Example: Solution to a Degenerated Diagonal SLE

(related to Section: Solving Simple Systems of Linear Equations)

We extend our previous example by some zero lines: The extended coefficient matrix has now the form

$$\left(\begin{array}{ccccccc|c}\alpha_{11}& 0&\ldots&0&0&\ldots&0&\beta_1\\ 0& \alpha_{22}&\ldots&0&\vdots&\vdots&\vdots&\beta_2\\ \vdots&\vdots&\ddots&\vdots&\vdots&\vdots&\vdots&\vdots\\ 0& 0 &\ldots&\alpha_{rr}&0&\ldots&0&\beta_r\\ \vdots& \vdots &\ldots&0&0&\ldots&0&\beta_{r+1}\\ \vdots& \vdots &\ldots&\vdots&\vdots&\ldots&\vdots&\vdots\\ 0& \ldots &\ldots&0&0&\ldots&0&\beta_m\\ \end{array}\right)$$

In this SLE, there exists an $r\in \{1,\ldots,\min(m,n)\},$ for which $\alpha_{jj}\neq 0$ for $j=1,\ldots,r,$ and $\alpha_{jj}=0$ for $j=r+1,\ldots,m.$

Like in the previous example, the solution of the SLE is given by $$x_j:=\frac{\beta_j}{\alpha_{jj}}$$ for $j=1,\ldots,r.$ For the remaining $x_j$ with $j=r+1,\ldots,n$ we can choose arbitrary values, if $\beta_j=0$ for all $j=r+1,\ldots m.$ Please note that the whole SLE has no (simultaneous) solution, if $\beta_j\neq 0$ for at least one of the "zero-lines" $j=r+1,\ldots m.$

Definitions: 1 2

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  1. Knabner, P; Barth, W.: "Lineare Algebra - Grundlagen und Anwendungen", Springer Spektrum, 2013