Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Positive definite matrices

  1. Jun 8, 2006 #1
    An Hermitian matrix [tex] H [/tex] is positive definite if all its eigenvalues are nonzero and positive. Assume that the matrices [tex] A,B [/tex] are positve definite, and that the difference [tex] A-B [/tex] is positve definite. Now, for which unitary matrices, [tex] U [/tex], is it true that the matrix [tex] A-UBU^{\dagger} [/tex] is positve definite.

    I haven't been able to solve this problems, and I'm not sure if it is because it is to difficult (i.e. the only way to solve it is to check for all [tex] U [/tex]) or because I'm to incompetent. Any suggestions would be appreciated.

  2. jcsd
  3. Jun 9, 2006 #2
    I haven't bothered to try this out fully so I may be going up a blind alley, but how about the definition of positive definite involving the inner product, i.e. [tex](Ax,x) > 0[/tex] for all [tex]x[/tex] in the vector space [tex]V[/tex]? Then, [tex]((A-B)x,x) > 0[/tex] if [tex]A - B[/tex] is to be positive definite.
  4. Jun 9, 2006 #3
    I deleted, and then resubmitted this post:

    Here is a start:

    A = [tex]\left[ \begin {array}{cc} 2 &0 \\ 0 &3 \end {array} \right][/tex]
    B = [tex]\left[ \begin {array}{cc} 1 &0 \\ 0 &2 \end {array} \right][/tex]
    U = [tex]\left[ \begin {array}{cc} 0 &1 \\ 1 &0 \end {array} \right][/tex]

    A is positive definite, B is positive definite, A - B is positive definite, but [tex]A - UBU^{-1}[/tex] is not positive definite.

    This points to what can go wrong in the general case.
  5. Jun 13, 2006 #4
    I think I have solved the problem for the [tex]2\times2 [/tex] case. A positive matrix [tex]A [/tex] can in this case be expressed as:


    A=\frac{\mbox{Tr}(A)}{2}(I+r_x \sigma_x+r_y\sigma_y + r_z \sigma_z),


    where [tex]\sigma_x, \sigma_y, \sigma_z[/tex] are the standard Pauli matrices, and [tex]\bar{r}_a=(r_x,r_y,r_z) [/tex] is a 3-vector of length less than one. This means that the difference between the matrices [tex]A [/tex] and [tex]UBU^{\dagger} [/tex] is positive iff


    \frac{\mbox{Tr}(A)-\mbox{Tr}(B)}{2} \geq \frac{|\mbox{Tr}(A)\bar{r}_a-


    where the angle between the vectors [tex]\bar{r}_a,\bar{r}_b [/tex] is given by the unitary [tex]U[/tex]. I'm, however, not sure if it is possible to solve the general problem using this approach.

    Last edited: Jun 14, 2006
  6. Jul 11, 2006 #5
    Davids got the right idea. For any unitary matrix, [itex] U^{\dagger} = U^-1 [/itex]. For any matrix A, [itex] Tr(A)= Tr(DAD^-1) [/itex].

    If A - B is positive definite, then Tr(A - B) > 0 .

    [tex] Tr(A - UBU^{\dagger}) = Tr(A - B) > 0 [/tex].

    Tr( A - B) > 0 is necessary, but it isn't sufficient. I'd start looking at how a unitary transform affects the eigenvalues of a a matrix.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook