# 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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