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I Can we retrieve the inverse of matrix A in this example?

  1. Dec 25, 2016 #1
    Suppose we have a product formed by a multiplication of a unitary matrix U and a diagonal matrix A, can we retrieve the inverse of A without knowing either U or A?
     
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  3. Dec 25, 2016 #2

    blue_leaf77

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    If you have UA, check if it's invertible. If it's not, there is no point trying to find the inverse of A because it does not exist.
     
  4. Dec 25, 2016 #3
    U and A are square matrices of the same rank, so UA is a square matrix too and it should be invertible. But even in this case, how to find A and its inverse from UA? In other words, can we decompose UA into U and A?
     
  5. Dec 25, 2016 #4

    blue_leaf77

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    Being square does not guarantee that it's invertible.
    I will give you a hint, what is the transpose conjugate of ##UA##?
     
  6. Dec 25, 2016 #5
    What if we find that UA is invertiabe, can we decompose it?
     
  7. Dec 25, 2016 #6

    blue_leaf77

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    Probably I should also emphasized that ##A## is always unique but not necessarily diagonal. Which means, if you sought for a diagonal ##A## it may be to no avail.
     
  8. Dec 25, 2016 #7
    I appreciate your contribution to answer but frankly I have no clear idea about your words. My question is clear from the beginning and still I have no answer on it, will I be able to decompose UA into U and A where is U is a unique unitary and A a unique diagonal or no?
     
  9. Dec 25, 2016 #8

    fresh_42

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    @blue_leaf77's question in post #4 contains the answer. What can you say about ##(UA)^\dagger = \overline{(UA)}^\tau## and what happens, if you multiply this by ##UA##?
     
  10. Dec 25, 2016 #9
    We will get ##A^2## because ##U^TU=1##, right
     
  11. Dec 25, 2016 #10

    fresh_42

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    If you interpret ##A^2=\overline{A}A##, then yes. You haven't said that ##A## is a real diagonal matrix, so ##\overline{A} \neq A## in general. This still doesn't give you a unique description of the diagonal elements, but some additional information. Maybe this can be used in ##1=(UA)(UA)^{-1}##.
     
  12. Dec 25, 2016 #11
    But why is A, if it is real diagonal, not unique? If I get A2 then each diagonal element in A is ##\sqrt {A^2}##.
     
  13. Dec 25, 2016 #12

    fresh_42

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    Again, you have only said ##A=diag(a_1,\ldots,a_n)##, so ##a_i \in \mathbb{C}## and ##(UA)^\dagger \cdot (UA)= \overline{A}\cdot A## is all you can conclude. Esp. this gives you ##n## equations ##c_j=\overline{a}_j \cdot a_j## which cannot be solved uniquely without additional information. Even in the real case there are two solutions for each ##j\, : \,\pm \, a_j##
     
  14. Dec 25, 2016 #13
    So in special case where all elements of A are real and positive, then no additional information is required and A is solved.
     
  15. Dec 25, 2016 #14

    fresh_42

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    Yes.
    If all diagonal elements of ##A=diag(a_j)## are positive real numbers, then you can compute ##\overline{(UA)}^\tau## if you know ##UA##, then multiply both to ##\overline{(UA)}^\tau\cdot (UA)=\overline{A}^\tau \cdot A=A \cdot A = diag(a_j^2)## which has to be diagonal, if the given conditions for ##U## and ##A## are correct. Thus you have ##n## equations ##c_j :=\left(\overline{(UA)}^\tau(UA) \right)_{jj}=a_j^2## which determine ##A## and with it ##A^{-1}## and ##U=(UA)A^{-1}##.
     
  16. Dec 25, 2016 #15

    micromass

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    Or more generally: https://en.wikipedia.org/wiki/Polar_decomposition
     
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