- #1

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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.

/David

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- Thread starter DavidK
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- #1

- 31

- 0

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.

/David

- #2

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- #3

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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.

- #4

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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:

[tex]

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

[/tex]

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

[tex]

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

\mbox{Tr}(B)\bar{r}_b|}{2},

[/tex]

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.

/David

[tex]

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

[/tex]

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

[tex]

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

\mbox{Tr}(B)\bar{r}_b|}{2},

[/tex]

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.

/David

Last edited:

- #5

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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.

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