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look at the text below:

A mapping $w = {{az + b} \over {cz + b}}$ is determined by a, b, c, d, actually by the ratios of three of these constants to the fourth because we can drop or introduce a common factor.

What does the author mean by the last sentence (we can drop or introduce a common factor)? I mean, what is this common factor? where does this common factor come from? The book mentions this But doesnt give more details.

This text is from the book of ''advanced engineering mathematics'' written by Erwin Kreyszig 10th edition page 746. This is a photo of the page: enter image description here

Arashrostami
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    The expression ${{az + b} \over {cz + b}}$ does not change if all coefficients $a, b, c, d$ are multiplied with the same factor. – Martin R Apr 10 '22 at 15:15
  • I know, but what creates this factor? For example, if we assume ${a \over d} = 1,{b \over d} = 2,{c \over d} = 3$ then it will produce a common factor? – Arashrostami Apr 10 '22 at 15:21
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    If (for example) $c \ne 0$ then ${{(a/c)z + (b/c)} \over {z + (b/c)}}$, which means that a linear transformation is determined by three parameters and not by four. – Martin R Apr 10 '22 at 15:24
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    @MartinR I think your comments basically constitute an answer – Mark S. Apr 10 '22 at 15:26
  • Oh. Now I understand. We can multiply all the constants in a common factor, so if we know, for example, the ratios of a, b, and c to d, we can divide all the constants by a factor d to get $w = {{({a \over c})z + ({b \over c})} \over {z + ({b \over c})}}$ Where all coefficients are known. – Arashrostami Apr 10 '22 at 15:36

2 Answers2

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The expression ${{az + b} \over {cz + d}}$ does not change if all coefficients $a, b, c, d$ are multiplied by a “common (nonzero) factor” $f$: $$ {{az + b} \over {cz + d}} = {{(af)z + (bf)} \over {(cf)z + (df)}} \, . $$ If (for example) $c \ne 0$ then we can choose $f=1/c$: $$ {{az + b} \over {cz + d}} = {{(a/c)z + (b/c)} \over {z + (d/c)}} \, . $$ This is what the author means by saying that the mapping is determined by the relative ratios of the coefficients. It shows that a linear transformation is determined by three parameters and not by four.

Martin R
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We can multiply all the constants in a common factor, so if we know, for example, the ratios of a, b, and c to d, we can divide all the constants by a factor d to get $w={{\left({{{a}\over{d}}}\right)z+\left({{{b}\over{d}}}\right)}\over{\left({{{c}\over{d}}}\right)z+1}}$ Where all coefficients are known.this means that a linear transformation is determined by three parameters and not by four.

superyamky
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