Temperament addition: Difference between revisions

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(WIP)

=== Sintel's proof of the linear-independence conjecture===

=== Sintel's proof of the linear-independence conjecture===

If A and B are mappings from Z^n to Z^m, with n > m, A, B full rank (using A and B as their rowspace equivalently):

====Sintel's original text====

<nowiki>If A and B are mappings from Z^n to Z^m, with n > m, A, B full rank (using A and B as their rowspace equivalently):

dim(A + B) - m = dim(ker(A) + ker(B)) - (n-m)

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m + m - m = n - n + m

m = m

m = m</nowiki>

====Douglas Blumeyer's interpretation====

We're going to take the strategy of beginning with what we're trying to prove, then reducing it to an obvious equivalence, which will show that our initial statement must be just as true.

So here's the statement we're trying to prove:

<math>\text{rank}(\text{union}(M_1, M_2)) - r = \text{nullity}(\text{union}(C_1, C_2)) - n</math>

<math>M_1</math> and <math>M_2</math> are mappings which both have dimensionality <math>d</math>, rank <math>r</math>, nullity <math>n</math>, and are full-rank, and <math>C_1</math> and <math>C_2</math> are their comma bases, respectively.

So, the left-hand side of this equation is a way to express the count of linearly independent basis vectors <math>L_{\text{ind}}</math> existing between <math>M_1</math> and <math>M_2</math>. The right-hand side tells you the same thing, but between <math>C_1</math> and <math>C_2</math>. The fact that these two things are equal is the thing we're trying to prove. So let's go!

Let's call the following Equation B. This makes sense because basis vectors between <math>M_1</math> and <math>M_2</math> are either going to be linearly dependent or linearly independent. The union is going to be all of <math>M_1</math>'s independent vectors, all of <math>M_2</math>'s independent vectors, and all of <math>M_1</math> and <math>M_2</math>'s dependent vectors but only one copy of them. While the intersection is going be all of <math>M_1</math> and <math>M_2</math>'s dependent vectors again — essentially the other copy of them. So they sum to the same thing.

<math>\text{rank}(M_1) + \text{rank}(M_2) = \text{rank}(\text{union}(M_1, M_2)) + \text{rank}(\text{intersection}(M_1, M_2))</math>

Then this is just Equation B, rearranged.

<math>\text{rank}(\text{union}(M_1, M_2)) = \text{rank}(M_1) + \text{rank}(M_2) - \text{rank}(\text{intersection}(M_1, M_2))</math>

This takes Equation B, solves it for <math>\text{rank}(\text{union}(M_1, M_2))</math>, then substitutes that result into Equation A, which is then flipped left/right, and then <math>r</math> is subtracted from both sides.

<math>\text{rank}(M_1) + \text{rank}(M_2) - \text{rank}(\text{intersection}(M_1, M_2)) - r = \text{nullity}(\text{union}(C_1, C_2)) - n</math>

By the "duality of the comma basis", this is Equation C:

<math>\text{nullity}(\text{union}(C_1, C_2)) = \text{nullity}(\text{nullspace}(\text{intersection}(M_1, M_2)))</math>

Now substitute in the right-hand side of Equation C for <math>\text{nullity}(\text{union}(C_1, C_2))</math> in Equation B.

<math>\text{rank}(M_1) + \text{rank}(M_2) - \text{rank}(\text{intersection}(M_1, M_2)) - r = \text{nullity}(\text{nullspace}(\text{intersection}(M_1, M_2))) - n</math>

This is the rank nullity theorem where <math>\text{intersection}(M_1, M_2)</math> is the temperament. Let's call it Equation D:

<math>\text{nullity}(\text{nullspace}(\text{intersection}(M_1, M_2))) + \text{rank}(\text{intersection}(M_1, M_2)) = d</math>

Now solve Equation D for <math>\text{nullity}(\text{nullspace}(\text{intersection}(M_1, M_2)))</math>, and substitute that result into Equation B:

<math>\text{rank}(M_1) + \text{rank}(M_2) - \text{rank}(\text{intersection}(M_1, M_2)) - r = d - \text{rank}(\text{intersection}(M_1, M_2)) - n</math>

Now realize that <math>\text{rank}(M_1)</math> and <math>\text{rank}(M_2)</math> are both equal to <math>r</math>.

<math>r + r - \text{rank}(\text{intersection}(M_1, M_2)) - r = d - \text{rank}(\text{intersection}(M_1, M_2)) - n</math>

Now cancel the <math>\text{rank}(\text{intersection}(M_1, M_2))</math> from both sides, and substitute in <math>(d - r)</math> for <math>n</math>.

Now cancel the <math>r</math>'s on the left and the <math>d</math>'s on the right:

So we know this is true.

