Matrix Divide: Difference between revisions

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{{Built-in|Matrix Divide|⌹}} is a [[dyadic function]] that performs [[wikipedia:matrix division|matrix division]] between two [[argument|arguments]]. It shares the [[glyph]] <source lang=apl inline>⌹</source> with the monadic function [[Matrix Inverse]].
{{Built-in|Matrix Divide|⌹}} is a [[dyadic function]] that performs [[wikipedia:matrix division|matrix division]] between two [[argument]]s of rank 2 or less. Some dialects automatically apply it to rank-2 [[subarray]]s of higher-rank arguments. It shares the [[glyph]] ''Quad Divide'' <syntaxhighlight lang=apl inline>⌹</syntaxhighlight> (often called ''Domino'') with the monadic function [[Matrix Inverse]].


== Examples ==
== Examples ==


The result of <source lang=apl inline>X⌹Y</source> is equal to <source lang=apl inline>(⌹Y)+.×X</source>, which is analogous to <source lang=apl inline>X÷Y</source> being equal to <source lang=apl inline>(÷Y)×X</source>. As a consequence, <source lang=apl inline>X≡Y+.×X⌹Y</source> is true for square matrices.
The result of <syntaxhighlight lang=apl inline>X⌹Y</syntaxhighlight> is equal to <syntaxhighlight lang=apl inline>(⌹Y)+.×X</syntaxhighlight>, which is analogous to <syntaxhighlight lang=apl inline>X÷Y</syntaxhighlight> being equal to <syntaxhighlight lang=apl inline>(÷Y)×X</syntaxhighlight>. As a consequence, <syntaxhighlight lang=apl inline>X≡Y+.×X⌹Y</syntaxhighlight> is true for square matrices.


<source lang=apl>
<syntaxhighlight lang=apl>
       ⎕←X←2 2⍴1 2 3 4
       ⎕←X←2 2⍴1 2 3 4
1 2
1 2
Line 20: Line 20:
       X≡Y+.×X⌹Y
       X≡Y+.×X⌹Y
1
1
</source>
</syntaxhighlight>


== Applications ==
== Applications ==
Line 28: Line 28:
The following example solves the system of equations <math>x + 2y = 5, 2x - y = 8</math>. The answer is <math>x=4.2, y=0.4</math>.
The following example solves the system of equations <math>x + 2y = 5, 2x - y = 8</math>. The answer is <math>x=4.2, y=0.4</math>.


<source lang=apl>
<syntaxhighlight lang=apl>
       ⎕←X←2 2⍴1 2 2 ¯1
       ⎕←X←2 2⍴1 2 2 ¯1
1  2
1  2
Line 35: Line 35:
       Y⌹X
       Y⌹X
4.2 0.4
4.2 0.4
</source>
</syntaxhighlight>


The following example solves the linear least squares over the five points <math>(1,5), (2,1), (3,4), (4,2), (5,8)</math>. The answer is <math>y=1.9 + 0.7x</math>.
The following example solves the linear least squares over the five points <math>(1,5), (2,1), (3,4), (4,2), (5,8)</math>. The answer is <math>y=1.9 + 0.7x</math>.


<source lang=apl>
<syntaxhighlight lang=apl>
       ⎕←X←1,⍪⍳5
       ⎕←X←1,⍪⍳5
1 1
1 1
Line 49: Line 49:
       Y⌹X
       Y⌹X
1.9 0.7
1.9 0.7
</source>
</syntaxhighlight>


When used with real vectors as both arguments, <source lang=apl inline>Y×X⌹Y</source> gives the [[wikipedia:Projection (linear algebra)#Finding projection with an inner product|projection]] of X onto a basis vector Y. The remaining component of X, namely <source lang=apl inline>R←X-Y×X⌹Y</source>, is [[wikipedia:Orthogonality#Euclidean vector spaces|orthogonal]] to Y (<source lang=apl inline>R+.×Y</source> is zero).
When used with real vectors as both arguments, <syntaxhighlight lang=apl inline>Y×X⌹Y</syntaxhighlight> gives the [[wikipedia:Projection (linear algebra)#Finding projection with an inner product|projection]] of X onto a basis vector Y. The remaining component of X, namely <syntaxhighlight lang=apl inline>R←X-Y×X⌹Y</syntaxhighlight>, is [[wikipedia:Orthogonality#Euclidean vector spaces|orthogonal]] to Y (<syntaxhighlight lang=apl inline>R+.×Y</syntaxhighlight> is zero).


