Temperament addition: Difference between revisions

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'''3. Defactor.''' Next, verify that both matrices are defactored. In this case, both matrices ''are'' enfactored, each by a factor of 30<ref>or you may prefer to think of this as three different (prime) factors: 2, 3, 5 (which multiply to 30)</ref>. So we'll use addabilization defactoring. Since there's only a single vector in the <math>L_{\text{dep}}</math>, therefore all we need to do is repeatedly add that one linearly dependent vector to the linearly independent vector until we find a vector with the target ~~GCF~~, which we can then simply divide out to defactor the matrix. Specifically, we add 11 times the linearly dependent vector. For the first matrix, {{map|1 0 -4 -13}} + 11⋅{{map|19 30 44 53}} = {{map|210 330 480 570}}, whose entries have a ~~GCF~~ = 30, so we can defactor the matrix by dividing that vector by 30, leaving us with {{map|7 11 16 19}}. Therefore the final matrix here is [{{map|19 30 44 53}} {{map|7 11 16 19}}⟩. The other matrix matrix happens to defactor in the same way: {{map|1 0 -4 17}} + 11⋅{{map|19 30 44 53}} = {{map|210 330 480 600}} whose ~~GCF~~ is also 30, reducing to {{map|7 11 16 20}}, so the final matrix is [{{map|19 30 44 53}} {{map|7 11 16 20}}⟩.

'''3. Defactor.''' Next, verify that both matrices are defactored. In this case, both matrices ''are'' enfactored, each by a factor of 30<ref>or you may prefer to think of this as three different (prime) factors: 2, 3, 5 (which multiply to 30)</ref>. So we'll use addabilization defactoring. Since there's only a single vector in the <math>L_{\text{dep}}</math>, therefore all we need to do is repeatedly add that one linearly dependent vector to the linearly independent vector until we find a vector with the target GCD, which we can then simply divide out to defactor the matrix. Specifically, we add 11 times the linearly dependent vector. For the first matrix, {{map|1 0 -4 -13}} + 11⋅{{map|19 30 44 53}} = {{map|210 330 480 570}}, whose entries have a GCD = 30, so we can defactor the matrix by dividing that vector by 30, leaving us with {{map|7 11 16 19}}. Therefore the final matrix here is [{{map|19 30 44 53}} {{map|7 11 16 19}}⟩. The other matrix matrix happens to defactor in the same way: {{map|1 0 -4 17}} + 11⋅{{map|19 30 44 53}} = {{map|210 330 480 600}} whose GCD is also 30, reducing to {{map|7 11 16 20}}, so the final matrix is [{{map|19 30 44 53}} {{map|7 11 16 20}}⟩.

'''4. Check negativity.''' The next thing we need to do is check the negativity of these two temperaments. If either of the matrices we're performing arithmetic on is negative, then we'll have to change it (if both are negative, then the problem cancels out, and we go back to being right). We check negativity by using the minors of these matrices. The first matrix's minors are (-1, -4, -10, -4, -13, -12) and the second matrix's minors are (-1, -4, 9, -4, 17, 32). What we're looking for here are their leading entries, because these are minors of a mapping (if we were looking at minors of comma bases, we'd be looking at the trailing entries instead). Specifically, we're looking to see if the leading entries are positive. They're not. Which tells us these matrices are both negative! But again, since they were ''both'' negative, the effect cancels out; no need to change anything (but if we wanted to, we could just take the linearly independent vector for each matrix and negate every entry in it).

