Page 100 - DMTH502_LINEAR_ALGEBRA

Linear Algebra

Notes
(a) R(i, j) = 0 if j < k 
i

(b) R(i, k ) =  ij  ...(1)
j

(c) k < ... < k
1 r 
Suppose  = (b ,...,b ) is a vector in the row space of R:
1 n
 = c  + ... + c  ...(2)
1 1 r r
Then we claim that c = b . For, by
j ki
r
k
b =  c R ( , )
i
kj i j
i 1
r
=  c  ij ...(3)
i
i 1
= c
j
In particular, if  = 0, i.e., if c  + ... + c  = 0, then c must be the k th coordinate of the zero vector
1 1 r r j j
so that c = 0, j = 1,..., r. Thus  ,..., are linearly independent.
j 1 r
n
Theorem 7: Let m and n be positive integers and let F be a field. Suppose W is a subspace of F and
dim W  m. Then there is precisely one m × n row-reduced echelon matrix over F which has W as
its row space.
Proof: There is at least one m × n row-reduced echelon matrix with row space W. Since dim
W  m, we can select some m vectors  ,..., in W which span W. Let A be the m × n matrix with
1 m
row vectors  ,..., and let R be a row-reduced echelon matrix which is row-equivalent to A.
1 m
Then the row space of R is W.
Now let R be any row-reduced echelon matrix which has W as its row space. Let  ,..., be the
1 r
non-zero row vectors of R and suppose that the leading non-zero entry of  occurs in column
i
k , i = 1,...,r. The vectors  ,..., form a basis for W. In the proof of Theorem, we observed that if
i 1 r
 = (b ,...,b ) is in W, then
1 n
 = c  + ... + c  ,
1 1 r r
and c = b ; in other words, the unique expression for  as a linear combination of  ,..., is
i ki 1 r
r
 =  b  ...(4)
ki i
i 1
Thus any vector  is determined if one knows the coordinates b , i = 1,..., r. For example,  is the
ki s
unique vector in W which has k th coordinate 1 and k th coordinate 0 for i  s.
s i
Suppose  is in W and   0. We claim the first non-zero coordinate of  occurs in one of the
columns k . Since
s
r
 =  b 
ki i
i 1
and   0, we can write

r
 =  b  i , b  0 ...(5)
ki
ks
i s


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