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Square root of matrix

  1. Jan 18, 2016 #1
    I'm doing an online course in quantum information theory, but it seems to require some knowledge of linear algebra that I don't have.
    A definition that popped up today was the definition of the absolute value of a matrix as:

    lAl = √(A*A) , where * denotes conjugate transpose.

    Now for a given matrix I can calculate the product A*A, but how is the square root of this defined? I have no idea, though I think it should be basis independent, so maybe square root of the trace.
    A*A is clearly positive semidefinite but I don't know if I can use that for anything.
     
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  3. Jan 18, 2016 #2

    PeroK

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    What's the definition of the square root of a number?
     
  4. Jan 18, 2016 #3
    Well I guess that the square root of a matrix A is then a matrix B such that B^2 = A. Well wouldn't it be better with a matrix B such that B*B = A. And is this definition basis independent.
     
  5. Jan 18, 2016 #4

    PeroK

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    That would be like requiring the square root of a complex number ##a## to satisfy ##b^*b = a##
     
  6. Jan 18, 2016 #5
    Ok well is the definition basis independent?
     
  7. Jan 18, 2016 #6

    PeroK

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    Let me rephrase that question. If a linear transformation T is represented by the matrix A in one basis and A' in another. Then is |A'| = |A|'?

    What do you think?
     
  8. Jan 18, 2016 #7
    I am not sure. I need to show that:

    (√(A))' = √(A')
    B' = √(A')

    Now if A' = UAU maybe I can use that somehow...
     
  9. Jan 18, 2016 #8

    PeroK

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    Or, think about diagonalizing A*A. It must be true if you want to accept that and get back to your quantum theory!
     
  10. Jan 18, 2016 #9
    Maybe something like this:

    A' = UAU
    B' = UBU

    Now B2 = A

    We want to show that:
    √A' = (√A)' = B'

    So is B'2 = A'?

    Well B'2 = UBUUBU = UB2U = UAU = A'

    But all this requires that A and A' and B and B' are related by a basis change with a unitary as above. When is this true?
     
  11. Jan 18, 2016 #10

    PeroK

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    That's essentially the proof. The only technicality is that B' must be the right square root. That probably depends on showing that a positive semi-definite matrix has a unique positive semi-definite square root.
     
  12. Jan 18, 2016 #11
    The square root of a positive semidefinite Hermitian matrix ##A## is the unique positive semidefinite matrix ##B## such that ##B^2=A##. You can look at Ch. VI s.3 of "Linear algebra done wrong" for an explanation of why such matrix exists and why it is unique.
     
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