Page 99 - DMTH502_LINEAR_ALGEBRA

Unit 5: Summary of Row-Equivalence

(the relations z = 0 for i > r, above). For example, if AX = Y is a system of linear equations in Notes
i
which the scalars y and A are real numbers, and if there is a solution in which x ,...,x are
k ij 1 n
complex numbers, then there is a solution with x ,...,x real numbers.
1 n
Self Assessment

3. Find all solutions to the following system of equations by row-reducing the coefficient
matrix:

3
x  2x  6x  0
1
2
3
3
 4x 1  5x  0
3
 3x  6x  13x  0
1 2 3
7 8
 x  2x  x  0
3 1 2 3 3
4. Find a row-reduced echelon matrix which is row-equivalent to
1  i  
 
A   2 2 

 i  1 i 
What are the solutions of AX = 0?
5.3 Summary of Row-Equivalence

In this section we shall utilize some elementary facts on bases and dimension in finite-dimensional
vector spaces to complete our discussion of row-equivalence of matrices. We recall that if A is an
n
m × n matrix over the field F the row vectors of A are the vectors  ,..., in F defined by
1 m
 = (A ,..., A )
i ij in
and that the row space of A is the subspace of F spanned by these vectors. The row rank of A is
n
the dimension of the row space of A.
If P is a k × m matrix over F, then the product B = PA is a k × n matrix whose row vectors  ,...,
1 k
are linear combinations
 = P + ... + P 
i i11 im m
of the row vectors of A. Thus the row space of B is a subspace of the row space of A. If P is an
m × m invertible matrix, then B is row-equivalent to A so that the symmetry of row-equivalence,
or the equation A = P B, implies that the row space of A is also a subspace of the row space of B.
–1
Theorem 5: Row-equivalent matrices have the same row space.
Thus we see that to study the row space of A we may as well study the row space of a
row-reduced echelon matrix which is row-equivalent to A. This we proceed to do.
Theorem 6: Let R be a non-zero row-reduced echelon matrix. Then the non-zero row vectors of
R form a basis for the row space of R.
Proof: Let  ,..., be the non-zero row vectors of R:
1 r
 = (R ,...,R )
i i1 in
Certainly these vectors span the row space of R; we need only prove they are linearly
independent. Since R is a row-reduced echelon matrix, there are positive integers k ,...,k such
1 r
that, for i  r

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