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Let $U$ be unitary and $D$ be diagonal. Can anything be stated about the structure of $$U^\dagger D U$$ then?

Also, would this change for an infinite matrix, i.e. a discrete linear operator?

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2 Answers 2

up vote 4 down vote accepted

If the entries of $D$ are real, there is a beautiful answer due to Schur. Let the entries of $D$ be $(d_1, d_2, \ldots, d_n)$. Then the possible diagonal entries of $U^* D U$ are the points in the convex hull of the $n!$ permutations of the $d_i$.

Another way to phrase this is the following: Let $d_1 \geq d_2 \geq \cdots \geq d_n$. Then $(c_1, c_2, \ldots, c_n)$ are possible diagonal entries for $U^* D U$ if and only if

$$\sum c_i = \sum d_i$$

and, for any $i_1 < i_2 < \cdots < i_k$, we have

$$c_{i_1} + c_{i_2} + \cdots + c_{i_k} \leq d_1 + d_2 + \cdots + d_k.$$

See section 5 of Bhatia's article Linear Algebra to Quantum Cohomology: The Story of Alfred Horn's Conjectures for a proof.


If $D$ is complex, $D=A+Bi$, then $\mathrm{diag}(U^* D U^*) = \mathrm{diag}(U^* A U) + \mathrm{diag}(U^* B U^*) i$. So the real and imaginary parts of the diagonal separately obey Schur's bounds with respect to $A$ and $B$. I'm not sure if you can do better than this.

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To help with reading: If the entries of D are real, then UDU is called Hermitian, and at the bottom UAU is called the Hermitian part, and iU*BU (or sometimes without the i, be careful) is the anti-Hermitian part. –  Jack Schmidt Jan 5 '11 at 16:43
    
Does your $*$ mean hermitian conjugation as the $\dagger$ means or only complex conjugation? –  Tobias Kienzler Jan 5 '11 at 18:45
    
Hermitian conjugate. –  David Speyer Jan 5 '11 at 19:36
    
In the last paragraph there are two extra * floating around. Hopefully it is clear: D=A+Bi, then diag(UDU) = diag(UAU) + diag(U*BU)i. –  Jack Schmidt Jan 6 '11 at 14:53

The n×n matrices of the form U* D U for D diagonal and U unitary are exactly the Normal matrices, that is the matrices that commute with their conjugate transpose. The extension to hilbert-space operators are called normal operators.

If by dagger you mean the plain transpose, then I believe Horn–Johnson's book discusses these matrices, but they are less common.

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Thank you. Is the notation really so confused? I always thought $U^*$ stands for complex conjugation, $U^\dagger$ for Hermitian conjugation and $U^T$ for transposition? But maybe that's because I'm coming from Physics... –  Tobias Kienzler Jan 6 '11 at 8:06
    
In math the * typically means Hermitian transpose (conjugate transpose). Some people use $U^H$. For plain complex conjugation, math people use a bar over the letter, like for numbers, $\bar U$. I've seen dagger used both for transpose and Hermitian transpose in math. –  Jack Schmidt Jan 6 '11 at 14:51

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