Mathematik-Online lexicon: Example: Pseudo-Inverse

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	Mathematik-Online lexicon:
	Example: Pseudo-Inverse

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overview

We solve the least squares problem for

$\displaystyle A = \left(\begin{array}{rrr} 2 & -4 & 5 \\ 6 & 0 & 3 \\ 2 & -... ...quad b = \left(\begin{array}{r} 1 \\ 3 \\ -1 \\ 3 \end{array}\right)\, .$

In order to compute the singular value decomposition, we first determine the eigenvalues and eigenvectors of

$\displaystyle A^{\operatorname t}A = \left(\begin{array}{rrr} 80 & -16 & 56 \\ -16 & 32 & -40 \\ 56 & -40 & 68 \end{array}\right)$

and obtain

$\displaystyle s_1^2 = 144,\, s_2^2 = 36,\, s_3^2=0,\quad V = \frac{1}{3} \lef... ...{array}{rrr} 2 & 2 & -1 \\ -1 & 2 & 2 \\ 2 & -1 & 2 \end{array}\right)\, .$

This implies

$\displaystyle S = \left(\begin{array}{rrr} 12 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0... ... & 0 \\ 6 & 3 & 0 \\ 6 & -3 & 0 \\ 6 & 3 & 0 \end{array}\right) = US\, .$

We obtain the first two columns of

by dividing the respective columns of

by the corresponding singular values. (The resulting vectors must have norm 1.) We find the other columns (not uniquely determined) by completing the first two columns to an orthonormal basis:

$\displaystyle U = \frac{1}{2} \left(\begin{array}{rrrr} 1 & -1 & -1 & 1 \\ 1 & 1 & -1 & -1 \\ 1 & -1 & 1 & -1 \\ 1 & 1 & 1 & 1 \end{array}\right).$

Thus, the pseudo-inverse (Moore-Penrose inverse) is

$\displaystyle A^+ = V \left(\begin{array}{cccc} \frac{1}{12} & 0 & 0 & 0 \\ ... ...} -2 & 6 & -2 & 6 \\ -5 & 3 & -5 & 3 \\ 4 & 0 & 4 & 0 \end{array}\right),$

and

$\displaystyle x = A^+b = \frac{1}{4}\left(\begin{array}{c} 2 \\ 1 \\ 0\end{array}\right)$

is the solution of minimal norm.

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automatisch erstellt am 20. 3. 2007