Section 1.5 - Exploiting partitioning of matrices and · PDF file Section 1.5 - Exploiting...
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Transcript of Section 1.5 - Exploiting partitioning of matrices and · PDF file Section 1.5 - Exploiting...
Section 1.5 - Exploiting partitioning of matrices and vectors
Maggie Myers Robert A. van de Geijn
The University of Texas at Austin
Practical Linear Algebra – Fall 2009
http://z.cs.utexas.edu/wiki/pla.wiki/ 1
Example: Partitioning vectors
Given a vector
x =
4 −1 3 1
one can think of this as its four elements
x =
χ0 χ1 χ2 χ3
where χ0 = 4, χ1 = −1, etc. Note
The parentheses are only there to delimit (outline) the vector. They have no particular other meaning.
http://z.cs.utexas.edu/wiki/pla.wiki/ 2
Example: Partitioning vectors
Given a vector
x =
4 −1 3 1
one can think of this as two subvectors:
x = ( x0 x1
) =
(
4 −1
) (
3 1
) =
4 −1 3 1
so that
x0 = (
4 −1
) and x1 =
( 3 1
)
http://z.cs.utexas.edu/wiki/pla.wiki/ 3
Example: Partitioning vectors
Given a vector
x =
4 −1 3 1
one can think of this as two subvectors:
x = ( x0 x1
) =
(
4 ) −13
1
=
4 −1 3 1
so that
x0 = (
4 )
and x1 =
−13 1
http://z.cs.utexas.edu/wiki/pla.wiki/ 4
Example: Partitioning vectors
Given a vector
x =
4 −1 3 1
one can think of this as two subvectors:
x = ( x0 x1
) =
( )
4 −1 3 1
=
4 −1 3 1
so that
x0 = ( )
and x1 =
4 −1 3 1
http://z.cs.utexas.edu/wiki/pla.wiki/ 5
Example: Inner product with partitioned vectors
Given vector
x = ( x0 x1
) =
4 −1 3 1
and y = ( y0y1 )
=
1 −2 3 −4
We find that
xT y = (4)× (1) + (−1)× (−2)︸ ︷︷ ︸ = (
4 −1
)T ( 1 −2
) = xT0 y0
+ (3)× (−3) + (1)×−4)︸ ︷︷ ︸ = (
3 1
)T ( 3 −4
) = xT1 y1︸ ︷︷ ︸
= xT0 y0 + x T 1 y1
http://z.cs.utexas.edu/wiki/pla.wiki/ 6
Theorem
Let x, y ∈ Rn and partition
x =
x0 x1 ...
xN−1
and y =
y0 y1 ...
yN−1
, where xi and yi have the same size, for i = 0, . . . , N − 1.
http://z.cs.utexas.edu/wiki/pla.wiki/ 7
Theorem (continued)
Then
xT y =
x0 x1 ...
xN−1
T
y0 y1 ...
yN−1
= ( xT0 x
T 1 · · · xTN−1
)
y0 y1 ...
yN−1
= xT0 y0 + x
T 1 y1 + · · ·+ xTN−1yN−1
http://z.cs.utexas.edu/wiki/pla.wiki/ 8
Example: axpywith partitioned vectors
Given
x = ( x0 x1
) =
4 −1 3 1
, y = ( y0y1 )
=
1 −2 3 −4
, and α = 4 We find that
αx+ y = 4
4 −1 3 1
+
1 −2 3 −4
=
(4)× (4) + (1) (4)× (−1) + (−2)
(4)× (3) + (3) (4)× (1) + (−4)
=
4 (
4 −1
) + (
1 −2
) 4 (
3 1
) + (
3 −4
) = ( αx0 + y0αx1 + y1
)
http://z.cs.utexas.edu/wiki/pla.wiki/ 9
Theorem
Let x, y ∈ Rn, α ∈ R, and partition
x =
x0 x1 ...
xN−1
and y =
y0 y1 ...
yN−1
, where xi and yi have the same size, for i = 0, . . . , N − 1.
http://z.cs.utexas.edu/wiki/pla.wiki/ 10
Theorem (continued)
Then
αx+ y = α
x0 x1 ...
xN−1
+
y0 y1 ...
yN−1
=
αx0 + y0 αx1 + y1
...
αxN−1 + yN−1
http://z.cs.utexas.edu/wiki/pla.wiki/ 11
Partitioning matrices
A =
A00 a01 A02aT10 α11 aT12 A20 a21 A22
= −1 2 4 1 0
1 0 −1 −2 1 2 −1 3 1 2 1 2 3 4 3 −1 −2 0 1 2
Pronounce a21 as a-two-one instead of a-twentyone, please.
Notice how the labels “A”, “a”, “α” are used.
