# 0.2 General solutions of simultaneous equations  (Page 2/4)

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• 1a. $M=N=r$ : One solution with no error, $\epsilon$ .
• 1b. $M=N>r$ : $\mathbf{b}\phantom{\rule{0.222222em}{0ex}}\in span\left\{\mathbf{A}\right\}$ : Many solutions with $\epsilon =\mathbf{0}$ .
• 1c. $M=N>r$ : $\mathbf{b}\phantom{\rule{0.222222em}{0ex}}not\in span\left\{\mathbf{A}\right\}$ : Many solutions with the same minimum error.
• 2a. $M>N=r$ : $\mathbf{b}\phantom{\rule{0.222222em}{0ex}}\in span\left\{\mathbf{A}\right\}$ : One solution $\epsilon =\mathbf{0}$ .
• 2b. $M>N=r$ : $\mathbf{b}\phantom{\rule{0.222222em}{0ex}}not\in span\left\{\mathbf{A}\right\}$ : One solution with minimum error.
• 2c. $M>N>r$ : $\mathbf{b}\phantom{\rule{0.222222em}{0ex}}\in span\left\{\mathbf{A}\right\}$ : Many solutions with $\epsilon =\mathbf{0}$ .
• 2d. $M>N>r$ : $\mathbf{b}\phantom{\rule{0.222222em}{0ex}}not\in span\left\{\mathbf{A}\right\}$ : Many solutions with the same minimum error.
• 3a. $N>M=r$ : Many solutions with $\epsilon =\mathbf{0}$ .
• 3b. $N>M>r$ : $\mathbf{b}\phantom{\rule{0.222222em}{0ex}}\in span\left\{\mathbf{A}\right\}$ : Many solutions with $\epsilon =\mathbf{0}$
• 3c. $N>M>r$ : $\mathbf{b}\phantom{\rule{0.222222em}{0ex}}not\in span\left\{\mathbf{A}\right\}$ : 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, $\mathbf{b}$ is necessarily in the span of $\mathbf{A}$ . 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.

$\left[\begin{array}{ccc}{a}_{11}& {a}_{12}& {a}_{13}\\ {a}_{21}& {a}_{22}& {a}_{23}\\ {a}_{31}& {a}_{32}& {a}_{33}\end{array}\right]\left[\begin{array}{c}{x}_{1}\\ {x}_{2}\\ {x}_{3}\end{array}\right]=\left[\begin{array}{c}{b}_{1}\\ {b}_{2}\\ {b}_{3}\end{array}\right]$

If the matrix has rank 3, then the $\mathbf{b}$ vector will necessarily be in the space spanned by the columns of $\mathbf{A}$ which puts it in case 1a. This can be solved for $\mathbf{x}$ by inverting $\mathbf{A}$ or using some more robust method. If the matrix has rank 1 or 2, the $\mathbf{b}$ 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 $\mathbf{A}$ is 4 x 3, then we have more equations than unknowns or the overspecified or overdetermined case.

$\left[\begin{array}{ccc}{a}_{11}& {a}_{12}& {a}_{13}\\ {a}_{21}& {a}_{22}& {a}_{23}\\ {a}_{31}& {a}_{32}& {a}_{33}\\ {a}_{41}& {a}_{42}& {a}_{43}\end{array}\right]\left[\begin{array}{c}{x}_{1}\\ {x}_{2}\\ {x}_{3}\end{array}\right]=\left[\begin{array}{c}{b}_{1}\\ {b}_{2}\\ {b}_{3}\\ {b}_{4}\end{array}\right]$

If this matrix has the maximum rank of 3, then we have case 2a or 2b depending on whether $\mathbf{b}$ is in the span of $\mathbf{A}$ or not. In either case, a unique solution $\mathbf{x}$ exists which can be found by [link] or [link] . For case 2a, we have a single exact solution with no equation error, $ϵ=\mathbf{0}$ 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 $\mathbf{A}$ is 3 x 4, then we have more unknowns than equations or the underspecified case.

$\left[\begin{array}{cccc}{a}_{11}& {a}_{12}& {a}_{13}& {a}_{14}\\ {a}_{21}& {a}_{22}& {a}_{23}& {a}_{24}\\ {a}_{31}& {a}_{32}& {a}_{33}& {a}_{34}\end{array}\right]\left[\begin{array}{c}{x}_{1}\\ {x}_{2}\\ {x}_{3}\\ {x}_{4}\end{array}\right]=\left[\begin{array}{c}{b}_{1}\\ {b}_{2}\\ {b}_{3}\end{array}\right]$

If this matrix has the maximum rank of 3, then we have case 3a and $\mathbf{b}$ must be in the span of $\mathbf{A}$ . For this case, many exact solutions $\mathbf{x}$ exist, all having zero equation error and a single one can be found with minimum solution norm $||\mathbf{x}||$ 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 $\epsilon$ be minimized and, if there is ambiguity (several solutions with the same minimum error), the ${l}_{2}$ norm squared of $\mathbf{x}$ also be minimized. A useful alternative tominimizing the norm of $\mathbf{x}$ is to require certain entries in $\mathbf{x}$ to be zero (sparse) or fixed to some non-zero value (equality constraints).

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