Revision as of 01:52, 8 January 2022 view source Cmloegcmluin (talk \| contribs) autopatrolled, emailconfirmed 5,160 edits canonicalize is more specific than reduce ← Older edit		Revision as of 18:20, 8 January 2022 view source Cmloegcmluin (talk \| contribs) autopatrolled, emailconfirmed 5,160 edits GCF back to GCD, for better distention from "greatest factor", and it's more popular anyway Newer edit →
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	'''3. Defactor.''' Next, verify that both matrices are defactored. In this case, both matrices ''are'' enfactored, each by a factor of 30<ref>or you may prefer to think of this as three different (prime) factors: 2, 3, 5 (which multiply to 30)</ref>. So we'll use addabilization defactoring. Since there's only a single vector in the <span style="color: #3C8031;"><math>L_{\text{dep}}</math></span>, therefore all we need to do is repeatedly add that <span style="color: #3C8031;">one linearly dependent vector</span> to the <span style="color: #B6321C;">linearly independent vector</span> until we find a vector with the target ~~GCF~~, which we can then simply divide out to defactor the matrix. Specifically, we add 11 times the <span style="color: #3C8031;">linearly dependent vector</span>. For the first matrix, {{map\|1 0 -4 -13}} + 11⋅<span style="color: #3C8031;">{{map\|19 30 44 53}}</span> = {{map\|210 330 480 570}}, whose entries have a ~~GCF~~ = 30, so we can defactor the matrix by dividing that vector by 30, leaving us with <span style="color: #B6321C;">{{map\|7 11 16 19}}</span>. Therefore the final matrix here is [<span style="color: #3C8031;">{{map\|19 30 44 53}}</span> {{map\|7 11 16 19}}⟩. The other matrix matrix happens to defactor in the same way: {{map\|1 0 -4 17}} + 11⋅<span style="color: #3C8031;">{{map\|19 30 44 53}}</span> = {{map\|210 330 480 600}} whose ~~GCF~~ is also 30, reducing to <span style="color: #B6321C;">{{map\|7 11 16 20}}</span>, so the final matrix is [<span style="color: #3C8031;">{{map\|19 30 44 53}}</span> <span style="color: #B6321C;">{{map\|7 11 16 20}}</span>⟩.		'''3. Defactor.''' Next, verify that both matrices are defactored. In this case, both matrices ''are'' enfactored, each by a factor of 30<ref>or you may prefer to think of this as three different (prime) factors: 2, 3, 5 (which multiply to 30)</ref>. So we'll use addabilization defactoring. Since there's only a single vector in the <span style="color: #3C8031;"><math>L_{\text{dep}}</math></span>, therefore all we need to do is repeatedly add that <span style="color: #3C8031;">one linearly dependent vector</span> to the <span style="color: #B6321C;">linearly independent vector</span> until we find a vector with the target GCD, which we can then simply divide out to defactor the matrix. Specifically, we add 11 times the <span style="color: #3C8031;">linearly dependent vector</span>. For the first matrix, {{map\|1 0 -4 -13}} + 11⋅<span style="color: #3C8031;">{{map\|19 30 44 53}}</span> = {{map\|210 330 480 570}}, whose entries have a GCD = 30, so we can defactor the matrix by dividing that vector by 30, leaving us with <span style="color: #B6321C;">{{map\|7 11 16 19}}</span>. Therefore the final matrix here is [<span style="color: #3C8031;">{{map\|19 30 44 53}}</span> {{map\|7 11 16 19}}⟩. The other matrix matrix happens to defactor in the same way: {{map\|1 0 -4 17}} + 11⋅<span style="color: #3C8031;">{{map\|19 30 44 53}}</span> = {{map\|210 330 480 600}} whose GCD is also 30, reducing to <span style="color: #B6321C;">{{map\|7 11 16 20}}</span>, so the final matrix is [<span style="color: #3C8031;">{{map\|19 30 44 53}}</span> <span style="color: #B6321C;">{{map\|7 11 16 20}}</span>⟩.

	'''4. Check negativity.''' The next thing we need to do is check the negativity of these two temperaments. If either of the matrices we're performing arithmetic on is negative, then we'll have to change it (if both are negative, then the problem cancels out, and we go back to being right). We check negativity by using the minors of these matrices. The first matrix's minors are (-1, -4, -10, -4, -13, -12) and the second matrix's minors are (-1, -4, 9, -4, 17, 32). What we're looking for here are their leading entries, because these are minors of a mapping (if we were looking at minors of comma bases, we'd be looking at the trailing entries instead). Specifically, we're looking to see if the leading entries are positive. They're not. Which tells us these matrices are both negative! But again, since they were ''both'' negative, the effect cancels out; no need to change anything (but if we wanted to, we could just take the <span style="color: #B6321C;">linearly independent vector</span> for each matrix and negate every entry in it).		'''4. Check negativity.''' The next thing we need to do is check the negativity of these two temperaments. If either of the matrices we're performing arithmetic on is negative, then we'll have to change it (if both are negative, then the problem cancels out, and we go back to being right). We check negativity by using the minors of these matrices. The first matrix's minors are (-1, -4, -10, -4, -13, -12) and the second matrix's minors are (-1, -4, 9, -4, 17, 32). What we're looking for here are their leading entries, because these are minors of a mapping (if we were looking at minors of comma bases, we'd be looking at the trailing entries instead). Specifically, we're looking to see if the leading entries are positive. They're not. Which tells us these matrices are both negative! But again, since they were ''both'' negative, the effect cancels out; no need to change anything (but if we wanted to, we could just take the <span style="color: #B6321C;">linearly independent vector</span> for each matrix and negate every entry in it).