Class Notes from January 25, 2010
The homework for Thursday is Section 1.6 and Section 2.1.
Today we ﬁnished Section 1.6 and started Section 2.1.
We reviewed the three basic cases of matrix multiplication using block decompo
sitions of the matrices. Then we considered the fourth case of matrix multiplication
using block decomposition. The goal is to simplify the computation of
AB
where
A
is an
m
×
n
matrix and
B
is an
n
×
r
matrix.The inverse can be expressed as the
product of elementary matrices. Decompose
A
into four blocks
A
=
±
A
11
A
12
A
21
A
22
²
(1)
and similarly decompose
B
B
=
±
B
11
B
12
B
21
B
22
²
.
(2)
Now we want to multiply so you must decompose in such a way that
A
11
and
A
12
have
s
columns and
B
11
and
B
12
have
s
rows. Then
A
21
and
A
22
have
n

s
columns and
B
21
and
B
22
have
n

s
rows. This means the products
A
11
B
11
, A
11
B
12
,A
12
B
21
, A
11
B
12
,
A
12
B
22
, A
21
B
11
, A
22
B
21
, A
21
B
12
and
A
22
B
22
all make sense. The block multiplication
formula in this case looks just like the formula for multiplying two 2
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 Winter '08
 staff
 Math, Multiplication, Matrices, Invertible matrix, Mikhail Botvinnik, Chess opening, Encyclopaedia of Chess Openings

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