Hey John,

I have been playing around with these questions in a particular context.

There is still something I am trying to figure out.

Specifically, I am thinking of this puzzle by Allison Burgers in [the comments in lecture 6](https://forum.azimuthproject.org/discussion/comment/17149/#Comment_17149)

Take any [linear map](https://en.wikipedia.org/wiki/Linear_map) \\(A : V \to W\\). This can be lifted to a monotone map \\(A_!\\) on the [complete lattices of vector subspaces](https://en.wikipedia.org/wiki/Linear_subspace#Lattice_of_subspaces) of \\(V\\) and \\(W\\).

I am trying to show

$$ A^! = A^+_! $$

where \\(A^+\\) is the [Moore-Penrose pseudo-inverse](https://en.wikipedia.org/wiki/Moore%E2%80%93Penrose_inverse). This is used all the time in [linear regression](https://en.wikipedia.org/wiki/Linear_regression).


First, \\(A_!\\) has arbitrary meets and joins over lattices of subspaces. This is because \\(A\\) is a linear operator. This suffices **Puzzle 18** and **Puzzle 19**, and moreover suffices [*the adjoint functor theorem for posets*](https://forum.azimuthproject.org/discussion/2031/lecture-16-chapter-1-the-adjoint-functor-theorem-for-posets).

Hence \\(A_!\\) has a left adjoint and right adjoint.

I will focus on \\(A_!\\)'s right adjoint, which is \\(A^!\\).

To show this is the Moore-Penrose inverse, we need to show the following axioms are obeyed (as per [Wikipedia](https://en.wikipedia.org/wiki/Moore%E2%80%93Penrose_inverse#Definition)):

- \\(A\circ A^+ \circ A = A\\)
- \\(A^+\circ A\circ A^+ = A^+\\)
- \\((A\circ A^+)^* = A\circ A^+\\)
- \\((A^+\circ A)^* = A^+\circ A\\)

Here \\(T^\ast\\) denotes the [Hermitian transpose](https://en.wikipedia.org/wiki/Hermitian_matrix) of \\(T\\).

**I can show a *weakened* version of the first two axioms, but I have no idea how to prove the second two axioms** :(

*Proposition*: \\(A_! \circ A^! \circ A_! = A_!\\)


Let \\(X \sqsubseteq V\\) be a subspace of \\(V\\) and \\(Y \sqsubseteq W\\) be a subspace of \\(W\\).

First, observe that:

$$ A_!(X) \sqsubseteq A_!(X) \tag{$\star$} $$

Hence by right adjointness:

$$ X \sqsubseteq (A^! \circ A_!) (X) $$

Since \\(A_!\\) is monotone we have:

$$ A_! (X) \sqsubseteq (A_! \circ A^! \circ A_!) (X) \tag{1} $$

Next, using \\((\star)\\) and \\(A^!\\) monotone, we have

$$ (A^! \circ A_!)(X) \sqsubseteq (A^! \circ A_!) (X) $$

Hence by right adjointness:

$$ (A_! \circ A^! \circ A_!)(X) \sqsubseteq A_!(X) \tag{2}$$

(1) and (2) suffice to show, for all \\(X\\), \\((A_! \circ A^! \circ A_!) (X) = A_! (X) \\). This suffices the proposition.

A rather similar proof gives \\(A^! \circ A_! \circ A^! = A^!\\).

I suspect that in general for *any* two adjoint functors \\(F \vdash G\\), then \\(F \circ G \circ F = F\\) and \\(G \circ F \circ G = G\\).

But this is as far as I've got with this. I've got the following problems:

- I am struggling to show the identity [\\( A^{++} = A\\)](https://en.wikipedia.org/wiki/Moore%E2%80%93Penrose_inverse#Basic_properties) holds. In this context this means \\(A^! \dashv A_! \dashv A^!\\). This property *usually* only holds for inverses.

- I am also not quite sure how to recover a linear map \\(T\\) where \\(A^! = T_!\\).

- Finally, I have now idea how to show \\(A_! \circ A^!\\) and \\(A^! \circ A_!\\) are Hermite, as per the four axioms.

It is interesting to note that \\(A \circ A^+\\) is a [projection](https://en.wikipedia.org/wiki/Projection_(linear_algebra)), and hence a monad. In fact it is the [orthogonal projection of \\(A\\)](http://www.math.lsa.umich.edu/~speyer/417/OrthoProj.pdf).