Jacobian transpose vs pseudoinverse

29 Jul 2016

Let’s derive both celebrated algorithms from first order principles to see how they differ from each other.

Jacobian transpose. Let $x_d$ be the desired state of the end-effector, $x(q)$ its current state corresponding to the joint configuration $q$, and let $J$ denote the Jacobian $\partial x / \partial q$. Naive minimization of the error in the Cartesian space

$$\begin{equation*} \mathcal{L}[q] = \frac{1}{2} (x_d - x(q))^T (x_d - x(q)) \end{equation*}$$

leads to the gradient descent algorithm

$$\begin{equation*} q_{\text{next}} = q + \alpha J^T (x_d - x(q)), \end{equation*}$$

where $\alpha$ is the step size.

Jacobian pseudoinverse. Consider a small change in the joint configuration $\Delta q$. It will lead to a small change in the state of the end-effector $\Delta x = J \Delta q$. Since we want the end-effector to move in the direction of $x_d$, let’s assume that $\Delta x = \alpha (x_d - x)$ with some positive $\alpha$. Question: “What is the smallest $\Delta q$ that accomplishes that?” It is well-known that the minimum norm solution to an underdetermined system of linear equations is given by the pseudoinverse. Thus, we arrive at the following gradient descent algorithm:

$$\begin{equation*} q_{\text{next}} = q + \alpha J^{+} (x_d - x(q)), \end{equation*}$$

where $J^{+} = J^T (JJ^T)^{-1}$ if $J$ has full row rank (which is ususally the case).

Comparison. Note that both algorithms have the form $\Delta q = J^T \Delta x$. The difference between them is in the choice of $\Delta x$. Jacobian transpose algorithm sets the step in the Cartesian space proportional to the distance to the goal $\Delta x = \alpha (x_d - x(q))$, whereas Jacobian pseudoinverse accounts for the curvature of the space in addition, $\Delta x = (JJ^T)^{-1} \alpha (x_d - x(q))$.