# Linearly independent subspaces

Let $E_1, \ldots, E_n$ be $n$ closed and linearly independent (but not necessarily orthogonal) subspaces of an Hilbert space $\cal{H}$, that is $(E_1 + \ldots + E_{i-1} + E_{i+1} + \ldots + E_n) \cap \, E_i = \{0\}$ for $i = 1, \ldots, n$, where, for example, $E_1 + E_2 = \{\Psi_1 + \Psi_2: \Psi_1 \in E_1, \Psi_2 \in E_2\}$. I think that the following two properties are true:

1) $(E_1 + \cdots + E_n)$ is a closed subspace;

2) if $(E_1 + \cdots + E_n)= \cal{H}$, then the operator $\hat E_1 + \ldots + \hat E_n$ is a bounded, iniective and surjective operator on $\cal{H}$, where $\hat E_i$ is the orthogonal projector on $E_i$.

I think I have proved these propositions, but the proofs, if correct, are rather long and complex. Maybe somebody may suggest me a simple proof or a book in which these properties are proved.

Update for user 108903. Thank you for the hint. Actually my proof of surjectivity is based on a conjecture that I have not proved. Let us consider only two closed subspaces $E_1, E_2$ of $\cal{H}$, such that $E_1 + E_2 = \cal{H}$ and $E_1 \cap E_2 = \{0\}$. A subset $M$ of $\cal{H}$ is said to be a linear manifold (hereafter manifold) if there exist a vector $\Psi$ and a subspace $E$ such that $M = \Psi + E$. The subspace $E$ is referred to as the generating subspace of $M$, and it is unique.

Conjecture. Let $M_1$ and $M_2$ be two manifolds such that $M_1 \cap M_2 = \emptyset$ (the empty set). Then the set $\{||\Psi_1 - \Psi_2|| \in \mathbb{R}: \Psi_1 \in M_1, \Psi_2 \in M_2\}$ admits a minimum $>0$, and the vector $\Psi_1 - \Psi_2$ corresponding to the minimum is orthogonal to both the generating subspaces of $M_1$ and $M_2$.

Proposition 1. Let $M_1$ and $M_2$ be two linear manifolds generated by $E^\perp_1$ and $E^\perp_2$, respectively. Then $M_1 \cap M_2 \neq \emptyset$.

Proof of proposition 1. Suppose $M_1 \cap M_2 = \emptyset$. Then, for the above proposition, there exists $\Psi \neq 0$ which is orthogonal to $E_1^\perp$ and $E_2^\perp$, that is $\Psi \in E_1 \cap E_2$, which contradicts the hypothesis. So $M_1 \cap M_2 \neq \emptyset$.

Proof of surjectivity. Any $\Psi \in \cal{H}$ is of the form $\Psi_1 + \Psi_2$, where $\Psi_i \in E_i$. Define $M_i = \Psi_i + E_i^\perp$, and let $\Phi \in M_1 \cap M_2$. Then $(\hat E_1 + \hat E_2) \Phi = \Psi_1 + \Psi_2$.

I am trying now to prove the conjecture, or to find further conditions under which it is valid. Maybe you can let me known your proof.

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The first is false for $n=2$; see Example 2.2 in www-personal.umich.edu/~rlsmith/… . But I wouldn't be surprised if the second is true. – user108903 Jan 19 '13 at 0:09
I see a simple proof that $\hat E_1+\dots+\hat E_n$ is bounded and injective with dense range, but I don't see why it should be surjective. I'd be curious to see your proof of surjectivity! – user108903 Jan 19 '13 at 9:10
@user108903: I made an update. – BGA Jan 19 '13 at 12:48
Thanks! I'm not sure about your conjecture; I don't see a counterexample at the moment, anyway. I now think my argument does give a positive answer to (2), so I've typed it up below. – user108903 Jan 19 '13 at 13:57

As pointed out above, (1) isn't true in general. For two subspaces, their sum being closed is equivalent to the angle between them being non-zero.

For (2), note first that $\hat E=\sum_{i=1}^n \hat E_i$ is a sum of finitely many bounded operators, so $\hat E$ is certainly bounded. We'll write $\hat E$ as a composition $\hat E=A\circ B$ of two bijective linear maps where $H\stackrel{B}{\to} E\stackrel{A}{\to} H$ and $E$ is the vector space $E=E_1\times\dots\times E_n$.

Let $A:E\to H$, $(e_1,\dots,e_n)\mapsto \sum_{i=1}^n e_i$. Then $A$ is surjective since $H=\sum_{j=1}^n E_i$ and it is injective by the independence of $E_1,\dots,E_n$.

Now $B:H\to E$, $h\mapsto (\hat E_1 h,\dots,\hat E_n h)$ has kernel consisting of the vectors orthogonal to each $E_i$, hence orthogonal to their sum $H$, so $\ker B=\{0\}$ and $B$ is injective.

So $\hat E=A\circ B$ is the composition of two injective maps, so is injective.

Give $E$ the inner product $(e,f)=\sum_{i=1}^n (e_i,f_i)_H$. Then $E$ is a Hilbert space (the Hilbert space direct sum of $E_1,\dots,E_n$) and it's easy to check that $B=A^*$. Since $H=\mbox{ran}(A)=\mbox{ran}(B^*)$ is closed, $\mbox{ran}(B)$ is also closed by the penultimate corollary on this page. So $\mbox{ran}(B)=\overline{\mbox{ran}(B)}=\ker(A)^\perp=E$ and $B$ is surjective.

Hence $\hat E=A\circ B$ is the composition of two bijections, so is a bijection.

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Very good, thanks! (Maybe in the fourth line, in the chain H->E->H, the operators A and B have to be interchanged?) – BGA Jan 19 '13 at 16:26
Thanks, I've swapped them. – user108903 Jan 19 '13 at 17:08