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:

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

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

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

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 .

$$Tr(A - UBU^{\dagger}) = Tr(A - B) > 0$$.

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.