# Matrix derivative of $\mbox{Tr} (\mathbf{AXB})$

The Matrix Cookbook says that:

$\frac{\partial}{\partial\mathbf{X}} Tr\{\mathbf{A}\mathbf{X}\mathbf{B}\} = \mathbf{A^T}\mathbf{B^T}$

I can't seem to get this. I know that:

$\frac{\partial}{\partial\mathbf{X}} Tr\{F(\mathbf{X})\} = f(\mathbf{A}\mathbf{X}\mathbf{B})^T$

Where $f$ is the scalar derivative of $F$.

So when I apply the rule: $\partial \mathbf{XY} = \partial \mathbf{X} \mathbf{Y} + \mathbf{X} \partial \mathbf{Y}$

I do:

Let: $\mathbf{C} = \mathbf{X}\mathbf{B}$

Then: $\mathbf{A}\mathbf{X}\mathbf{B} = \mathbf{A}\mathbf{C}$.

$\frac{\partial}{\partial\mathbf{X}} \mathbf{A}\mathbf{C} = \frac{\partial}{\partial\mathbf{X}} \mathbf{A} \mathbf{XB} + \mathbf{A} \frac{\partial}{\partial\mathbf{X}}\mathbf{X}\mathbf{B} = \mathbf{AB}$

But then:

$(\mathbf{AB})^{T} = \mathbf{B^TA^T} \neq \mathbf{A^{T}B^{T}}$

Am I confusing the notion of scalar derivative and matrix derivative? How can I verify the Cookbook's claim?

• As a means of checking your answer, note that $$Tr(AXB) = Tr(BAX)$$ – Omnomnomnom Mar 3 '17 at 23:19

The matrix inner product (denoted by a colon) is equivalent to the trace $$A^T:B = {\rm tr}(AB)$$
Therefore \eqalign{ f &= {\rm tr}(AXB) \cr &= {\rm tr}(BAX) \cr &= (BA)^T:X \cr &= A^TB^T:X \cr \cr df &= A^TB^T:dX \cr \cr \frac{\partial f}{\partial X} &= A^TB^T \cr }
I assume the domain is the space of $n\times n$-matrices $M(n,n)$. The function $f(X)=Tr(AXB)$ is a linear function $M(n,n)\rightarrow R$ since $f(X+Y)=Tr(A(X+Y)B)=Tr(AXB)+Tr(AYB)$. It is thus equal to its differential. To compute its matrix in the basis $e_{ij}$ where $e_{ij}$ is the matrix which has $1$ at the $(i,j)$-place ($i$ is the line and $j$ the column) and all other coefficients are zero.
$Tr(Ae_{ij}B)=Tr(ABe_{ij})$. The unique non zero coefficient on the diagonal of $ABe_{ij}$ is the $(j,j)$ coefficient which is the coefficient $(j,i)$ of $AB$ which is $\sum_ka_{jk}b_{ki}$ and this is the coefficient $(i,j)$ of $(AB)^T=B^TA^T$.