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Show A = UDU(dagger) can be written as f(A) = Uf(D)U(dagger)

  1. May 8, 2013 #1
    1. The problem statement, all variables and given/known data

    A diagonalisable square matrix A can be written [itex]A = UDU\dagger[/itex], where U is a unitary matrix and D is diagonal. Show that nay function of A defined by a power series,

    [itex]f(A) = f_{0}I + f_{1}A + f_{2}A^{2} + ... + f_{n}A^{n} + ... [/itex]

    can be expressed as [itex]f(A) = Uf(D)U\dagger[/itex]

    3. The attempt at a solution

    Not sure where to start. I know one can repeat a linear operator to get results such as [itex]A^{0} = 1, A^{1} = A, A^{2} = AA[/itex] but how do you show that for D? Are we simply doing the same? [itex]D^{0} = 1, D^{1} = D, D^{2} = DD [/itex]

    So [itex]f(A) = U * f_{0}I + f_{1}D + f_{2}D^{2} + ... + f_{n}D^{n} + ... * U\dagger[/itex] ???

    Thanks
     
  2. jcsd
  3. May 8, 2013 #2

    Fredrik

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    That last line is messed up in some way. If you meant to start the right-hand side with ##U(\dots## and end it with ##\dots)U^\dagger##, then you have just written down the equality you're supposed to prove. But yes, that thing in the middle is ##f(D)##. So it can't be a bad idea to try to use that (the definition of ##f(D)##) to rewrite ##Uf(D)U^\dagger## in some way.
     
  4. May 8, 2013 #3
    Should it look like this,

    [itex]f(A) = U*f_{0}I*U\dagger + U*f_{1}D*U\dagger + U*f_{2}D^{2}*U\dagger + ... + U*f_{n}D^{n}*U\dagger + ... [/itex] ?
     
  5. May 8, 2013 #4

    Fredrik

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    I assumed that you you were trying to write down the equality ##f(A)=Uf(D)U^\dagger##, with ##f(D)=1+f_0 D+\cdots+f_n D^n+\cdots##. You obviously can't assume that the former equality holds, since that's what you want to prove, but you can start
    $$Uf(D)U^\dagger=U\left(1+f_0 D+\cdots+f_n D^n+\cdots\right)U^\dagger,$$ to see if you can end up with ##f(A)##. The right-hand side above can be written the way you just wrote it, but I wouldn't use ##*## for multiplication. It's standard to not use any symbol at all, and if you want to use one, it should be ##\circ##, since multiplication of linear operators is just composition of functions.
     
  6. May 8, 2013 #5
    Ok got it. Thanks for your help, Fredrik
     
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