==Multivector approach==

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 (WIP)
-=== Sintel's proof of the linear-independence conjecture===
+=== Sintel's proof of the <span style="color: #B6321C;">linear-independence</span> conjecture===
-If A and B are mappings from Z^n to Z^m, with n > m, A, B full rank (using A and B as their rowspace equivalently):
+====Sintel's original text====
+ <nowiki>If A and B are mappings from Z^n to Z^m, with n > m, A, B full rank (using A and B as their rowspace equivalently):
 dim(A + B) - m = dim(ker(A) + ker(B)) - (n-m)
@@ Line 602: / Line 604: @@
 m + m - m = n - n + m
-m = m
+m = m</nowiki>
+====Douglas Blumeyer's interpretation====
+We're going to take the strategy of beginning with what we're trying to prove, then reducing it to an obvious equivalence, which will show that our initial statement must be just as true.
+So here's the statement we're trying to prove:
+<math>\text{rank}(\text{union}(M_1, M_2)) - r = \text{nullity}(\text{union}(C_1, C_2)) - n</math>
+<math>M_1</math> and <math>M_2</math> are mappings which both have dimensionality <math>d</math>, rank <math>r</math>, nullity <math>n</math>, and are full-rank, and <math>C_1</math> and <math>C_2</math> are their comma bases, respectively.
+So, the left-hand side of this equation is a way to express the count of <span style="color: #B6321C;">linearly independent basis vectors <math>L_{\text{ind}}</math></span> existing between <math>M_1</math> and <math>M_2</math>. The right-hand side tells you the same thing, but between <math>C_1</math> and <math>C_2</math>. The fact that these two things are equal is the thing we're trying to prove. So let's go!
+Let's call the following Equation B. This makes sense because basis vectors between <math>M_1</math> and <math>M_2</math> are either going to be <span style="color: #3C8031;">linearly dependent</span> or <span style="color: #B6321C;">linearly independent</span>. The union is going to be all of <math>M_1</math>'s <span style="color: #B6321C;">independent vectors</span>, all of <math>M_2</math>'s <span style="color: #B6321C;">independent vectors</span>, and all of <math>M_1</math> and <math>M_2</math>'s <span style="color: #3C8031;">dependent vectors</span> but only one copy of them. While the intersection is going be all of <math>M_1</math> and <math>M_2</math>'s <span style="color: #3C8031;">dependent vectors</span> again — essentially the other copy of them. So they sum to the same thing.
+<math>\text{rank}(M_1) + \text{rank}(M_2) = \text{rank}(\text{union}(M_1, M_2)) + \text{rank}(\text{intersection}(M_1, M_2))</math>
+Then this is just Equation B, rearranged.
+<math>\text{rank}(\text{union}(M_1, M_2)) = \text{rank}(M_1) + \text{rank}(M_2) - \text{rank}(\text{intersection}(M_1, M_2))</math>
+This takes Equation B, solves it for <math>\text{rank}(\text{union}(M_1, M_2))</math>, then substitutes that result into Equation A, which is then flipped left/right, and then <math>r</math> is subtracted from both sides.
+<math>\text{rank}(M_1) + \text{rank}(M_2) - \text{rank}(\text{intersection}(M_1, M_2)) - r = \text{nullity}(\text{union}(C_1, C_2)) - n</math>
+By the "duality of the comma basis", this is Equation C:
+<math>\text{nullity}(\text{union}(C_1, C_2)) = \text{nullity}(\text{nullspace}(\text{intersection}(M_1, M_2)))</math>
+Now substitute in the right-hand side of Equation C for <math>\text{nullity}(\text{union}(C_1, C_2))</math> in Equation B.
+<math>\text{rank}(M_1) + \text{rank}(M_2) - \text{rank}(\text{intersection}(M_1, M_2)) - r = \text{nullity}(\text{nullspace}(\text{intersection}(M_1, M_2))) - n</math>
+This is the rank nullity theorem where <math>\text{intersection}(M_1, M_2)</math> is the temperament. Let's call it Equation D:
+<math>\text{nullity}(\text{nullspace}(\text{intersection}(M_1, M_2))) + \text{rank}(\text{intersection}(M_1, M_2)) = d</math>
+Now solve Equation D for <math>\text{nullity}(\text{nullspace}(\text{intersection}(M_1, M_2)))</math>, and substitute that result into Equation B:
+<math>\text{rank}(M_1) + \text{rank}(M_2) - \text{rank}(\text{intersection}(M_1, M_2)) - r = d - \text{rank}(\text{intersection}(M_1, M_2)) - n</math>
+Now realize that <math>\text{rank}(M_1)</math> and <math>\text{rank}(M_2)</math> are both equal to <math>r</math>.
+<math>r + r - \text{rank}(\text{intersection}(M_1, M_2)) - r = d - \text{rank}(\text{intersection}(M_1, M_2)) - n</math>
+Now cancel the <math>\text{rank}(\text{intersection}(M_1, M_2))</math> from both sides, and substitute in <math>(d - r)</math> for <math>n</math>.
+<math>r + r - r = d - d + r</math>
+Now cancel the <math>r</math>'s on the left and the <math>d</math>'s on the right:
+<math>r = r</math>
+So we know this is true.
 ==Multivector approach==