Action of a projection on a sequence of pairwise orthogonal subspaces.

Question

Let us consider a sequence of finite-dimensional subspaces $\{H_n\}$ with $dim H_n=dim H_m$ and $H_n\perp H_m$ ($m\neq n$) in the Hilbert space $H$. Does there exist any projection $P$ satisfying in the following relations: $$0\neq \overline{P(H_n)}=\overline{P(H_m)}=\overline{P(\cup_{j\geq1} H_j)}$$ where $m\neq n$.

Answer 1

Have you tried anything?

Try the simple case $H = \mathbb{C}^n$ and $H_i = \{(0,0, \dots, 0, z, 0, \dots) : z \in \mathbb{C}\}$ is the span of the $i$th standard basis vector. (Try $n = 2$ and $n = 3$.)

Can you find a projection with the above properties in this case? If not, then why not?

Can you extend this to the general case?

Answer 2

My answer was wrong. The mistake that invalidates the whole argument happened between $(5) $ and $(6)$, where $$ \lambda_{jk}P_{jj}=\lambda_{jk}\lambda_{kj}P_{kk} $$ should have been $$ \lambda_{jk}P_{jj}=\lambda_{jk}\lambda_{kj}^2P_{kk}, $$ which gives no new information whatsoever.

Answer 3

If $P$ is assumed to be self-adjoint (and hence the term "orthogonal projection" is redundant.) then Martin Argerami has given an answer to this.

Either way, my response below is that such a $P$ does exist in any case.

For each $H_j$, pick a unit vector $e_j \in H_j$. Since the $H_j$ are orthogonal, this means that $\{e_j\}$ is an orthonormal set. Let $P$ be the orthogonal projection onto the vector $v = \sum_{j=1}^\infty \frac{1}{j^2}e_j$, i.e. for each $w \in H$, $P(w) = \langle w, v\rangle \frac{v}{\|v\|}$. Note that $v \neq 0$ since the $e_j$ are orthogonal so $\|v\|^2 = \sum_{j=1}^\infty \frac{1}{j^4} > 0$, by the (infinite) Pythagorean theorem.

We then see that $P(H_j) = \operatorname{span}(v)$ since $e_j \in H_j$ and $P(e_j) = \frac{1}{j^2}v$. It then follows that $P(H_j)$ is a one dimensional subspace and hence closed so $\overline{P(H_j)} = P(H_j) = \operatorname{im} P$, the image of $P$. Note that the image of $P$ is closed. We then see that $$P\left(\bigcup_{j \geq 1} H_j\right) = \bigcup_{j \geq 1} P(H_j) = \bigcup_{j \geq 1} \operatorname{im} P = \operatorname{im} P$$ and so $P$ satisfies all the required properties.