Embeddings, projections, adjoints, and pseduoinverses

I just revised a post from a week ago about rotations. The revision makes explicit the process of embedding a 3D vector into the quaternions, then pulling it back out.

The 3D vector is embedded in the quaternions by making it the vector part of a quaternion with zero real part:

(p₁, p₂, p₃) → (0, p₁, p₂, p₃)

and the quaternion is returned to 3D by cutting off the real part:

(p₀, p₁, p₂, p₃) → (p₁, p₂, p₃).

To give names to the the process of moving to and from quaterions, we have an embedding E : ℝ³ to ℝ⁴ and a projection P from ℝ⁴ to ℝ³.

We can represent E as a 4 × 3 matrix

$E = \begin{bmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix}$

and P by a 3 × 4 matrix

$P = \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix}$

We’d like to say E and P are inverses. Surely P undoes what E does, so they’re inverses in that sense. But E cannot undo P because you lose information projecting from ℝ⁴ to ℝ³ and E cannot recover information that was lost.

The rest of this post will look at three generalizations of inverses and how E and P relate to each.

Left and right inverse

Neither matrix is invertible, but PE equals the identity matrix on ℝ³ , and so P is a left inverse of E and E is a right inverse of P.

$PE = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix}$

On the other hand, EP is not an identity matrix, and so E is not a left inverse of E, and neither is P a right inverse of E.

$EP = \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix}$

Adjoint

P is the transpose of E, which means it is the adjoint of E. Adjoints are another way to generalize the idea of an inverse. More on that here.

Pseudo-inverse

The Moore-Penrose pseudo-inverse acts a lot like an inverse, which is somewhat uncanny because all matrices have a pseudo-inverse, even rectangular matrices.

Pseudo-inverses are symmetric, i.e. if A⁺ is the pseudo-inverse of A, then A is the pseudo-inverse of A⁺.

Given an m by n matrix A, the Moore-Penrose pseudoinverse A⁺ is the unique n by m matrix satisfying four conditions:

A A⁺ A = A
A⁺ A A⁺ = A⁺
(A A⁺)* = A A⁺
(A⁺ A)* = A⁺ A

To show that A⁺ = P we have to establish

EPE = E
PEP = A⁺
(EP)* = EP
(PE)* = PE

We calculated EP and PE above, and both are real and symmetric, so properties 3 and 4 hold.

We can also compute

$EPE = \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} \begin{bmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} = E$

and

$PEP = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} =P$

showing that properties 1 and 2 hold as well.

Embeddings, Projections, and Inverses

Left and right inverse

Adjoint

Pseudo-inverse

Related posts

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