Solve: A*A'=B for A

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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)?
 

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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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No. For example, if ##B## is the identity matrix, then ##A## could be any rotation or reflection matrix.
 
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  • #4
feynman1
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No. For example, if ##B## is the identity matrix, then ##A## could be any rotation or reflection matrix.
then should we conclude the derivative dA/dB doesn't in general exist?
 
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  • #5
martinbn
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then should we conclude the derivative A'(B) doesn't in general exist?
What derivative! ##A'## was your notation for the transpose.
 
  • #6
feynman1
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What derivative! ##A'## was your notation for the transpose.
changed the notation now
 
  • #7
fresh_42
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then should we conclude the derivative dA/dB doesn't in general exist?
This cannot be concluded so easily. You have to define what you mean by your notation.

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.
 
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  • #8
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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?

changed the notation now
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?

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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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?

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.
dA/dB means a tensor valued derivative
 
  • #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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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.
 
  • #12
feynman1
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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.
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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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.
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.
 
  • #14
feynman1
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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.
Just wanted to extend a scaler property to a tensor property without other thoughts, sorry for having no background info
 
  • #15
Gaussian97
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Well, you should really define exactly what everything means. Naively I would interpret everything as follows:
##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
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It is ok to give information about the question. There is no need for us to pry it out of you.
Amen to this!
 
  • #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.
 
  • #19
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Thread closed, since we are no closer to understanding what is being asked than we were in the first post.
 

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