Temperament addition: Difference between revisions

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Temperament arithmetic for <math>g_{\text{min}}>1</math> temperaments (again, that's with both <math>r>1</math> and <math>n>1</math>) can also be done using matrices, and it works in essentially the same way — entry-wise addition or subtraction — but there are some complications that make it significantly more involved than it is with multivectors. There's essentially five steps:

# Find the linear dependence basis <math>L_{\text{dep}}</math>

# Find the linear-dependence basis <math>L_{\text{dep}}</math>

# Put the matrices in a form with the <math>L_{\text{dep}}</math>

# Check for enfactoring, and perform an addabilization defactor (if necessary)

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And it's wrong not only because it is clearly enfactored (at least one factor of 2, that is visible in the first vector). The full explanation of why this is the wrong answer is beyond the scope of this example. However, if we now follow through with the instructions described above, we can find the correct answer.

'''1. Find the linear dependence basis.''' We know where to start: first find the <math>L_{\text{dep}}</math> and put each of these two mappings into a form that includes it explicitly. In this case, their <math>L_{\text{dep}}</math> consists of a single vector: {{ket|{{map|19 30 44 53}}}}.

'''1. Find the linear-dependence basis.''' We know where to start: first find the <math>L_{\text{dep}}</math> and put each of these two mappings into a form that includes it explicitly. In this case, their <math>L_{\text{dep}}</math> consists of a single vector: {{ket|{{map|19 30 44 53}}}}.

'''2. Reproduce the original temperament.''' The original matrices had two vectors, so as our second step, we pad out these matrices by drawing from vectors from the original matrices, starting from their first vectors, so now we have [{{map|19 30 44 53}} {{map|1 0 -4 -13}}⟩ and [{{map|19 30 44 53}} {{map|1 0 -4 17}}⟩. We could choose any vectors from the original matrices, as long as they are linearly independent from the ones we already have; if one is not, skip it and move on. In this case the first vectors are both fine, though.

@@ Line 698: / Line 698: @@
 Temperament arithmetic for <math>g_{\text{min}}>1</math> temperaments (again, that's with both <math>r>1</math> and <math>n>1</math>) can also be done using matrices, and it works in essentially the same way — entry-wise addition or subtraction — but there are some complications that make it significantly more involved than it is with multivectors. There's essentially five steps:
-# Find the linear dependence basis <span style="color: #3C8031;"><math>L_{\text{dep}}</math></span>
+# Find the linear-dependence basis <span style="color: #3C8031;"><math>L_{\text{dep}}</math></span>
 # Put the matrices in a form with the <span style="color: #3C8031;"><math>L_{\text{dep}}</math></span>
 # Check for enfactoring, and perform an addabilization defactor (if necessary)
@@ Line 781: / Line 781: @@
 And it's wrong not only because it is clearly enfactored (at least one factor of 2, that is visible in the first vector). The full explanation of why this is the wrong answer is beyond the scope of this example. However, if we now follow through with the instructions described above, we can find the correct answer.
-'''1. Find the linear dependence basis.''' We know where to start: first find the <span style="color: #3C8031;"><math>L_{\text{dep}}</math></span> and put each of these two mappings into a form that includes it explicitly. In this case, their <span style="color: #3C8031;"><math>L_{\text{dep}}</math></span> consists of a single vector: <span style="color: #3C8031;">{{ket|{{map|19 30 44 53}}}}</span>.
+'''1. Find the linear-dependence basis.''' We know where to start: first find the <span style="color: #3C8031;"><math>L_{\text{dep}}</math></span> and put each of these two mappings into a form that includes it explicitly. In this case, their <span style="color: #3C8031;"><math>L_{\text{dep}}</math></span> consists of a single vector: <span style="color: #3C8031;">{{ket|{{map|19 30 44 53}}}}</span>.
 '''2. Reproduce the original temperament.''' The original matrices had two vectors, so as our second step, we pad out these matrices by drawing from vectors from the original matrices, starting from their first vectors, so now we have [<span style="color: #3C8031;">{{map|19 30 44 53}}</span> {{map|1 0 -4 -13}}⟩ and [<span style="color: #3C8031;">{{map|19 30 44 53}}</span> {{map|1 0 -4 17}}⟩. We could choose any vectors from the original matrices, as long as they are <span style="color: #B6321C;">linearly independent</span> from the ones we already have; if one is not, skip it and move on. In this case the first vectors are both fine, though.