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

More at this blog:

More:

1. […] Some examples of the Moore-Penrose inverse […]

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Pingback by Moore-Perose inverse for light-cone vectors « Eikonal Blog — 2010.02.26 @ 16:54

2. […] by Some examples of the Moore-Penrose inverse « Eikonal Blog — 2010.10.28 @ 14:22 […]

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Pingback by Weighted Moore-Penrose inverse « Eikonal Blog — 2010.10.28 @ 14:25

3. A simpler Moore Penrose pseudo inverse for square matrices is:
$A=\begin{pmatrix}2&0 \\ 0 & 0\end{pmatrix}$ and $A^+=\begin{pmatrix}0.5 & 0 \\ 0 & 0 \end{pmatrix}$, am I right?

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Comment by P. Cooke — 2011.01.20 @ 01:40

• Sure, yes for such a simple example, where the matrix $A$ is proportional to a projector.

But it is not obvious what would such a generalized inverse be for more elaborated examples like this singular quadratic matrix $A=\begin{pmatrix} 1 & 2 & 0 \\ 0 & 0& 1 \\ 0 & 0 & 1 \end{pmatrix}$, or a non-quadratic matrix $A=\begin{pmatrix} 2 & 0 \\ 1 & 3 \\ 0 & -2 \end{pmatrix}$. The Moore-Penrose inverse is one specific prescription to get the unique “inverse” as the solution of the defining conditions 1-4. For some simple classes of matrices $A$ one can get some simplified expressions for their MP inverses (as displayed in the main body of this posting), but for the generic cases one has to do homework and solve conditions 1-4 afresh.

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Comment by sandokan65 — 2011.01.21 @ 13:04