Page 145 - A Course in Linear Algebra with Applications

5.2: The Row and Column Spaces of a Matrix 129

Consequently there is a non-trivial solution C of the linear
system AC = 0 such that Cj =fi 0 for j = i%,..., i r. Using the
equation B = EA, we find that BC = EAC = EO = 0. This
i
means that columns i , . . . , i r of B are also linearly dependent.
j
Therefore, if columns i , . . . , a of B are linearly independent,
j
then so are columns ji, • • • ,j 3 of A. Hence the dimension of
the column space of B does not exceed the dimension of the
column space of A.
l
Since A = E~ B, this argument can be applied equally
well to show that the dimension of the column space of A does
not exceed that of B. Therefore these dimensions are equal.
The truth of the corresponding statement for row spaces
can be quickly deduced from what has just been proved. Let
T
B = AE where E is an elementary matrix. Then B =
T T T T
(AE) — E A . Now E is also an elementary matrix, so
by the last paragraph the column spaces of A T and B T have
the same dimension. But obviously the column space of A T
and the row space of A have the same dimension, and there is
a similar statement for B: the required result follows at once.

We are now in a position to connect row and column
spaces with normal form and at the same time to clarify a
point left open in Chapter Two.

Theorem 5.2.4
If A is any matrix, then the following integers are equal:
(i) the dimension of the row space of A;
(ii) the dimension of the column space of A;
(iii) the number of 1's in a normal form of A.
Proof
By applying elementary row and column operations to A, we
can reduce it to normal form, say