7
$\begingroup$

Given a real rectangular matrix $X$, and two scalar-valued matrix functions, $f(X)$ and $g(X)$, does the product rule for differentiation of a product of scalar valued functions, hold when differentiating the product, $f(X)g(X)$ w.r.t $X$ ? If not, what would be the corresponding product rule? Let's assume that the product $f(X)g(X)$ gives a real valued scalar, and is well-defined in terms of the dimensions.

Note: $f(.)$ and $g(.)$ can be the matrix trace function for example.

$\endgroup$
9
$\begingroup$

Yes, the standard product rule applies. The gradient of the product is $$f(X)\nabla_X g(X)+g(X)\nabla_X f(X).$$ The dimensions of the gradients, of course, are the same as those of $X$ itself.

You might find The Matrix Cookbook useful here.

$\endgroup$
0
19
$\begingroup$

The product rule holds in very great generality. Let $X,Y,Z,W$ be Banach spaces with open subset $U \subset X$, and suppose $f: U \rightarrow Y$ and $g: U \rightarrow Z$ are Frechet differentiable. If $B(\cdot, \cdot): Y \times Z \rightarrow W$ is a continuous bilinear map, then for any $\xi \in X$,

$$ \frac{d}{dx}[ B(f(x), g(x))](\xi) = B(f'(x)\xi, g(x)) + B(f(x), g'(x)\xi)$$

where all the derivatives in question are Frechet derivatives. To apply to your case, we take $U = X = \mathbb{R}^{n \times n}$, $Y =Z = W = \mathbb{R}$, and $B(y,z) = yz$.

$\endgroup$
2
  • $\begingroup$ This notation is a little imprecise, since we write $B(f'(x), g(x))$ where $f'(x)$ is not an element of $Y$ but rather of $L(X,Y)$. So I guess we are meant to understand $B(f'(x), g(x))$ as the linear operator $\xi \mapsto B(f'(x) \xi, g(x))$ in $L(X,W)$. $\endgroup$ – Nate Eldredge Apr 25 '20 at 15:06
  • $\begingroup$ @NateEldredge Thank you, you are absolutely correct. I'll make a correction! $\endgroup$ – Christopher A. Wong Apr 30 '20 at 2:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.