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_!\$$

**Proof.**

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.
\$$\Box\$$

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).