Divergence as transpose of gradient?

Question

In his online lectures on Computational Science, Prof. Gilbert Strang often interprets divergence as the "transpose" of the gradient, for example [here](http://youtu.be/gYME3EbIqV4?t=31m50s) (at 32:30), however he does not explain the reason. How is it that the divergence can be interpreted as the transpose of the gradient?

Answer 1

This is simply a result of vector/matrix multiplication. If you have a vector field, the dot product of the del operator and the vector field (both the vector field and del operator are column vectors) gives you the divergence. However, you would get the same result by doing matrix/vector multiplication if you transposed the del operator and made it a 1X3 row vector instead of a column vector. That way the (1X3)*(3X1) gives you a 1x1 so it gives you the scalar and is the same thing as the dot product.

X.*Y = X_transpose*Y