# Intuition behind "Transpose Matrix"

I have come across the differential $$\frac{\partial \bf{w}}{\partial \bf{w}^T}$$ many times now, and I notice that it is equivalent to the transpose operator. That is, if we have something of the form $$\bf{A}\frac{\partial w}{\partial w^T}$$, we can rewrite it as $$\bf{A}^T$$ whenever the dimensions of $$\bf{A}$$ and $$\frac{\partial \bf{w}}{\partial \bf{w}^T}$$ agree.

Intuitively, the change of $$\bf{w}$$ with respect to $$\bf{w}^T$$ is a transpose. Is my intuition valid and is there a way to prove this?

$$\bf{EDIT1}$$: To clarify, $$\bf{A}$$ is a matrix while $$\bf{w}$$ and $$\bf{w}^T$$ are vectors. The vector $$\bf{w}$$ is taken to be a column vector of dimension $$n$$ and $$\bf{A}$$ is taken to be a $$n \times n$$ matrix. From this we have that $$\frac{\partial \bf{w}}{\partial \bf{w}^T}$$ is a $$n \times n$$ matrix $$\bf{W}$$ where $$W_{ij} = \frac{\partial \bf{w}_i}{\partial \bf{w}^T_j}$$. The question that follows now is, why does $$\bf{W}$$ act as a transpose operator on $$\bf{A}$$?

$$\textbf{EDIT2:}$$ To give an example, when computing $$\nabla_\bf{w}$$MSE$$_{\text{train}}$$ in this post (last response), the author has a step that goes from $$\bf{wX^TX}\frac{\partial \bf{w}}{\partial \bf{w}^T}$$ to $$(\bf{wX^TX})^T$$

• Maybe you tell us what is meant by ${\partial{\bf w}\over\partial{\bf w}^T}$. Jun 5 '19 at 18:35
• The question has been updated with (hopefully) sufficient information. Jun 5 '19 at 18:55

Sadly, there is no intuition to be learned, only that matrix differential calculus has inconsistent notation.

I wouldn't see $$\frac{\partial w}{\partial w}$$ as something that "does transposition": the result you mention follows from the linearity of differentiation and from using the denominator convention.

By linearity of differentiation, $$A \frac{\partial w}{\partial w} = \frac{\partial A w}{\partial w}$$

If we follow numerator convention, this is the Jacobian matrix of the linear function $$Aw$$, that is a matrix whose $$i,j$$th element is $$\frac{\partial A_i^T w}{\partial w_j} ,$$ where $$A_i^T$$ is the $$i$$th row of $$A$$. Then, $$\left[\frac{\partial A w}{\partial w}\right]_{i,j} =\frac{\partial }{\partial w_j} \sum_{k=1}^m A_{i,k}w_k=A_{i,j},$$ where $$A_i^T$$ is the $$i$$th row of $$A$$, so $$A \frac{\partial w}{\partial w} = A.$$

If we instead follow denominator convention, then your expression means the gradient of the linear function $$Ax$$, that is the matrix whose $$i,j$$ element is

$$\frac{\partial A_j^T w}{\partial w_i}$$ Then $$\left[\frac{\partial A w}{\partial w}\right]_{i,j} =\frac{\partial }{\partial w_i} \sum_{k=1}^m A_{j,k}w_k = A_{j,i},$$ so $$A \frac{\partial w}{\partial w} = A^T.$$

The fact that the author uses a transposed vector at the denominator seems to indicate the numerator convention, however the transposed result hints at a denominator convention. Confusing indeed!

• I would say that the notation $A \frac{\partial w}{\partial w^T} = A^T$ seems to be inappropriate to begin with. Indeed, there is no matrix $X$ that satisfies $AX=A^T$ for all $A$. The right way to factor out $A$ from the $\partial$-sign in the denominator notation is $\frac{\partial Au}{\partial x}=\frac{\partial u}{\partial x}A^T$ (as mentioned here).
– A.Γ.
Jun 6 '19 at 17:03
• I agree with you on all points, it is a very sloppy argument. The only conclusion seems to be that you have to be careful with matrix differential calculus (and that some ML people have no regard for correctness XD) Jun 6 '19 at 17:17
• This has helped me clean up my notation, as well as standardize a specific way I carry out matrix calculus now. Both Riccardo's answer and the comments following helped me understand when to look for intuition and when to look for sloppy notation. Jun 7 '19 at 13:34