Why do we use the label “aT10”?
http://z.cs.utexas.edu/wiki/pla.wiki/ 12
Example: blocked matrix-vector multiplication
Consider
A =
A00 a01 A02aT10 α11 aT12 A20 a21 A22
= −1 2 4 1 0
1 0 −1 −2 1 2 −1 3 1 2 1 2 3 4 3 −1 −2 0 1 2
,
x =
x0χ1 x2
=
1 2 3 4 5
, and y = y0ψ1
y2
,
where y0, y2 ∈ R2.
http://z.cs.utexas.edu/wiki/pla.wiki/ 13
Example (continued)
Then Ax =0BBBB@ −1 2 4 1 0
1 0 −1 −2 1 2 −1 3 1 2 1 2 3 4 3 −1 −2 0 1 2
1CCCCA 0BBBB@
1 2
3
4 5
1CCCCA = 0BBBB@
(−1)× (1) + (2)× (2) + (4)× (3) + (1)× (4) + (0)× (5) (1)× (1) + (0)× (2) + (−1)× (3) + (−2)× (4) + (1)× (5) (2)× (1) + (−1)× (2) + (3)× (3) + (1)× (4) + (2)× (5) (1)× (1) + (2)× (2) + (3)× (3) + (4)× (4) + (3)× (5)
(−1)× (1) + (−2)× (2) + (0)× (3) + (1)× (4) + (2)× (5)
1CCCCA
http://z.cs.utexas.edu/wiki/pla.wiki/ 14
Example (continued)
0BBBB@ (−1)× (1) + (2)× (2) + (4)× (3) + (1)× (4) + (0)× (5)
(1)× (1) + (0)× (2) + (−1)× (3) + (−2)× (4) + (1)× (5) (2)× (1) + (−1)× (2) + (3)× (3) + (1)× (4) + (2)× (5) (1)× (1) + (2)× (2) + (3)× (3) + (4)× (4) + (3)× (5)
(−1)× (1) + (−2)× (2) + (0)× (3) + (1)× (4) + (2)× (5)
1CCCCA
=
0BBBBBB@
„ −1 2
1 0
«„ 1 2
« +
„ 4 −1
« 3 +
„ 1 0 −2 1
«„ 4 5
« `
2 −1 ´„ 1
2
« + (3)3 +
` 1 2
´„ 4 5
« „
1 2 −1 −2
«„ 1 2
« +
„ 3 0
« 3 +
„ 4 3 1 2
«„ 4 5
«
1CCCCCCA
=
0BBBB@ „
3 1
« +
„ 12 −3
« +
„ 4 −3
« 0 + 9 + 14„
5 −5
« +
„ 9 0
« +
„ 31 14
« 1CCCCA =
0BBBB@ 19 −5 23 45 9
1CCCCA
http://z.cs.utexas.edu/wiki/pla.wiki/ 15
Blocked matrix-vector multiplication
Let A ∈ Rm×n, x ∈ Rn, and y ∈ Rn. Let m = m0 +m1 + · · ·mM−1, mi ≥ 0 for i = 0, . . . ,M − 1; and n = n0 + n1 + · · ·nN−1, nj ≥ 0 for j = 0, . . . , N − 1; and
Partition
A =
0BBB@ A0,0 A0,1 · · · A0,N−1 A1,0 A1,1 · · · A1,N−1
... ...
. . . ...
AM−1,0 AM−1,1 · · · AM−1,N−1
1CCCA ,
x =
0BBB@ x0 x1 ...
xN−1
1CCCA , and y = 0BBB@
y0 y1 ...
yM−1
1CCCA with Ai,j ∈ Rmi×nj , xj ∈ Rnj , and yi ∈ Rmi .
http://z.cs.utexas.edu/wiki/pla.wiki/ 16
Theorem (continued)
Then0BBB@ y0 y1 ...
yM−1
1CCCA = 0BBB@
A0,0 A0,1 · · · A0,N−1 A1,0 A1,1 · · · A1,N−1
... ...
. . . ...
AM−1,0 AM−1,1 · · · AM−1,N−1
1CCCA 0BBB@
x0 x1 ...
xN−1
1CCCA
=
0BBB@ A0,0x0 +A0,1x1 + · · ·+A0,N−1xN−1 A1,0x0 +A1,1x1 + · · ·+A1,N−1xN−1
...
AM−1,0x0 +AM−1,1x1 + · · ·+AM−1,N−1xN−1
1CCCA
In other words...
yi = N−1∑ j=0
Ai,jxj .
http://z.cs.utexas.edu/wiki/pla.wiki/ 17
Example (revisited)
Consider
A =
0@ A00 a01 A02aT10 α11 aT12 A20 a21 A22
1A = 0BBBB@ −1 2 4 1 0
1 0 −1 −2 1 2 −1 3 1 2 1 2 3 4 3 −1 −2 0 1 2
1CCCCA ,
x =
0@ x0χ1 x2
1A = 0BBBB@
1 2
3
4 5
1CCCCA , and y = 0@ y0ψ1
y2
1A ,
where y0, y2 ∈ R2.
http://z.cs.utexas.edu/wiki/pla.wiki/ 18
Example (continued)
Then
y =
0@ y0ψ1 y2
1A = 0@ A00 a01 A02aT10 α11 aT12
A20 a21 A22
1A0@ x0χ1 x2
1A =
0@ A00x0 + a01χ1 +A02x2aT10x0 + α11χ1 + aT12x2 A20x0 + a21χ1 +A22x2
1A
=
0BBBBBB@
„ −1 2
1 0
«
