Could you explain the logic going from the expression at the top of the slide to the equations on the bottom? Not sure I understand why the top expression is optimized when the bottom equations are true.
motoole2
With respect to why the pseudoinverse is a closed-form solution to the least squares problem, please see this slide from earlier in the semester.
Next, let me redefine $\nabla I \frac{\partial W}{\partial p}$ a little. Instead of being a $1\times 6$ vector for a specific $x$ value, let me reinterpret this as a $N\times 6$ matrix, where every row corresponds to a different $x$. This might make the math a little more clear.
We then have a matrix $A = \nabla I \frac{\partial W}{\partial p}$, a vector $x = \Delta p$, and a vector $b = T(x) - I(W(x;p))$. Then, using our closed-form solution to the least squares problem, $\Delta p = (A^T A)^{-1} A^T b$ where $A^T A = H = (\nabla I \frac{\partial W}{\partial p})^T (\nabla I \frac{\partial W}{\partial p})$, and $A^T b = (\nabla I \frac{\partial W}{\partial p})^T (T(x) - I(W(x;p))$.
Could you explain the logic going from the expression at the top of the slide to the equations on the bottom? Not sure I understand why the top expression is optimized when the bottom equations are true.
With respect to why the pseudoinverse is a closed-form solution to the least squares problem, please see this slide from earlier in the semester.
Next, let me redefine $\nabla I \frac{\partial W}{\partial p}$ a little. Instead of being a $1\times 6$ vector for a specific $x$ value, let me reinterpret this as a $N\times 6$ matrix, where every row corresponds to a different $x$. This might make the math a little more clear.
We then have a matrix $A = \nabla I \frac{\partial W}{\partial p}$, a vector $x = \Delta p$, and a vector $b = T(x) - I(W(x;p))$. Then, using our closed-form solution to the least squares problem, $\Delta p = (A^T A)^{-1} A^T b$ where $A^T A = H = (\nabla I \frac{\partial W}{\partial p})^T (\nabla I \frac{\partial W}{\partial p})$, and $A^T b = (\nabla I \frac{\partial W}{\partial p})^T (T(x) - I(W(x;p))$.