# 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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## 2010.02.19

### Weighted Moore-Penrose inverse

Filed under: mathematics — Tags: — sandokan65 @ 15:13

This is a generalization of the original concept of Moore-Penrose inverse (MPI). The weighted MPI $A^{+(N,M)}$ of a matrix $A \in {\Bbb F}^{n\times m}$ is defined by the following four properties:

• (A): $A \cdot A^{+(N,M)} \cdot A = A$,
• (B): $A^{+(N,M)} \cdot A \cdot A^{+(N,M)} = A^{+(N,M)}$,
• (C)_N: $(M \cdot A \cdot A^{+(N,M)})^c = M \cdot A \cdot A^{+(N,M)}$,
• (D)_M: $(A^{+(N,M)}\cdot A \cdot N)^c = A^{+(N,M)}\cdot A \cdot N$.

where the weighting matrices $M$ and $N$ are of the orders $n\times n$ and $m\times m$.

When weighting matrices are equal to the corresponding identities, the above definition reduces to ordinary MPI $A^c$.

Source: R. B. Bapat, S. K. Jain and S. Pati “Weighted Moore-Penrose Inverse of a Boolean Matrix”; Linear Algebra and Its Applications 225:267-279 (1997); North-Holland; pg. 692-704. http://www.math.ohiou.edu/~jain/077.pdf.

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## 2010.02.17

### Some examples of the Moore-Penrose inverse

Filed under: mathematics — Tags: — sandokan65 @ 16:49

Source: T2009.02.12

Definition: For rectilinear matrix $A \in {\Bbb F}^{n\times m}$ there exist a unique matrix $A^+ \in {\Bbb F}^{m\times n}$ (called the Moore-Penrose inverse [MPI]) s/t:

• (A): $A A^+ A = A$,
• (B): $A^+ A A^+ = A^+$,
• (C): $(A A^+)^c = A A^+$,
• (D): $(A^+ A)^c = A^+ A$.

where $M^c$ is the appropriate conjugation defined on the field ${\Bbb F}$, i.e. $(M^c)^c = M$ ($\forall M$).

In particular, for ${\Bbb F} = {\Bbb R}$:

• (A): $A A^+ A = A$,
• (B): $A^+ A A^+ = A^+$,
• (C): $(A A^+)^T = A A^+$,
• (D): $(A^+ A)^T = A^+ A$,

where $A^T$ is the transposition of the matrix $A$.

and for ${\Bbb F} = {\Bbb C}$:

• (A): $A A^+ A = A$,
• (B): $A^+ A A^+ = A^+$,
• (C): $(A A^+)^\dagger = A A^+$,
• (D): $(A^+ A)^\dagger = A^+ A$,

where $A^\dagger :\equiv (A^*)^T$ is the Hermitian conjugation of a matrix $A$.

## Properties

• The matrix equation $A \cdot \underline{y} = \underline{x}$ has infinitely many solutions
$\underline{y} = A^{+} \underline{x} + ({\bf 1}_n - A^{+} A) \underline{q}$
parametrized by an arbitrary \$\underline{q}, provided that the consistency condition $A A^{+} \underline{x} = \underline{x}$ is satisfied.
• If $B$ is a square positive semidefinite matrix, then for each $\underline{z}$ the following inequality holds:
$(A\underline{z}-\underline{x})^T \cdot B \cdot (A\underline{z}-\underline{x}) \ge \underline{x}^T C \underline{x}$.
Here $C:\equiv B - B A (A^T B A)^{+} A^T B^T$.
That inequality gets to be an equality for $\underline{z}=(A^T B A)^{+} A^T B \underline{x} + [{\bf 1}_n - (A^T B A)^{+}(A^T B A)]\underline{q}$.
• $\Pi_1:\equiv A^+ A$ and $\Pi_2:\equiv A A^{+}$ are projectors: $\Pi_{1,2}^2 = \Pi_{1,2}$.

## The LU method

• 1) Separate $A$ to an LU-product $A=L_0 \cdot U_0$.
• 2) Then trim $U_0$ by dropping the empty rows (getting $U$), and trim $L_0$ by trimming the corresponding columns (getting $L$). Note that it is still valid that $A=L \cdot U$.
• 3) Finally calculate $\Phi:\equiv U^T (U U^T)^{-1}$ and $\Psi:\equiv (L^T L)^{-1} L^T$, to get $A^{+} = \Phi \cdot\Psi$.

## Examples

Example 1:

• $A=\begin{pmatrix} 1&1&0\\ 0&1&1 \end{pmatrix}$,
• $A^{+}=\begin{pmatrix} \frac23&-\frac13\\ \frac13&\frac13 \\ -\frac13&\frac23 \end{pmatrix}$,
• $A\cdot A^{+} = {\bf 1}_2$,
• $A^{+} \cdot A = \begin{pmatrix} \frac23&\frac13&-\frac13 \\ \frac13&\frac23&\frac13 \\ -\frac13&\frac13&\frac23 \end{pmatrix}.$.

Example 2:

• $A=\begin{pmatrix} a\\ b \end{pmatrix}$,
• $A^{+}=\begin{pmatrix} \frac{a}{a^2+b^2}& \frac{b}{a^2+b^2} \end{pmatrix}$,
• $A^{+}\cdot A = {\bf 1}_2$,
• $A\cdot A^{+} =\begin{pmatrix} \frac{a^2}{a^2+b^2}& \frac{ab}{a^2+b^2 } \\ \frac{ab}{a^2+b^2}& \frac{b^2}{a^2+b^2} \end{pmatrix}.$.

Example 3:

• $A=\begin{pmatrix} \underline{a} \end{pmatrix}$,
• $A^{+}=\begin{pmatrix} \frac1{\underline{a}^T\cdot\underline{a}}\underline{a}^T \end{pmatrix} = \frac1{\underline{a}^T\cdot\underline{a}} A^T$,
• $A^{+}\cdot A = 1$,
• $A\cdot A^{+} = \underline{a} \underline{a}^T$.

Example 4:

• $A = \left(U | \underline{v}\right)$,
• $A^{+} = \begin{pmatrix} U^+ - \frac1{\underline{r}^T\cdot\underline{r}} U^{+}\underline{v}\underline{v}^T (1-U^{+T}U^T) \\ \frac1{\underline{r}^T\cdot\underline{r}}\underline{v}^T (1-U^{+T}U^T) \end{pmatrix}$, where $\underline{r} :\equiv (1- U U^{+}) \underline{v}$,
• $A^{+}\cdot A = \begin{pmatrix} U^{+} U & 0 \\ 0 & 1 \end{pmatrix}$,
• $A\cdot A^{+} = U U^{+} + \frac1{\underline{r}^T\cdot\underline{r}}\underline{r}\underline{r}^T.$.

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