11
$\begingroup$

Let $\mathbf F : \mathbb R^p \to \mathbb R^s$ and $\phi : \mathbb R^p \to \mathbb R$ be differentiable functions. Let the function $\mathbf G$ be defined as follows: $$\mathbf G : \mathbb R^p \to \mathbb R^s \qquad \mathbf G(\mathbf y) = \phi(\mathbf y)\mathbf F(\mathbf y)$$

Furthermore, let $y_0$ be a point in $\mathbb R^p$. Then the Jacobian of $\mathbf G$ and of $\mathbf F$ at $y_0$, denoted respectively $D\mathbf G(y_0)$ and $D\mathbf F(y_0)$ are $s \times p$ matrices, whereas the Jacobian of $\phi$ at $y_0$, denoted $D\phi(y_0)$, is a row vector $p$ entries long and may thus be turned into a gradient: $$\nabla \phi(y_0) \doteq D\phi(y_0)^\top$$

Now the question is, how can I express $D\mathbf G(y_0)$ in terms of the other two Jacobians? I tried recklessly applying the product rule, $$D\mathbf g(y_0) \stackrel{?}{=} \phi(y_0) D\mathbf F(y_0) + D\phi(y_0) \mathbf F(y_0) $$ but the dimensions of the matrices do not match up correctly. What am I doing wrong?

$\endgroup$

2 Answers 2

11
$\begingroup$

There are two ways of looking at this that sort it out: we can use indices, or we can understand everything as linear maps and work it out explicitly.

  • In index notation, we have functions $G_i = \phi(y) F_i(y)$. All of these are scalars, so the usual product rule for scalar functions applies: $$ \frac{\partial G_i}{\partial y_j} = \phi \frac{\partial F_i}{\partial y_j} + \frac{\partial \phi}{\partial y_j} F_i. $$ Since $\phi$ is a scalar, to write down the matrix corresponding to this, we can reverse the order to put the $j$ terms on the right, i.e. $$ (DG)_{ij} = \phi (DF)_{ij} + F_i (\nabla\phi)_j = (\phi DF + F \otimes \nabla \phi)_{ij}, $$ $\otimes$ being the dyadic product $(A \otimes B)_{ij} = A_i B_j$.

Approaching the problem in a coordinate-free way, the derivative $DG(y)$ is a linear map $\mathbb{R}^p \to \mathbb{R}^s$ given uniquely by $$ G(y+h) = G(y) + DG(y)(h) + o(\lVert h \rVert). $$ Written this way, it doesn't matter how $h$ is incorporated providing that the expression ends up in $\mathbb{R}^s$. One can show that the product rule (or a clever use of the chain rule) in this formalism gives you $$ DG(y)(h) = [\phi(y)] DF(y)(h) + [D\phi(y)(h)] F(y), $$ where the terms in brackets are both scalars (and hence we can push them about to end up with all the $h$s on the right if we wish).

The important thing to understand is that since the derivative is a linear map, it has to have an argument fed into it somewhere. If it divides up into terms that are some form of products of derivatives and parts of the original function, the argument must be fed into the derivatives, not the other parts of the function (this is a good reason to think about functions $\mathbb{R}^n \supset U \to \mathbb{R}^m$, since then the argument of the function is restricted to a subset, but the tangent vectors that you feed into the derivative are not.

$\endgroup$
7
$\begingroup$

Let $F_1,...,F_s$ denote the components of $F$. Then $G_i = \phi F_i$, and so $$(DG)_{ij}=\frac{\partial G_i}{\partial x_j} = \frac{\partial \phi}{\partial x_j}F_i+\phi\frac{\partial F_i}{\partial x_j} = \frac{\partial \phi}{\partial x_j}F_i+\phi (DF)_{ij}$$ The $s\times p$ matrix $A$ such that $A_{ij}=\frac{\partial \phi}{\partial x_j}F_i$ is equal to $F\nabla\phi$, where here, $F$ is thought of as a $s$-dimensional column vector and $\nabla \phi$ is a $p$-dimensional row vector. Therefore, we have $$DG = D(\phi F) = F\nabla \phi+\phi DF$$ Keep in mind that on the RHS, the first product is matrix multiplication, and the second is scalar multiplication of the matrix $DF$ by the scalar function $\phi$.

$\endgroup$
2
  • $\begingroup$ Ah, so basically the order in which I take the matrix multiplication is reversed somehow (with respect to what I've written in the question) so that the gradient of the scalar function slides after the vector function. Is there a particular reason as to why that would be? $\endgroup$
    – giobrach
    Jun 20, 2017 at 1:45
  • $\begingroup$ That's just what needs to happen for the dimensions to work out. $\endgroup$
    – florence
    Jun 20, 2017 at 1:48

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .