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feynman1

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feynman1

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- #2

anuttarasammyak

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You can write 2X2 matrices on paper easily. Why do not you write them down and see whether formula satisfying your condition determine unique matrix components ?

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- #3

Infrared

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- #4

feynman1

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then should we conclude the derivative dA/dB doesn't in general exist?

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- #5

martinbn

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What derivative! ##A'## was your notation for the transpose.then should we conclude the derivative A'(B) doesn't in general exist?

- #6

feynman1

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changed the notation nowWhat derivative! ##A'## was your notation for the transpose.

- #7

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This cannot be concluded so easily. You have to define what you mean by your notation.then should we conclude the derivative dA/dB doesn't in general exist?

Derivatives are always directional. Thus dA/dB indicates a function A considered in direction B. A needs to vary if B does, but how? If it does not, then dA/dB = 0. You have two problems with this notation: How does A depend on B, and how is this dependence locally unique, i.e. locally a function? If there are more than óne possibility, which one should we take?

Another possibility is to consider AA'-B=0 as an algebraic variety, i.e. a geometric object. All matrix elements of A and B (and eventually A' whatever it means) become variables in this perspective. The variety is embedded in a Euclidean vector space so you can consider changes in coordinate directions. But this doesn't give you an explanation for dB.

Long story short: Try to figure out what dA/dB could mean and you will find the difficulties yourself.

- #8

Mark44

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then should we conclude the derivative dA/dB doesn't in general exist?

It seems obvious to me that you don't understand this problem. At first you wrote that A' meant the transpose of A, then you decided that it meant the derivative of A. Is that your final answer?changed the notation now

Also, if A' is the derivative, which derivative is it? It doesn't make much sense to me to talk about the derivative of one matrix with respect to another; e.g., dA/dB, but it doe make sense to talk about the derivative of a matrix with respect to some variable, say t; e.g., dA/dt. Since you are so uncertain about this problem, it seems reasonable to assume that you aren't certain which derivative is meant.

If we have ##A(t) = \begin{bmatrix} a(t) & b(t) \\ c(t) & d(t) \end{bmatrix}##, then ##A'(t) = \frac{dA(t)}{dt}## would be ##\begin{bmatrix} a'(t) & b'(t) \\ c'(t) & d'(t) \end{bmatrix}##. In this case, the equation AA' = B could mean ##A \frac{dA}{dt} = B##, and this differential equation could be solved, although not for a unique solution.

- #9

feynman1

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dA/dB means a tensor valued derivativeIt seems obvious to me that you don't understand this problem. At first you wrote that A' meant the transpose of A, then you decided that it meant the derivative of A. Is that your final answer?

Also, if A' is the derivative, which derivative is it? It doesn't make much sense to me to talk about the derivative of one matrix with respect to another; e.g., dA/dB, but it doe make sense to talk about the derivative of a matrix with respect to some variable, say t; e.g., dA/dt. Since you are so uncertain about this problem, it seems reasonable to assume that you aren't certain which derivative is meant.

If we have ##A(t) = \begin{bmatrix} a(t) & b(t) \\ c(t) & d(t) \end{bmatrix}##, then ##A'(t) = \frac{dA(t)}{dt}## would be ##\begin{bmatrix} a'(t) & b'(t) \\ c'(t) & d'(t) \end{bmatrix}##. In this case, the equation AA' = B could mean ##A \frac{dA}{dt} = B##, and this differential equation could be solved, although not for a unique solution.

- #10

Infrared

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then should we conclude the derivative dA/dB doesn't in general exist?

I don't quite know what ##dA/dB## means, but if it presupposes that ##A## can be solved for as a function of ##B##, then yes I don't think it would make sense without extra information.

- #11

martinbn

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- #12

feynman1

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dA/dB is the derivative of some component of A w.r.t some component of B, with multiple elements. It is tensor valued because both A and B are second order tensors, matrices.

- #13

martinbn

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This doesn't actually answer my question. Anyway, can you at least tell us where the question came from? It is ok to give information about the question. There is no need for us to pry it out of you.dA/dB is the derivative of some component of A w.r.t some component of B, with multiple elements. It is tensor valued because both A and B are second order tensors, matrices.

- #14

feynman1

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Just wanted to extend a scaler property to a tensor property without other thoughts, sorry for having no background infoThis doesn't actually answer my question. Anyway, can you at least tell us where the question came from? It is ok to give information about the question. There is no need for us to pry it out of you.

- #15

Gaussian97

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##A## and ##B## are rank-2 tensors, with components ##A_{\mu\nu}## and ##B_{\mu\nu}##. The derivative is defined as the object with components ##\frac{dA_{\mu\nu}}{dB_{\alpha\beta}}##.

Furthermore, the ##*## operation would be a 1-index contraction (as usual when multiplying matrices).

Therefore the equation

$$A*A'=B$$

is equating a 4-index object in the left with a 2-index object in the right. This makes no sense to me.

So, if you want some help. Either you wait for someone who understands your notation, or you give us an explicit definition of everything involved here.

- #16

feynman1

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A*A'=B where * is a normal dot product. Aik*Ajk=B. A, B are matrices.

- #17

Mark44

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Amen to this!It is ok to give information about the question. There is no need for us to pry it out of you.

- #18

Office_Shredder

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A*A'=B where * is a normal dot product. Aik*Ajk=B. A, B are matrices.

This makes it look like ' is just transposing again, there are no derivatives in this post.

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Mark44

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