Which Unitary Matrices Keep A - UBU† Positive Definite?

[DavidK]

An Hermitian matrix H is positive definite if all its eigenvalues are nonzero and positive. Assume that the matrices A,B are positve definite, and that the difference A-B is positve definite. Now, for which unitary matrices, U, is it true that the matrix A-UBU^{\dagger} 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 U) or because I'm to incompetent. Any suggestions would be appreciated.

/David

 

[Lonewolf]

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. (Ax,x) > 0 for all x in the vector space V? Then, ((A-B)x,x) > 0 if A - B is to be positive definite.

 

[Jimmy Snyder]

I deleted, and then resubmitted this post:

Here is a start:

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

A is positive definite, B is positive definite, A - B is positive definite, but A - UBU^{-1} is not positive definite.

This points to what can go wrong in the general case.

 

[DavidK]

I think I have solved the problem for the 2\times2 case. A positive matrix A can in this case be expressed as:

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

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

<br /> <br /> \frac{\mbox{Tr}(A)-\mbox{Tr}(B)}{2} \geq \frac{|\mbox{Tr}(A)\bar{r}_a-<br /> \mbox{Tr}(B)\bar{r}_b|}{2}, <br /> <br />

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

/David

 

[BoTemp]

Davids got the right idea. For any unitary matrix, U^{\dagger} = U^-1. For any matrix A, Tr(A)= Tr(DAD^-1).

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

Tr(A - UBU^{\dagger}) = Tr(A - B) &gt; 0.

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.