<source lang=apl>
<syntaxhighlight lang=apl>
       (X Y)←(2 7)(3 1)
       (X Y)←(2 7)(3 1)
       X⌹Y
       X⌹Y
Line 63: Line 63:
       ⎕CT>|Y+.×X-Y×X⌹Y  ⍝ ∧ is orthogonal to Y (with negligible error)
       ⎕CT>|Y+.×X-Y×X⌹Y  ⍝ ∧ is orthogonal to Y (with negligible error)
1
1
</source>
</syntaxhighlight>


== External links ==
== External links ==
Line 73: Line 73:
=== Documentation ===
=== Documentation ===


* [http://help.dyalog.com/latest/#Language/Primitive%20Functions/Matrix%20Divide.htm Dyalog]
* [https://help.dyalog.com/latest/#Language/Primitive%20Functions/Matrix%20Divide.htm Dyalog]
* [http://microapl.com/apl_help/ch_020_020_280.htm APLX]
* [http://microapl.com/apl_help/ch_020_020_280.htm APLX]
* [http://wiki.nars2000.org/index.php/Matrix_Inverse/Divide NARS2000]
* [http://wiki.nars2000.org/index.php/Matrix_Inverse/Divide NARS2000]
* J [https://www.jsoftware.com/help/dictionary/d131.htm Dictionary], [https://code.jsoftware.com/wiki/Vocabulary/percentdot#dyadic NuVoc] (as <source lang=j inline>%.</source>)
* J [https://www.jsoftware.com/help/dictionary/d131.htm Dictionary], [https://code.jsoftware.com/wiki/Vocabulary/percentdot#dyadic NuVoc] (as <syntaxhighlight lang=j inline>%.</syntaxhighlight>)


{{APL built-ins}}[[Category:Primitive functions]]
{{APL built-ins}}[[Category:Primitive functions]]

Revision as of 22:16, 10 September 2022

Matrix Divide () is a dyadic function that performs matrix division between two arguments of rank 2 or less. Some dialects automatically apply it to rank-2 subarrays of higher-rank arguments. It shares the glyph Quad Divide (often called Domino) with the monadic function Matrix Inverse.

Examples

The result of X⌹Y is equal to (⌹Y)+.×X, which is analogous to X÷Y being equal to (÷Y)×X. As a consequence, X≡Y+.×X⌹Y is true for square matrices.

      ⎕←X←2 2⍴1 2 3 4
1 2
3 4
      ⎕←Y←2 2⍴5 6 7 8
5 6
7 8
      X⌹Y
 5  4
¯4 ¯3
      (⌹Y)+.×X
 5  4
¯4 ¯3
      X≡Y+.×X⌹Y
1

Applications

From the properties of Moore-Penrose inverse (which Matrix Inverse uses), Matrix Divide can not only be used to solve a system of linear equations, but also to find the linear least squares solution to an overdetermined system.

The following example solves the system of equations . The answer is .

      ⎕←X←2 2⍴1 2 2 ¯1
1  2
2 ¯1
      Y←5 8
      Y⌹X
4.2 0.4

The following example solves the linear least squares over the five points . The answer is .

      ⎕←X←1,⍪⍳5
1 1
1 2
1 3
1 4
1 5
      Y←5 1 4 2 8
      Y⌹X
1.9 0.7

When used with real vectors as both arguments, Y×X⌹Y gives the projection of X onto a basis vector Y. The remaining component of X, namely R←X-Y×X⌹Y, is orthogonal to Y (R+.×Y is zero).

      (X Y)←(2 7)(3 1)
      X⌹Y
1.3
      Y×X⌹Y  ⍝ Projection of X onto Y
3.9 1.3
      X-Y×X⌹Y  ⍝ The remaining component in X
¯1.9 5.7
      ⎕CT>|Y+.×X-Y×X⌹Y  ⍝ ∧ is orthogonal to Y (with negligible error)
1

External links

Lesson

Documentation


APL built-ins [edit]
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Computational MatchNot MatchMembershipFindNub SieveEncodeDecodeMatrix InverseMatrix DivideFormatExecuteMaterialiseRange
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