Table of Contents

Topic 13. Adjoint Operators¶

Author: Christian Sorensen c.sorensen2002@gmail.com

Introduction¶

Adjoint Operators The adjoint of an operator (also known as Hilbert adjoint, not to be confused with the adjugate or adjoint matrix) is a linear operator that, “undoes” a corresponding operator. If operator A takes a vector from space S1 and maps it to space S2, the adjoint maps a vector from S2 back to S1.

The definition is given as <Ax,y>=<x,A*y> where A* is the adjoint of the operator A. While most engineering students are used to seeing matrix adjoints (transposes), any linear operator can have an adjoint. A given operator may not have an adjoint. Adjoints are important because their range and null spaces make up 2 of the 4 fundamental subspaces of the operator.

Explanation of the theory¶

Mathematical Definition:

$< Ax,y > = < x,A^{*}y > ;\ \ A^{*}\text{ is the adjoint of A}$

Properties of the adjoint (in a Hilbert space):

1. $\left( A_{1} + A_{2} \right)^{*} = A_{1}^{*} + A_{2}^{*}$

2. $\left( \text{aA} \right)^{*} = \overline{a}A^{*}\ (where\ \overline{a}\ is\ the\ complex\ conjugate\ of\ a)$

3. $\left( A_{2}\ A_{1} \right)^{*} = A_{1}^{*}A_{2}^{*}\ (The\ order\ has\ been\ reversed,\ similar\ to\ inversion)$

4. $\text{If }\exists\ A^{- 1},\ \left( A^{- 1} \right)^{*} = \left( A^{*} \right)^{- 1}$

An operator is said to be self adjoint (Hermetian) if the following is true:

$$A = A^{*}$$

While all operators may have an adjoint, from here on out the adjoint will be applied strictly to matrices, although some of the definitions may apply more generally. (Everything up to this point is applicable to all linear operators and their adjoints) For matrices the following are true:

$$\forall\ A\ \epsilon\ R^{\text{NxN}},\ A^{*} = A^{T} and\ \forall \ A \ \epsilon \ C^{\text{NxN}},A^{*} = A^{H}$$ $$\left( A^{T} \right)^{- 1} = \left( A^{- 1} \right)^{T}\text{ or }\left( A^{H} \right)^{- 1} = \left( A^{- 1} \right)^{H}$$

Simple Numerical Examples¶

Simple Numerical Example:

$$\text{For }A = \ \left( \begin{matrix}1 & -1 \\ -2 & 3 \end{matrix} \right), \text{ find } A^{*}$$ $$< Ax,y > = < x,A^{*}y >$$ $$y^{T}\text{Ax} = {{(A}^{*}y)}^{T}x$$ $$\left( y_{1}\ y_{2} \right)\left( \begin{matrix}1 & -1 \\ -2 & 3 \end{matrix} \right)x = {{(A}^{*}\left( \begin{matrix}y_{1} \\ y_{2} \end{matrix} \right))}^{T}x$$ $$\left( y_{1}\ y_{2} \right)\left( \begin{matrix}1 & -1 \\ -2 & 3 \end{matrix} \right) = {{(A}^{*}\left( \begin{matrix}y_{1} \\ y_{2} \end{matrix} \right))}^{T}$$ $$\left( \begin{matrix} y_{1} - 2y_{2}& - y_{1} + 3y_{2} \end{matrix}\right) = \left( \begin{matrix} a_{1}^{*} & a_{2}^{*} \\ a_{3}^{*} & a_{4}^{*} \end{matrix} \right) \left( \begin{matrix} y_{1} \\ y_{2} \end{matrix} \right)$$ $$\left( \begin{matrix} y_{1} - 2y_{2} & - {y}_{1} + 3y_{2} \end{matrix} \right) = \left( \begin{matrix} a_{1}^{*}y_{1} + a_{2}^{*}y_{2} \\ a_{3}^{*}y_{1} + a_{4}^{*}y_{2} \end{matrix} \right)^{T}$$ $$\left( \begin{matrix} y_{1} - 2y_{2} & - {y}_{1} + 3y_{2} \end{matrix} \right) = \left( \begin{matrix} a_{1}^{*}y_{1} + a_{2}^{*}y_{2} & a_{3}^{*}y_{1} + a_{4}^{*}y_{2} \end{matrix} \right)^{T}$$ $$\left( \begin{matrix} a_{1}^{*} & a_{3}^{*} \\ a_{2}^{*} & a_{4}^{*} \end{matrix} \right) = \left( \begin{matrix}1 & -2 \\ -1 & 3 \end{matrix} \right) = A^{T}$$

