Compatible Hilbert space subspaces - need help understanding a statement made in a book

Question

A book I'm reading has the following in a section on lattices formed by subspaces of a Hilbert space :

Two subspaces $M$ and $N$ are compatible if there exist three mutually disjoint subspaces $M_1$, $N_1$, and $K$, such that

$M=M_1 \vee K $ and $N = N_1 \vee K$

If follows then that $K = M \wedge N $ and $M_1 = M \wedge K^{\bot}$, $N_1 = N \wedge K^{\bot}$

(Here two subspaces $M$ and $N$ are defined as 'disjoint' if $M \subset N^{\bot}$, and $M^{\bot}$ stands for the set of vectors orthogonal to all of $M$)

I'm not sure how the last line follows. I can see that $K = M \wedge N$, but then if so, $K^{\bot}$ will be the set of vectors orthogonal to $M$ (and $N$ as well), and so $M \wedge K^{\bot}$ would be $0$. Is there a printing error?

Answer 1

No, the book doesn't have any typo.

First let me give you an example to have an imagination of what is going on. The simplest example to imagine is the following. Let $M_1$ be the $x$-axis in $\mathbb{R}^3$, $N_1$ be the $y$-axis, $K$ be the $z$-axis. Obviously $M_1$, $N_1$ and $K$ are mutually disjoint subspaces of $\mathbb{R}^3$, now you can see that $M$ is the $x$o$z$ plane and $N$ is the $y$o$z$ plane. And all those three relations $K=M\cap N$, $M_1=M\cap K^{\perp}$ and $N_1=N\cap K^{\perp}$.

But now for proving the two last that you have doubt about them. I will do for the $M_1$ and one for the $N_1$ will follow with same story.

First consider that because $M_1$ and $K$ are mutually disjoint, and also as $M=M_1+K$; $$\left.\begin{array}{l} M_1\subseteq M\\ M_1\subseteq K^{\perp}\end{array}\right\}\Longrightarrow M^{\perp}\subseteq M\cap K^{\perp}$$

Now for the opposite inclusion, chose and arbitrary element in $M\cap K^{\perp}$, say $v$. Since $v\in M=M_1+K$ we can right it as $v=a+b$ where $a\in M_1$ and $b\in K$. Since $v\in K^{\perp}$ we have for every vector in $K$, say $k$, the inner product of $v$ and $k$ should be zero. $$0=\langle a+b,k\rangle=\langle a,k\rangle+\langle b,k\rangle=0+\langle b,k\rangle$$ Inner product of $a$ and $k$ is zero because $a$ is in $M_1$ and $M_1$ is subset of orthogonal completion of $K$. And no element of a space except the trivial vector $0$ is normal to that space so for the equation $\langle b,k\rangle=0$ be hold, we have to have $b=0$ and this shows $v=a\in M_1$. So we showed $M\cap K^{\perp}\subseteq M_1$.

Answer 2

No, there is not printing error,

First of all you should consider any subspace of a Hilbert space like "$H$" is a Hilbert space and any Hilbert space has a unique decomposition like: $H=M+M^{\perp}$.

Then according to uniqueness of decomposition, $M_1$ and $K$ are decomposition of Hilbert space $M$.

And $N_1$ and $K$ are decomposition of Hilbert space $N$ too.

Then we have obviously $M_1=M\cap K^{\perp}$

And $M_1\cap K^{\perp}=0$.

For $N$ is similar.

Answer 3

I can see that $K = M \wedge N$, but then if so, $K^{\bot}$ will be the set of vectors orthogonal to $M$

No. From $K = M \wedge N$ follows only that $K^{\bot}$ is orthogonal to those vectors of $M$ that belong also to $N$. So $K^{\bot}$ is not necesserily orthogonal to all vectors of $M$. See AmirHosein's example, where $M$ is the $xz$ plane, $N$ is the $yz$ plane, so $K = M \wedge N$ is the $z$ axis and $K^{\bot}$ is the $xy$ plane.