Page 104 - DMTH502_LINEAR_ALGEBRA
P. 104
Linear Algebra
Notes Let A be the m × n matrix with row vectors :
i
= (A ,...,A )
i i1 in
Perform a sequence of elementary row operations, starting with A and terminating with a row-
reduced echelon matrix R. We have previously described how to do this. At this point, the
dimension of W (the row space of A) is apparent, since this dimension is simply the number of
non-zero row vectors of R. If ,..., are the non-zero row vectors of R, then = { ,..., } is a basis
1 r 1 r
for W. If the first non-zero coordinate of is the k th one, then we have for i r
i i
(a) R(i, j) = 0, if j < k
i
(b) R(i, k ) =
j ij
(c) k < ... < k
1 r
The subspace W consists of all vectors
= c + ... + c
1 1 r r
r
= c i (R 1 i ,...,R in )
i 1
The coordinates b ,...,b of such a vector are then
1 n
r
b = c R ...(1)
j i ij
i 1
In particular, b = c , and so if = (b ,...,b ) is a linear combination of the , it must be the
ki j 1 n i
particular linear combination.
r
= b ki i ...(2)
i 1
The conditions on that (2) should hold are
r
n
b = b R j 1,..., . ...(3)
j ki ij
i 1
Now (3) is the explicit description of the subspace W spanned by ,..., , that is, the subspace
1 m
n
consists of all vectors in F whose coordinates satisfy (3). What kind of description is (3)? In the
first place it describes W as all solutions = (b ,...,b ) of the system of homogeneous linear
1 n
equations (3). This system of equations is of a very special nature, because it expresses (n – r) of
the coordinates as linear combinations of the r distinguished coordinates b ,...,b . One has
k1 kr
complete freedom of choice in the coordinates b , that is, if c ,...,c are any r scalars, there is one
ki 1 r
and only one vector in W which has c as its k th coordinate.
i i
The significant point here is this: Given the vectors , row-reduction is a straightforward method
i
of determining the integers r, k ,...,k and the scalars R which give the description of the subspace
1 r ij
spanned by ,..., . One should observe that every subspace W of F has a description of the type
n
1 m
(3). We should also point out some things about question (2). We have already stated how one
can find an invertible m × m matrix P such that R = PA. The knowledge of P enables one to find
the scalars x ,...,x such that
1 m
= x + ... + x
1 1 m m
when this is possible. For the row vectors of R are given by
m
= P
i ij j
j 1
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