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Background

I was recently thinking about matrices and had an idea about a new definition of derivatives which reduces to the old one under certain circumstances. The idea is illustrated by an example:

Let A be a $2 \times 2$ matrix and the function $Z$ be defined by: $$ Z(A) = A^2 $$ $$\implies Z(A+\hat \epsilon)= (A+\hat \epsilon)^2 $$ Where $\epsilon$ is is a $2 \times 2$ matrix such that $\epsilon^2 = 0$

$$ \hat \epsilon= \begin{bmatrix} 0 & 0 \\ h & 0 \\ \end{bmatrix}$$

Now we compute $Z(A+\hat \epsilon)- Z(A) = (A+\hat \epsilon)^2 -A^2 = A \epsilon + \epsilon A$

We now give an explicit definition of $A$:

$$ A= \begin{bmatrix} a & b \\ c & d \\ \end{bmatrix}$$

Hence, $$ A \epsilon = \begin{bmatrix} b h & 0 \\ dh & 0 \\ \end{bmatrix}$$

Similarly,
$$ \epsilon A = \begin{bmatrix} 0 & 0 \\ ah & bh \\ \end{bmatrix}$$

Adding the above we get:

$$\lim_{h \to 0}\frac{Z(A+\hat \epsilon(h))- Z(A)}{h} = \begin{bmatrix} b & 0 \\ a+d & b \\ \end{bmatrix}$$

Similarly we could have taken $\epsilon^T $ for the derivative: $$ \lim_{h \to 0}\frac{Z(A+\hat \epsilon^T(h))- Z(A)}{h} = \begin{bmatrix} c & a+d \\ 0 & c \\ \end{bmatrix}$$

Or any combination ...

Questions

Does taking derivatives of matrices like this already exist? If so what is their applications? What is the inverse of this"matrix derivative"? Is there a possible limit of a sum interpretation of the inverse of a matrix derivative?

drewdles
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1 Answers1

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I think this is related to the differential of your matrix valued function. In particular, for your function $Z(A)=A^2$, the differential is \begin{equation} \text{d}Z=A\text{d}A+\text{d}A A, \end{equation} so that you can approximate any change in Z linearly by \begin{equation} Z=A+h (A\text{d}A+\text{d}A A), \end{equation} if $A=A+h \text{d}A$. The differential is then the limit \begin{equation} \text{d}Z=\lim_{h\rightarrow 0}\frac{Z(A+h\text{d}A)-Z(A)}{h}. \end{equation} Now consider choosing \begin{equation} h\text{d}A=\hat{\epsilon} \end{equation} as defined in the question. Your derivative is in fact the differential d$Z$ applied to this choice of differential d$A$, which can be readily checked (this is true of the first definition of derivative you made, for the second definition you would choose the transpose of $\hat{\epsilon}$ for the differential of A, which would again give the correct result).

For the second half of the question, you are interested in an inverse to this derivative. You can certainly define the integral \begin{equation} \int_\gamma \text{d}Z, \end{equation} for some path $\gamma$ in matrix space. This integral is independent of path, and satisfies a generalized fundamental theorem of calculus \begin{equation} \int_\gamma \text{d}Z=Z(A\rvert_{\gamma_1})-Z(A\rvert_{\gamma_0}) \end{equation} where $\gamma_{0,1}$ are the endpoints of the path.

  • I am probably misunderstanding you answer, but are you suggesting that the derivative of $Z(A)=A^2$ w.r.t. $A \in \mathbb{R}^{2 \times 2}$ is a matrix? – Ted Black Mar 30 '24 at 11:08
  • No, the appearance of dA in the formula implies that the differential is a map from matrices to matrices. Let's call our space of matrices M. Z is a map from M into M, $Z : M \rightarrow M$. The pushforward of our map is $dZ : T_A M \rightarrow T_{Z(A)} M$, which is a map from matrices to matrices. This can be seen from the formula for dZ, because dA denotes an element of the tangent space $T_A M$. – Teddy Baker Sep 11 '24 at 18:26