# Multiplication Between Matrices Expressed in Row of Yue-Xian Li February, 2014 ```Multiplication Between Matrices Expressed in Row of
Column Vectors or Column of Row Vectors
University of British Columbia, Vancouver
Yue-Xian Li
February, 2014
1
A
Matrix expressed as a row of column vectors
An m &times; n matrix can be expressed as a row of n column vectors. Rules of matrix multiplication
hold for such expressions.
In the following text, no proof will be provided. Instead, examples will be provided to clarify the
points.
Example A1:
A=
1 1
0 1
= [a1 a2 ] ,
where a1 =
1
0
,
a2 =
1
1
.
Thus,
A
2
3
= [a1 a2 ]
2
3
= 2a1 + 3a2 = 2
1
0
+3
1
1
=
5
3
Let’s us verify the result using normal matrix multiplication:
A
2
3
=
1 1
0 1
2
3
=
2
(1)(2) + (1)(3)
(0)(2) + (1)(3)
=
5
3
.
.
Summary:
(1) ∀ m &times; n matrix A, one can express it as a row of column vectors
aj ∈ Rm
A = [a1 a2 &middot; &middot; &middot; an ],
(1 ≤ j ≤ n).
(2) Let x = [x1 x2 &middot; &middot; &middot; xn ]T be a column vector in Rn , then


x1
 x2 


Ax = [a1 a2 &middot; &middot; &middot; an ]  ..  = x1 a1 + x2 a2 + &middot; &middot; &middot; + xn an
 . 
xn
is a linear combination of column vectors of A.
(3) Let B be a l &times; m matrix, then
BA = B[a1 a2 &middot; &middot; &middot; an ] = [Ba1 Ba2 &middot; &middot; &middot; Ban ].
Since Baj (1 ≤ j ≤ n) is a linear combination of columns of B, each column of BA is a linear
combination of columns of B.
Example A2: If A is identical as in Example A1 and B =
BA = [Ba1 Ba2 ] = [[b1 b2 ]a1 [b1 b2 ]a2 ] = [[b1 b2 ]
1
0
1 2
1 2
[b1 b2 ]
= [b1 b2 ], then
1
1 3
] = [b1 b1 + b2 ] =
.
1
1 3
Notice that the first column of BA is equal to b1 while the second is equal to b1 + b2 .
Conclusion: Columns of the matrix product Tn Tn−1 &middot; &middot; &middot; T1 are linear combinations of columns of
the leftmost (last) matrix Tn .
3
B
Matrix expressed as a column of row vectors
An m &times; n matrix can be expressed as a column of m row vectors. Rules of matrix multiplication
hold for such expressions.
Example B1:
A=
1 1
0 1
=
a1
a2
,
a1 = [1 1] ,
a2 = [0 1] .
Thus,
[3 4] A = [3 4]
a1
a2
= 3a1 + 4a2 = 3 [1 1] + 4 [0 1] = [3 7] .
Let’s us verify the result using normal matrix multiplication:
[3 4] A = [3 4]
1 1
0 1
= [(3)(1) + (4)(0) (3)(1) + (4)(1)] = [3 7] .
Summary:
(1) ∀ m &times; n matrix A, one can express it as a column of row vectors


a1
 a2 


aTi ∈ Rn (1 ≤ i ≤ m).
A =  ..  ,
 . 
am
(2) Let x = [x1 x2 &middot; &middot; &middot; xm ] be a row vector in Rm , then


a1
 a2 


xA = [x1 x2 &middot; &middot; &middot; xm ]  ..  = x1 a1 + x2 a2 + &middot; &middot; &middot; + xm am
 . 
am
is a linear combination of row vectors of A.
4
(3) Let B be a n &times; l matrix, then



AB = 


a1
a2
..
.





B = 


am
a1 B
a2 B
..
.



.

am B
Since ai B (1 ≤ i ≤ m) is a linear combination of rows of B, each row of AB is a linear
combination of rows of B.
Example B2: If A is identical as in Example B1 and B =
 
b
1
 a1 b2   [1 1]
a1
a1 B
=
AB =
B=
=

b1  
a2
a2 B
a2
[0 1]
b2

1 1
3 3
=
b1
, then
b2

b1
b2 
 = b1 + b2 = 4 4 .
b1 
b2
3 3
b2
Notice that the first row of AB is equal to b1 + b2 while the second is equal to b2 .
Conclusion: Rows of the matrix product Tn Tn−1 &middot; &middot; &middot; T1 are linear combinations of rows of the
rightmost (first) matrix T1 .
5
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