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Solbing equation A(u)=B(v) for square matrices A and B

  1. Apr 1, 2012 #1


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    Dear all,

    this is perhaps a trivial question, so I apologise in advance. Any help is greatly appreciated nonetheless.

    ==The Equation==
    The equation under consideration is:


    where A and B are n times n matrices, while u and v are n-dimensional vectors.

    ==The Question==
    From the above equation, I would like to determine u as a function of v.
    Question: is the only way of determining u as a function of v to require that A is invertible?
    In other words, is it correct to say that the only way of solving the above equation for u is u=A^{-1}B(v) ?

    Thanks a lot,
  2. jcsd
  3. Apr 1, 2012 #2

    Well, more than "saying that the only way...", I'd say that requiring A to be invertible is the only way to guarantee that there's a solution to your problem, otherwise we'd have to look at each case in particular.

  4. Apr 2, 2012 #3


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    Thanks for your reply, DonAntonio.

    You are right, my question was a bit too vague. But you understood what I meant.

    Perhaps a more meaningful way to put the question would be:

    -given two n times n matrices, A and B
    -let u and v be two n-dimensional vectors


    For each choice of v, the above equation has a UNIQUE solution u if and only if det(A) is non-zero.


  5. Apr 2, 2012 #4


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    Science Advisor

    There is a unique solution if and only if A is invertible and in that case u= A-1Bv.
  6. Apr 6, 2012 #5


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    Thanks everybody,

    glad to see things got clearer in my head. Problem solved.

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