mathbf{B}^{-1}(<b_1-P_{\mathcal{P}_3}(b_1),h>,\ldots,<b_d-P_{\mathcal{P}_3}(b_d),h>)^T, \end{equation*} \begin{equation*} [\mathbf{B}]_{i,j} = <b_i-P_{\mathcal{P}_3}(b_i),b_j-P_{\mathcal{P}_3}(b_j)>,\quad i,j = 1,\ldots,d \end{equation*} and where $<\cdot,\cdot>$ is the inner product of $\mathcal{H}$.

Is this correct and consistent with your answer @David Gao?