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Matrices of the form: $$\begin{pmatrix} iz+iy&-iz+w&-iy-w\\ -iz-w&iz+ix&-ix+w\\ -iy+w&-ix-w&iy+ix\end{pmatrix}$$ where $x,y,z,w$ may be assumed to be real, form a Lie algebra. That is, such matrices are closed under addition and multiplication, and include inverses, zero, and unity. Each row or column sums to zero. Can the exponentiation of this matrix be put into closed form?

That is, can we solve: $$\begin{pmatrix} u_{11}&u_{12}&u_{13}\\ u_{21}&u_{22}&u_{23}\\ u_{31}&u_{32}&u_{33}\end{pmatrix} = \exp \begin{pmatrix} iz+iy&-iz+w&-iy-w\\ -iz-w&iz+ix&-ix+w\\ -iy+w&-ix-w&iy+ix\end{pmatrix}$$ in closed form for all the $u_{jk}$?

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Do you have a meaningful description of that algebra? –  Mariano Suárez-Alvarez Apr 6 '11 at 21:39
    
Looks complicated. I can only think about diagonalizing $M$ and then exponentiating. Note that $M$ has one 0 eigenvalue... –  Fabian Apr 6 '11 at 21:42
    
@Mariano; It's obviously isomorphic (or whatever the word mathematicians use) to u(2), i.e. the Hermitian 2x2 matrices. Note that there's a difference in how math and physics define these things so I could be off by a factor of $i$. –  Carl Brannen Apr 6 '11 at 22:33
    
If it were obviously isomorphic to that, I would not have asked... In any case, if you know it is you should probably add that information to the question. –  Mariano Suárez-Alvarez Apr 7 '11 at 6:05
    
@Mariano; To show equivalence to u(2), use a Householder transformation, i.e. $u\to HuH^{-1}$ where H is a Householder matrix that block diagonalizes the above matrices. Sorry, that's an abuse of "obvious". –  Carl Brannen Apr 8 '11 at 21:11

1 Answer 1

First of all, to be a Lie algebra it has to be closed under commutator, not matrix multiplication. This one is not closed under matrix multiplication (if you require the x,y,z,w real), although it is closed under commutator. As for the matrix exponential, that can be computed in closed form, e.g. by Maple using the MatrixExponential command. The result, however, is very complicated, too big to be copied here.

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Okay, if you make x,y,z,w imaginary it will be closed under multiplication. There's this factor of $i$ that math and physics does differently. –  Carl Brannen Apr 6 '11 at 23:07

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