Jtdylktuy

I googled around and searched inside the forum but I'm still confused about a problem.

I have 2 matrix functions $f,g : mathbb{R}^{n times n} times mathbb{R}^{a times b} rightarrow mathbb{R}^{n times n}$. Starting from this, I have the following expression:

$$ t(Q, X, Y) = text{tr}(f(g(Q, X),Y))$$

where $text{tr}$ is the trace operator and $X, Y in mathbb{R}^{a times b}$ and $Q in mathbb{R}^{n times n}$.

How do I evaluate $frac{partial t(Q,X, Y)}{partial X}$ and $frac{partial t(Q,X, Y)}{partial Y}$?

I mean, I would like to know how to correctly apply the chain rule.

* Addition *

I will try to give more information about my problem.
Suppose that $a = b = n$ and that $f(A,B) = AB$ and $g(A,B) = BA + AB$ (actually this is only an example of possible functions $f$ and $g$).
Then I have that:

$$f(g(Q,X),Y) = f(XQ + QX, Y) = XQY + QXY$$

Then, using matrix calculus (hoping there are no error!), I have that:

$$ frac{partial t(Q,X,Y)}{partial X} = QY + YQ\
frac{partial t(Q,X,Y)}{partial Y} = XQ + QX$$

I can easily compute the result if I know the form of $f$ and $g$. Notice that the derivatives I obtained are in a matrix form.
But actually I need to deal with generic functions. And for this reason I need to use the chain rule.
The problem is that the chain rule formulas I know are helpful to derive the derivative with respect to a certain element of the matrix $X$ (or $Y$). In this case, I'm not able to have a matrix form of the derivatives.

So, my question is... there is a chain rule formula I'm missing which let me describe these derivatives in a matrix form?

* Addition 2 *

The chain rule formulas that I know are reported here http://en.wikipedia.org/wiki/Matrix_calculus#Scalar-by-matrix_identities (see the 7th row of the table)

Let's use uppercase letters for the matrix variables, so they're easy to distinguish from the lowercase scalars
$$eqalign{
G &= G(Q,X) cr
F &= F(G,Y) cr
t &= {rm tr}(F) = I:F cr
dt &= I:dF cr
}$$
First, let's calculate the differential and gradient wrt $Y$
$$eqalign{
dt &= I:Big(frac{partial F}{partial Y}:dYBig) cr
frac{partial t}{partial Y} &= I:frac{partial F}{partial Y} cr
}$$
And now wrt $X$
$$eqalign{
dt &= I:Big(frac{partial F}{partial G}:frac{partial G}{partial X}:dXBig) cr
frac{partial t}{partial X} &= I:frac{partial F}{partial G}:frac{partial G}{partial X} crcr
}$$
Note that the matrix-by-matrix gradients are 4th order tensors. For example, here is one of the gradients in component form
$$eqalign{
Big(frac{partial G}{partial X}Big)_{ijkl} = frac{partial G_{ij}}{partial X_{kl}}crcr
}$$

Also note that colons are used to denote the double-contraction product, e.g.
$$Big(frac{partial F}{partial G}:frac{partial G}{partial X}Big)_{ijkl}
= frac{partial F_{ij}}{partial G_{mn}},frac{partial G_{mn}}{partial X_{kl}}$$