Linear Algebra Exercises-n-Answers.pdf

Answers to Exercises 115
while
and the two are equal.
trace(HG) = (h 1,1 g 1,1 + h 1,2 g 2,1 + · · · + h 1,n g n,1 )
+ (h 2,1 g 1,2 + h 2,2 g 2,2 + · · · + h 2,n g n,2 )
+ · · · + (h n,1 g 1,n + h n,2 g 2,n + · · · + h n,n g n,n )
Three.IV.3.39 A matrix is upper triangular if and only if its i, j entry is zero whenever i > j.
Thus, if G, H are upper triangular then h i,j and g i,j are zero when i > j. An entry in the product
p i,j = g i,1 h 1,j + · · · + g i,n h n,j is zero unless at least some of the terms are nonzero, that is, unless for
at least some of the summands g i,r h r,j both i ≤ r and r ≤ j. Of course, if i > j this cannot happen
and so the product of two upper triangular matrices is upper triangular. (A similar argument works
for lower triangular matrices.)
Three.IV.3.40
Three.IV.3.41
and
will do.
The sum along the i-th row of the product is this.
p i,1 + · · · + p i,n = (h i,1 g 1,1 + h i,2 g 2,1 + · · · + h i,n g n,1 )
+ (h i,1 g 1,2 + h i,2 g 2,2 + · · · + h i,n g n,2 )
+ · · · + (h i,1 g 1,n + h i,2 g 2,n + · · · + h i,n g n,n )
= h i,1 (g 1,1 + g 1,2 + · · · + g 1,n )
+ h i,2 (g 2,1 + g 2,2 + · · · + g 2,n )
+ · · · + h i,n (g n,1 + g n,2 + · · · + g n,n )
= h i,1 · 1 + · · · + h i,n · 1
= 1
Matrices representing (say, with respect to E 2 , E 2 ⊂ R 2 ) the maps that send
⃗β 1
⃗β 1
h
↦−→ β ⃗ 1 β2 ⃗
h
↦−→ ⃗0
g
↦−→ β ⃗ g
2 β2 ⃗ ↦−→ ⃗0
Three.IV.3.42 The combination is to have all entries of the matrix be zero except for one (possibly)
nonzero entry in each row and column. Such a matrix can be written as the product of a permutation
matrix and a diagonal matrix, e.g.,
⎛
⎝ 0 4 0
⎞ ⎛
2 0 0 ⎠ = ⎝ 0 1 0
⎞ ⎛
1 0 0⎠
⎝ 4 0 0
⎞
0 2 0 ⎠
0 0 −5 0 0 1 0 0 −5
and its action is thus to rescale the rows and permute them.
Three.IV.3.43 (a) Each entry p i,j = g i,1 h 1,j + · · · + g 1,r h r,1 takes r multiplications and there are
m · n entries. Thus there are m · n · r multiplications.
(b) Let H 1 be 5×10, let H 2 be 10×20, let H 3 be 20×5, let H 4 be 5×1. Then, using the formula
from the prior part,
this association uses this many multiplications
((H 1 H 2 )H 3 )H 4 1000 + 500 + 25 = 1525
(H 1 (H 2 H 3 ))H 4 1000 + 250 + 25 = 1275
(H 1 H 2 )(H 3 H 4 ) 1000 + 100 + 100 = 1200
H 1 (H 2 (H 3 H 4 )) 100 + 200 + 50 = 350
H 1 ((H 2 H 3 )H 4 ) 1000 + 50 + 50 = 1100
shows which is cheapest.
(c) This is reported by Knuth as an improvement by S. Winograd of a formula due to V. Strassen:
where w = aA − (a − c − d)(A − C + D),
(
a
) (
b A
)
B
c d C D
(
)
aA + bB
w + (c + d)(C − A) + (a + b − c − d)D
=
w + (a − c)(D − C) − d(A − B − C + D) w + (a − c)(D − C) + (c + d)(C − A)
takes seven multiplications and fifteen additions (save the intermediate results).