R matrix gradient demonstration

Hello guys,

Can someone please give a well explained demonstration for the R matrix gradient equation?

Below, we have chatGPT demonstration. This is fantastic, but I don’t know if it is correct.

Prompt was: “given the following gradient equation $$\frac{d}{dR}L(X,Y,R)=\frac{d}{dR}\Big(\frac{1}{m}| X R -Y|_{F}^{2}\Big) = \frac{2}{m}X^{T} (X R - Y)$$, where X, Y, R are matrices, can you explain to me the demonstration to this equation?”


To begin, the matrix norm |A|_F is defined as the square root of the sum of the squares of the elements of the matrix A, and is known as the Frobenius norm. It is given by the formula:

|A|*F = \sqrt{\sum*{i=1}^m \sum_{j=1}^n |a_{ij}|^2}

where m and n are the dimensions of the matrix A.

The goal is to find the gradient of the loss function L(X, Y, R) = \frac{1}{m}|XR - Y|_F^2 with respect to the matrix R. We can do this using the chain rule.

First, we can rewrite the loss function as:

L(X, Y, R) = \frac{1}{m}\left(\sum_{i=1}^m \sum_{j=1}^n |(XR)*{ij} - Y*{ij}|^2\right)

Next, we can use the chain rule to find the gradient:

\frac{d}{dR}L(X, Y, R) = \frac{d}{dR}\left(\frac{1}{m}\left(\sum_{i=1}^m \sum_{j=1}^n |(XR)*{ij} - Y*{ij}|^2\right)\right)

= \frac{1}{m}\left(\sum_{i=1}^m \sum_{j=1}^n 2((XR)*{ij} - Y*{ij})\frac{d}{dR}((XR)_{ij})\right)

= \frac{1}{m}\left(\sum_{i=1}^m \sum_{j=1}^n 2((XR)*{ij} - Y*{ij})\frac{d}{dR}\left(\sum_{k=1}^n X_{ik}R_{kj}\right)\right)

= \frac{1}{m}\left(\sum_{i=1}^m \sum_{j=1}^n 2((XR)*{ij} - Y*{ij})\sum_{k=1}^n X_{ik}\frac{d}{dR}(R_{kj})\right)

= \frac{1}{m}\left(\sum_{i=1}^m \sum_{j=1}^n 2((XR)*{ij} - Y*{ij})\sum_{k=1}^n X_{ik}\delta_{kj}\right)

= \frac{1}{m}\left(\sum_{i=1}^m \sum_{j=1}^n 2((XR)*{ij} - Y*{ij})X_{ij}\right)

= \frac{2}{m}\left(\sum_{i=1}^m \sum_{j=1}^n X_{ij}(XR)*{ij} - Y*{ij}\right)

= \frac{2}{m}X^T(XR - Y)

This is the final result.

Wow! That looks pretty plausible to me. The LaTeX is a little sketchy in places (they are a bit inconsistent in the notation for the subscripts), but the math looks mostly plausible. The only sort of questionable and “handwavey” bit is the way they handle the “second layer” of the Chain Rule to get from

\frac {\partial}{\partial R}(XR)_{ij}

To the end there. What is \delta_{ij} and how do we end up with transpose of X. Seems like the third level of summation just magically disappeared about three lines from the end and then magically reappears as the matrix multiplication by X^T, but maybe I’m just not thinking hard enough yet to “grok” what is really happening there. :laughing:

Still it’s a pretty amazing performance for a piece of software. I’m seriously impressed.

I asked it later what was \delta_{ij}, and it is the Kronecker delta.

The biggest problem to me, is that the summation, on the next-to-last line, seems to result in a scalar instead of a matrix. Am I correct?