# Significance of the following Matrix?

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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Did you miss a minus sign on $u_2$ in the second row? –  Kartik Audhkhasi Aug 24 '12 at 14:46
Jennifer, thank you. Kartik, $u_2$ in the second column is positive.The matrix is correct as posted. –  Antillar Maximus Aug 24 '12 at 14:55
The only properties I notice are that it is traceless, invertible (unless $u_1=0$, of course), and the determinant only depends on $u_1$, not on $u_2$. However, it seems that the product of two such matrices also give a matrix of that form. Oh, and the factor $\sqrt{2}$ can be included in $u_1$ to make the matrix simpler. –  celtschk Aug 24 '12 at 15:03
celtschk, thank you. I'd like to add a +1, but I don't have enough rep to do so. :). Can something more be said if it is broken into 2x2 blocks? –  Antillar Maximus Aug 24 '12 at 15:27
@AntillarMaximus, what are you doing with these matrices? –  Will Jagy Aug 31 '12 at 18:19

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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The overline denotes complex conjugation, btw. –  rschwieb Aug 31 '12 at 18:10
The characteristic polynomial is $$\lambda^4 + (4 u_1^2 - u_2^2) \lambda^2 + 4 u_1^4$$ so we can solve for $\lambda^2$ with the quadratic formula. –  Will Jagy Aug 31 '12 at 18:38
Thanks for the summary! Keep 'em coming. I am particularly interested in the Lie Algebra connection you brought up. At this point, I am on a fishing expedition. :) –  Antillar Maximus Aug 31 '12 at 19:04
@rschwieb, one can make a Lie algebra out of these. It's nice, the nonzero entry is a little 2 by 2 determinant. –  Will Jagy Sep 1 '12 at 21:17
Marked this as an answer. I also found the Lie Algebra based answer very useful, so the bounty goes there. Thanks to all who responded. –  Antillar Maximus Sep 2 '12 at 15:51

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 $$[[U,V],W] + [[V,W],U] + [[W,U],V] = 0.$$

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