Eikonal Blog

2010.02.26

Moore-Penrose inverse for light-cone vectors

Filed under: mathematics — Tags: — sandokan65 @ 16:54

This is continuation of previous post https://eikonal.wordpress.com/2010/02/17/some-examples-of-the-moore-penrose-inverse/


Example (added 2010.02.25 Fri): D=1+1 light-cone vector (i.e. R_\mu R^\mu \equiv 0) R_\mu (\mu=0,1) can be cast via D=1+1 Dirac matrices \left\{\gamma^0 = \begin{pmatrix}1 & 0 \\ 0 & -1 \end{pmatrix}, \gamma^1 = \begin{pmatrix}0 & 1  \\ -1 & 0 \end{pmatrix}\right\} into the nilpotent (i.e. \not{R}^2 \equiv 0) matrix \not{R}:\equiv R_\mu \gamma^\mu = R_0 \begin{pmatrix}1 & \epsilon \\ -\epsilon & -1 \end{pmatrix}, where R_1 = \epsilon R_0 (and \epsilon=\pm 1).

To find the MPI of that matrix, start with the transposed matrix \not{R}^T  = \begin{pmatrix}R_0 & -R_1 \\ -R_1 & -R_0 \end{pmatrix}. It is also nilpotent (i.e. (\not{R}^T)^2 = 0) but products following (singular) product are non-zero:

  • \not{R}^T \cdot \not{R}= 2R_0^2 \begin{pmatrix}1 & \epsilon \\ \epsilon & 1 \end{pmatrix},
  • \not{R} \cdot \not{R}^T = 2R_0^2 \begin{pmatrix}1 & -\epsilon \\ -\epsilon & 1 \end{pmatrix}.

Now calculate \not{R}^+ via following limiting process:

    \not{R}^+ = \lim_{\delta\rightarrow 0} (\not{R}^T \not{R} + \delta {\bf 1})^{-1} \not{R}^T
    = \frac1{2 R_0^2} \lim_{\delta\rightarrow 0}\begin{pmatrix}1+\delta & \epsilon \\ \epsilon & 1+\delta\end{pmatrix}^{-1}  \not{R}^T
    = \frac1{2 R_0^2} \lim_{\delta\rightarrow 0} \frac1{2\delta+\delta^2} \begin{pmatrix}1+\delta & -\epsilon \\ -\epsilon & 1+\delta\end{pmatrix}  \not{R}^T
    = \frac1{2 R_0} \lim_{\delta\rightarrow 0} \frac1{2\delta(1+\frac12\delta)} \begin{pmatrix}\delta & -\epsilon\delta \\ \epsilon\delta & -\delta\end{pmatrix}
    = \frac1{4 R_0}  \begin{pmatrix}1 & -\epsilon \\ \epsilon & -1\end{pmatrix}  = \frac1{4R_0^2} \not{R}^T.

Direct check verifies that all four defining properties of MPI are satisfied.

Note that \not{R}^+ corresponds to the D=1+1 vector (R^+)_\mu = \left\{\frac1{4R_0}, -\frac{\epsilon}{4R_0}\right\} = \frac1{4R_0^2} \left\{R_0, -R_1\right\} = \frac1{4R_0^2} \hat{P}R_\mu\hat{P}, where \hat{P} is the space-parity operator.

The two products \Pi_1:\equiv \not{R}\cdot \not{R}^+ = \frac12 \begin{pmatrix}1 & -\epsilon \\ -\epsilon & 1\end{pmatrix} and \Pi_2:\equiv \not{R}^+ \cdot \not{R}  = \frac12 \begin{pmatrix}1 & +\epsilon \\ +\epsilon & 1\end{pmatrix} are projectors: \Pi_a \Pi_b = \delta_{a,b} \Pi_b (a, b \in \{1,2\}), \Pi_1 + \Pi_2 = {\bf 1}. They project on two light cone directions defined by vectors R_\mu and R^+_\mu.


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