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feynman1
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A and B are 2*2 matrices. A' is the transpose of A. Will the solution of A*A'=B for A yield a unique A(B)?
then should we conclude the derivative dA/dB doesn't in general exist?No. For example, if ##B## is the identity matrix, then ##A## could be any rotation or reflection matrix.
What derivative! ##A'## was your notation for the transpose.then should we conclude the derivative A'(B) doesn't in general exist?
changed the notation nowWhat derivative! ##A'## was your notation for the transpose.
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?
A and B are 2*2 matrices. A' is the transpose of A. Will the solution of A*A'=B for A yield a unique A(B)?
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
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.
then should we conclude the derivative dA/dB doesn't in general exist?
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.After all this, what is the original questions? What is the equation? Is it with ##A'## being the transpose or is it some kond of derivative? I think it would be helpfull if you gave us more information. Also you shouldn't assume that "tensor valued derivative" is something everyone knows. Give as reference or a definition. The notation you use dA/dB is unclear as well. Is it ##dAdB^{-1}## or ##dB^{-1}dA##? After all matrices do not commute in general, so a fraction is ambiguous.
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.
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.
Amen to this!It is ok to give information about the question. There is no need for us to pry it out of you.
A*A'=B where * is a normal dot product. Aik*Ajk=B. A, B are matrices.