Linear Programming Lecture Notes - Penn State Personal Web Server

which is a contradiction.
Case 2: Suppose that the size of the largest set of linearly independent rows is k. Denote
such a set by S = {as1, . . . , ask }. There are several possibilities: (i) Both ai and aj are in
this set. In this case, we simply replace our argument above with constants α1 through αk
and the result is the same.
(ii) ai and aj are not in this set, in which case we know that there are αs1, . . . , αsk and
βs1, . . . , βsk so that:
αs1a1 + · · · + αsk ak = ai
βs1a1 + · · · + βsk ak = aj
But this implies that αai + aj can also be written as a linear combination of the elements of
S and thus the rank of A ′ is no larger than the rank of A.
(iii) Now suppose that ai in the set S. Then there are constants αs1, . . . , αsk
αs1a1 + · · · + αsk ak = aj
Without loss of generality, suppose that ai = as1 then:
αai + αs1ai + αs2as2 + · · · + αsk ak = αai + aj
so that:
Again, this implies that αai + aj is still a linear combination of the elements of S and so we
cannot have increased the size of the largest linearly independent set of vectors, nor could
we have decreased it.
(iv) Finally, suppose that aj ∈ S. Again let aj = as1. Then there are constants
αs1, . . . , αsk
αs1aJ + αs2as2 + · · · + αskask = ai
Apply Lemma 3.38 to replace aj in S with some other row vector al. If l = i, then we reduce
to sub-case (iii). If l = i, then we reduce to sub-case (ii).
Finally, suppose we multiply a row by α. This is reduces to the case of multiplying row
i by α − 1 and adding it to row i, which is covered in the above analysis. This completes
the proof.
We have the following theorem, whose proof is again outside the scope of this course.
There are very nice proofs available in [Lan87].
Theorem 3.42. If A ∈ R m×n is a matrix, then the row rank of A is equal to the column
rank of A. Further, rank(A) ≤ min{m, n}.
Lastly, we will not prove the following theorem, but it should be clear from all the work
we have done up to this point.
Theorem 3.43. If A ∈ R m×m (i.e., A is a square matrix) and rank(A) = m, then A is
invertible.
Definition 3.44. Suppose that A ∈ R m×n and let m ≤ n. Then A has full row rank if
rank(A) = m.
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