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

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

leads to the gradient descent algorithm

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

where $\alpha$ is the step size.

Jacobian pseudoinverse. Consider a small change in the joint configuration $\mathrm{\Delta }q$. It will lead to a small change in the state of the end-effector $\mathrm{\Delta }x=J\mathrm{\Delta }q$. Since we want the end-effector to move in the direction of $x_d$, let’s assume that $\mathrm{\Delta }x=\alpha (x_d-x)$ with some positive $\alpha$. Question: “What is the smallest $\mathrm{\Delta }q$ 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:

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

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

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