Information Technology Reference
In-Depth Information
Proof.
Let
A
beamatrix
n
m
. Let us denote
by
A
i
the
i
-row vectors of a matrix
A
and by
X
j
the
j
-column vectors of
X
. With this
notation we evaluate both members of Eq. (3.25). The following matrix product:
×
h
,
X
amatrix
h
×
k
,and
B
amatrix
k
×
×
×
A
X
B
(3.26)
is equal to:
⎛
⎞
X
1
X
2
X
k
A
1
×
A
1
×
...
A
1
×
⎝
⎠
×
X
1
X
2
X
k
A
2
×
A
2
×
...
A
2
×
B
.
...
... ... ...
X
1
X
2
X
k
A
n
×
A
n
×
...
A
n
×
Therefore, if
A
i
X
j
X
j
abbreviates the scalar product
A
i
×
of the row vector
A
i
with
the column vector
X
j
, then the product
A
×
X
×
B
is equal to:
⎛
⎝
⎞
⎠
.
b
1
,
1
A
1
X
1
+
b
2
,
1
A
1
X
2
+
...
+
b
k
,
1
A
1
X
k
b
1
,
m
A
1
X
1
+
b
2
,
m
A
1
X
2
+
...
+
b
k
,
m
A
1
X
k
...
...
...
...
b
1
,
1
A
2
X
1
+
b
2
,
1
A
2
X
2
+
...
+
b
k
,
1
A
2
X
k
.........
b
1
,
m
A
2
X
1
+
b
2
,
m
A
2
X
2
+
...
+
b
k
,
m
A
2
X
k
.........
b
1
,
1
A
n
X
1
+
b
2
,
1
A
n
X
2
+
...
+
b
k
,
1
A
n
X
k
b
1
,
m
A
n
X
1
+
b
2
,
m
A
n
X
2
+
...
+
b
k
,
m
A
n
X
k
(3.27)
The right member of Eq. (3.25) is:
B
T
(
⊗
A
)
×
vec
(
X
)
(3.28)
which is equal to:
⎛
⎝
⎞
⎠
×
⎛
⎝
⎞
⎠
X
1
X
2
...
X
k
b
1
,
1
Ab
2
,
1
A
...
b
k
,
1
A
b
1
,
2
Ab
2
,
2
A
b
k
,
2
A
... ... ... ...
b
1
,
m
Ab
2
,
m
A
...
...
b
k
,
m
A
that is, by making explicit the rows of
A
:
⎛
⎛
⎞
⎛
⎞
⎞
X
1
X
2
...
X
k
b
1
,
1
A
1
b
2
,
1
A
1
...
b
k
,
1
A
1
⎝
⎠
×
⎝
⎠
⎝
⎠
b
1
,
1
A
2
b
2
,
1
A
2
...
b
k
,
1
A
2
... ... ... ...
b
1
,
1
A
n
b
2
,
1
A
n
...
b
k
,
1
A
n
⎛
⎝
⎞
⎠
×
⎛
⎝
⎞
⎠
X
1
X
2
...
X
k
b
1
,
2
A
1
b
2
,
2
A
1
...
b
k
,
2
A
1
b
1
,
2
A
2
b
2
,
2
A
2
...
b
k
,
2
A
2
... ... ... ...
b
1
,
2
A
n
b
2
,
2
A
n
...
b
k
,
2
A
n
..........................................
⎛
⎞
⎛
⎞
X
1
X
2
...
X
k
b
1
,
m
A
1
b
2
,
m
A
1
...
b
k
,
m
A
1
⎝
⎠
×
⎝
⎠
b
1
,
m
A
2
b
2
,
m
A
2
...
b
k
,
m
A
2
... ... ... ...
b
1
,
m
A
n
b
2
,
m
A
n
...
b
k
,
m
A
n
