# Eikonal Blog

## 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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### PanoptiClick

Filed under: surveillance — sandokan65 @ 15:46

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