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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Filed under: surveillance — sandokan65 @ 15:46

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