Matlab Implementation

The matlab implementation of the adjoint for a matrix is simple. Let A be a matrix, then the adjoint of that matrix can be calculated by A'

An Engineering Application¶

Engineering application:

Inverse kinematics of a robotic manipulator with redundant degrees of freedom.

In robotics the Jacobian is the matrix that describes the transformation between Joint and task space. For an n DOF robot it will look something like the following:

$$\begin{pmatrix} \dot{x} \\ \dot{y} \\ \dot {z} \\ \dot{\alpha} \\ \dot{\beta} \\ \dot{\gamma} \\ \end{pmatrix} = \begin{pmatrix} J_{11} & \cdots & J_{1n} \\ \vdots & \ddots & \vdots \\ J_{61} & \cdots & J_{6n} \\ \end{pmatrix}\begin{pmatrix} \dot{q_{1}} \\ \dot{q_{2}} \\ \vdots \\ \dot{q_{n}} \\ \end{pmatrix}$$

Where the LHS of the equation are the cartesian velocities of the end effector of the robot, and the $\dot{q}$ vector is the robots joint velocities.

In the case of a 7+ DOF robot, the robot is kinematically redundant, there are more joint states then cartesian states, meaning there are multiple joint configurations that give the same cartesian pose.

The inverse kinematics problem is as follows—given a desired cartesian pose determine the joint configuration of the robot. This problem is stated as:

$$\dot{\xi} = J\dot{q}$$

Due to the redundancy, there is no matrix inverse $J^{- 1}$ that can be used to solve the problem exactly. This is where the adjoint of the Jacobian comes into play.

We need to choose a specific $\dot{q}$ vector, and it is convenient to choose the $\dot{q}$ vector that with minimum norm, because the energy associated with a movement increases with the square of the $\dot{q}$ vector.

Expressing the constraint mathematically leads to the following solveable problem:

Choose a vector $\dot{q}$ s. t. $\dot{\xi} = J\dot{\text{q }}$and $\left\| \dot{\text{q}} \right\|_{2}$ is minimized.

Let $n \in N\left( J \right)$ and let $\dot{q_{0}}$ be a solution of $\dot{\xi} = J\dot{q}$. This means that $\dot{q_{0}} + n$ is also a solution of $\dot{\xi} = J\dot{q}$. Because we are operating inside of a Hilbert space, there is a unique vector that satisfies the equation with minimum norm.

That solution will have $n = 0$ or in otherwords will be orthogonal to $N\left( J \right)$.

Understanding the fundamental theorem of linear algebra (see the associated chapter in the jupyter notebook titled 4 fundamental subspaces) gives that $\left\lbrack N\left( J \right) \right\rbrack^{\bot} = R\left( J^{*} \right) = R(J^{T})$ .

This means the minimum-norm solution of our equation is

$$\dot{q} \in R\left( J^{T} \right) = J^{T}z\ $$

Where z is given as the solution of $\left( JJ^{T} \right)z = \dot{\xi}$

If we take the inverse of $\left( JJ^{T} \right)$, whose existance can be proven as long as the jacobian isn’t singular, we can solve for $\dot{q}$.

$$\dot{q} = J^{T}\ \left( JJ^{T} \right)^{- 1}\dot{\xi}$$

Challenge Problem¶

Add a homework assignment that might take 10 minutes to complete. Make sure you can work the problem yourself, but you do not need to submit a solution to the problem.