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  • 1a. M = N = r : One solution with no error, ε .
  • 1b. M = N > r : b s p a n { A } : Many solutions with ε = 0 .
  • 1c. M = N > r : b n o t s p a n { A } : Many solutions with the same minimum error.
  • 2a. M > N = r : b s p a n { A } : One solution ε = 0 .
  • 2b. M > N = r : b n o t s p a n { A } : One solution with minimum error.
  • 2c. M > N > r : b s p a n { A } : Many solutions with ε = 0 .
  • 2d. M > N > r : b n o t s p a n { A } : Many solutions with the same minimum error.
  • 3a. N > M = r : Many solutions with ε = 0 .
  • 3b. N > M > r : b s p a n { A } : Many solutions with ε = 0
  • 3c. N > M > r : b n o t s p a n { A } : Many solutions with the same minimum error.

Figure 1. Ten Cases for the Pseudoinverse.

Here we have:

  • case 1 has the same number of equations as unknowns ( A is square, M = N ),
  • case 2 has more equations than unknowns, therefore, is over specified ( A is taller than wide, M > N ),
  • case 3 has fewer equations than unknowns, therefore, is underspecified ( A is wider than tall N > M ).

This is a setting for frames and sparse representations.

In case 1a and 3a, b is necessarily in the span of A . In addition to these classifications, the possible orthogonality of thecolumns or rows of the matrices gives special characteristics.

Examples

Case 1: Here we see a 3 x 3 square matrix which is an example of case 1 in Figure 1 and 2.

a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 x 1 x 2 x 3 = b 1 b 2 b 3

If the matrix has rank 3, then the b vector will necessarily be in the space spanned by the columns of A which puts it in case 1a. This can be solved for x by inverting A or using some more robust method. If the matrix has rank 1 or 2, the b may or may not lie in the spanned subspace, so the classification will be 1b or 1c and minimization of | | x | | 2 2 yields a unique solution.

Case 2: If A is 4 x 3, then we have more equations than unknowns or the overspecified or overdetermined case.

a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 a 41 a 42 a 43 x 1 x 2 x 3 = b 1 b 2 b 3 b 4

If this matrix has the maximum rank of 3, then we have case 2a or 2b depending on whether b is in the span of A or not. In either case, a unique solution x exists which can be found by [link] or [link] . For case 2a, we have a single exact solution with no equation error, ϵ = 0 just as case 1a. For case 2b, we have a single optimal approximate solution with the least possible equation error. If the matrix hasrank 1 or 2, the classification will be 2c or 2d and minimization of | | x | | 2 2 yelds a unique solution.

Case 3: If A is 3 x 4, then we have more unknowns than equations or the underspecified case.

a 11 a 12 a 13 a 14 a 21 a 22 a 23 a 24 a 31 a 32 a 33 a 34 x 1 x 2 x 3 x 4 = b 1 b 2 b 3

If this matrix has the maximum rank of 3, then we have case 3a and b must be in the span of A . For this case, many exact solutions x exist, all having zero equation error and a single one can be found with minimum solution norm | | x | | using [link] or [link] . If the matrix has rank 1 or 2, the classification will be 3b or 3c.

Solutions

There are several assumptions or side conditions that could be used in order to define a useful unique solution of [link] . The side conditions used to define the Moore-Penrose pseudo-inverse are that the l 2 norm squared of the equation error ε be minimized and, if there is ambiguity (several solutions with the same minimum error), the l 2 norm squared of x also be minimized. A useful alternative tominimizing the norm of x is to require certain entries in x to be zero (sparse) or fixed to some non-zero value (equality constraints).

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Source:  OpenStax, Basic vector space methods in signal and systems theory. OpenStax CNX. Dec 19, 2012 Download for free at http://cnx.org/content/col10636/1.5
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