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I am unfamiliar with advanced Matrix theory (nor am I a mathematician), so please bear with me.

Is there anything significant about the following Matrix structure? Are there any special symmetries or conserved quantities that can be extracted from this?

$ \pmatrix{-u_2 & 0 & -\sqrt{2} u_1 & 0 \\ 0 & u_2 & 0 & -\sqrt{2} u_1 \\ \sqrt{2} u_1 & 0 & 0 & 0 \\ 0 & \sqrt{2} u_1 & 0 & 0} $

where $u_1,u_2$ are real and have a maximum value of $1$.

Edit:

I was reading something about the correspondence between the group of traceless matrices and the SU(2) group (are they isomorphic?). Am I on the right track? Anything about the blocks that stand out? I was also reading about trace conserving "volume", but I am not sure what that means.

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    @WillJagy, this is a transformation Matrix that transforms quantum field operators. They will be used in an equation of the form DX=MX, where D is the differential operator and X is a vector that contains quantum field operators.2012-08-31

2 Answers 2

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To complile some of the obvious results (which hopefully don't contain mistakes):

It has determinant $4(u_1)^4$. So, it's going to be nonsingular when $u_1>0$. Even when $u_1=0$, as long as it's nonzero, it's still not nilpotent.

If $\lambda$ is an eigenvalue, then the rest are $\overline{\lambda}$, $-\lambda$ and $-\overline{\lambda}$.

It has trace $0$, whence Lie algebra singles it out as an "infinitesimal volume preserving transformation". (That last point requires a bit of background in Lie algebra to understand.) It can be reexpressed as $AB-BA$ for two other matrices $A$ and $B$.

This matrix is always diagonalizable.

It is close but not quite anti-symmetric if $u_1>0$. It is easy to see the symmetric part is in the upper left $2\times 2$ block, and the antisymmetric part is everything else.

I started on the singular values, but confirmed in WolframAlpha that they are complicated.

WolframAlpha also confirms that the QR, LU and SVD decmopositions are complicated.

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    @AntillarMaximus They're your points to spend, but I'll just say as far as your posted question goes (about invariants and such), I am really surprised you concluded the other answer more than 6 times more useful. 100 points is a fine consolation prize :P2012-09-02
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You do get a Lie algebra if you ignore the bounds you briefly mentioned. First, define $ L \; = \; \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right). $ and $ R \; = \; \left( \begin{array}{cccc} 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \end{array} \right). $

Next, define a Lie bracket for two of your matrices $V,W$ as $ [V,W ] = L (V W - W V) R. $ This gives a Lie algebra, as the various closure properties hold (addition, multiplication by a scalar) as well as anticommutativity and the Jacobi identity $ [[U,V],W] + [[V,W],U] + [[W,U],V] = 0. $