Matrix echelon forms: Difference between revisions

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The multiplication of rows by whatever integer is required to clear the denominators of the RREF form is the action which is responsible for causing enfactoring, whenever the IRREF produces an enfactored mapping from a defactored one.

The multiplication of rows by whatever integer is required to clear the denominators of the RREF form is the action which is responsible for causing [[enfactoring]], whenever the IRREF produces an enfactored mapping from a defactored one.

It is not possible for an RREF to be IREF without also being IRREF.

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So HNF has a lot in common with IRREF, which is the IREF you find by converting the RREF, but it is not always the same as IRREF.

~~Defactored~~ [[Canonical form]]~~, or DCF,~~ is closely related to HNF, because the second step of finding the DCF is taking the HNF. So the ~~DCF~~ is always ''a'' HNF, and therefore it has all the same properties of being echelon, integer, and normalized, and in turn therefore it also provides a unique representation. However it is not necessary ''the'' same HNF of the original mapping, due to the first step being defactoring; it is the same as as HNF except when the original mapping is enfactored.

[[Canonical form]] is closely related to HNF, because the second step of finding the DCF is taking the HNF. So the canonical form is always ''a'' HNF, and therefore it has all the same properties of being echelon, integer, and normalized, and in turn therefore it also provides a unique representation. However it is not necessary ''the'' same HNF of the original mapping, due to the first step being defactoring; it is the same as as HNF except when the original mapping is enfactored.

In the below example, <math>p_{ij}</math> represents any positive integer, and <math>a_{ij}</math> represents any nonnegative integer less than the <math>p</math> in its column.

{| class="wikitable" style="text-align: center;"

|+HNF (and ~~DFC~~)

|+HNF (and canonical form)

|style="width: 25px; background-color: #e69138;"|p₁₁

|style="width: 25px; background-color: #ffe599;"|a₁₂

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[[File:Cases for temperament mapping forms3.png|300px|thumb|right]]

Considering only full-rank, integer mappings, we find three cases for a given temperament which is not enfactored. In all three cases, HNF is the same as DCF:

Considering only full-rank, integer mappings, we find three cases for a given temperament which is not enfactored. In all three cases, HNF is the same as canonical form (here abbreviated as DCF, for defactored canonical form):

# The RREF, IRREF, and HNF are all ''different''. Example: [[Porcupine_family#Porcupine|porcupine]] with RREF of {{ket|{{map|1 0 <math>-\frac13</math>}} {{map|0 1 <math>\frac53</math>}}}}, IRREF of {{ket|{{map|3 0 -1}} {{map|0 3 5}}}}, and HNF of {{ket|{{map|1 2 3}} {{map|0 3 5}}}}.

# The RREF, IRREF, HNF are all ''the same''. Example: [[Meantone_family#Meantone_.2812.2619.2C_2.3.5.29|meantone]] with all equal to {{ket|{{map|1 0 -4}} {{map|0 1 4}}}}. This case is quite rare.

@@ Line 132: / Line 132: @@
 |}
-The multiplication of rows by whatever integer is required to clear the denominators of the RREF form is the action which is responsible for causing enfactoring, whenever the IRREF produces an enfactored mapping from a defactored one.
+The multiplication of rows by whatever integer is required to clear the denominators of the RREF form is the action which is responsible for causing [[enfactoring]], whenever the IRREF produces an enfactored mapping from a defactored one.
 It is not possible for an RREF to be IREF without also being IRREF.
@@ Line 142: / Line 142: @@
 So HNF has a lot in common with IRREF, which is the IREF you find by converting the RREF, but it is not always the same as IRREF.
-Defactored [[Canonical form]], or DCF, is closely related to HNF, because the second step of finding the DCF is taking the HNF. So the DCF is always ''a'' HNF, and therefore it has all the same properties of being echelon, integer, and normalized, and in turn therefore it also provides a unique representation. However it is not necessary ''the'' same HNF of the original mapping, due to the first step being defactoring; it is the same as as HNF except when the original mapping is enfactored.
+[[Canonical form]] is closely related to HNF, because the second step of finding the DCF is taking the HNF. So the canonical form is always ''a'' HNF, and therefore it has all the same properties of being echelon, integer, and normalized, and in turn therefore it also provides a unique representation. However it is not necessary ''the'' same HNF of the original mapping, due to the first step being defactoring; it is the same as as HNF except when the original mapping is enfactored.
 In the below example, <span><math>p_{ij}</math></span> represents any positive integer, and <span><math>a_{ij}</math></span> represents any nonnegative integer less than the <span><math>p</math></span> in its column.
 {| class="wikitable" style="text-align: center;"
-|+HNF (and DFC)
+|+HNF (and canonical form)
 |style="width: 25px; background-color: #e69138;"|p₁₁
 |style="width: 25px; background-color: #ffe599;"|a₁₂
@@ Line 195: / Line 195: @@
 [[File:Cases for temperament mapping forms3.png|300px|thumb|right]]
-Considering only full-rank, integer mappings, we find three cases for a given temperament which is not enfactored. In all three cases, HNF is the same as DCF:
+Considering only full-rank, integer mappings, we find three cases for a given temperament which is not enfactored. In all three cases, HNF is the same as canonical form (here abbreviated as DCF, for defactored canonical form):
 # The RREF, IRREF, and HNF are all ''different''. Example: [[Porcupine_family#Porcupine|porcupine]] with RREF of {{ket|{{map|1 0 <span><math>-\frac13</math></span>}} {{map|0 1 <span><math>\frac53</math></span>}}}}, IRREF of {{ket|{{map|3 0 -1}} {{map|0 3 5}}}}, and HNF of {{ket|{{map|1 2 3}} {{map|0 3 5}}}}.
 # The RREF, IRREF, HNF are all ''the same''. Example: [[Meantone_family#Meantone_.2812.2619.2C_2.3.5.29|meantone]] with all equal to {{ket|{{map|1 0 -4}} {{map|0 1 4}}}}. This case is quite rare.