I think the reason that this matrix is symmetric is because the partial derivative operation is commutative, i.e. the order of performing the derivatives does not make a difference, so d2f/dx1dx2 is the same as d2f/dx2dx1 and so the corresponding elements when transposed remain the same. Supposedly there are certain conditions on the function for this to be true: https://en.wikipedia.org/wiki/Symmetry_of_second_derivatives.
sponge
Could I get an explanation on what it geometrically means to take a derivative with respect to one coordinate and then take the derivative of that with respect to another coordinate in this 3D situation. I understood the curvature argument, but still not really able to visualize this.
keenan
@sponge Not sure if this is the answer you're looking for, but:
I often think of the gradient of a function $\nabla f$ as a vector $\nabla f$ such that the inner product $\langle \nabla f, X \rangle$ tells me how quickly $f$ is changing in the $X$-direction.
I often think of the Hessian of a function $\nabla f$ as a matrix $\nabla^2 f$ such that $(\nabla^2 f) X$ tells me how much the gradient changes when I move in the $X$-direction.
frog
I believe the symmetry comes from Clairaut's? (don't remember his name) Theorem. I know that you can continue to extend this to notion of taking terms of the Taylor series into higher and higher orders (say, the third derivative or the cubic term), but these won't fit be able to be represented by matrices (even if the function is R^n -> R). We'd need higher dimensional representations for these transformations, no? How do we represent these?
I think the reason that this matrix is symmetric is because the partial derivative operation is commutative, i.e. the order of performing the derivatives does not make a difference, so d2f/dx1dx2 is the same as d2f/dx2dx1 and so the corresponding elements when transposed remain the same. Supposedly there are certain conditions on the function for this to be true: https://en.wikipedia.org/wiki/Symmetry_of_second_derivatives.
Could I get an explanation on what it geometrically means to take a derivative with respect to one coordinate and then take the derivative of that with respect to another coordinate in this 3D situation. I understood the curvature argument, but still not really able to visualize this.
@sponge Not sure if this is the answer you're looking for, but:
I believe the symmetry comes from Clairaut's? (don't remember his name) Theorem. I know that you can continue to extend this to notion of taking terms of the Taylor series into higher and higher orders (say, the third derivative or the cubic term), but these won't fit be able to be represented by matrices (even if the function is R^n -> R). We'd need higher dimensional representations for these transformations, no? How do we represent these?