Page 203 - Matrix Analysis & Applied Linear Algebra

198 Chapter 4 Vector Spaces

Subspace Dimension
For vector spaces M and N such that M⊆N, the following state-
ments are true.
• dim M≤ dim N. (4.4.5)
• If dim M = dim N, then M = N. (4.4.6)

Proof. Let dim M = m and dim N = n, and use an indirect argument to
prove (4.4.5). If it were the case that m>n, then there would exist a linearly
independent subset of N (namely, a basis for M ) containing more than n vec-
tors. But this is impossible because dim N is the size of a maximal independent
subset of N. Thus m ≤ n. Now prove (4.4.6). If m = n but M = N, then
there exists a vector x such that x ∈N but x /∈M. If B is a basis for M,
then x /∈ span (B) , and the extension set E = B∪{x} is a linearly independent
subset of N —recall (4.3.15). But E contains m +1 = n + 1 vectors, which is
impossible because dim N = n is the size of a maximal independent subset of
N. Hence M = N.
Let’s now find bases and dimensions for the four fundamental subspaces
of an m × n matrix A of rank r, and let’s start with R (A). The entire set
of columns in A spans R (A), but they won’t form a basis when there are
dependencies among some of the columns. However, the set of basic columns in
A is also a spanning set—recall (4.2.8)—and the basic columns always constitute
a linearly independent set because no basic column can be a combination of other
basic columns (otherwise it wouldn’t be basic). So, the set of basic columns is a
basis for R (A), and, since there are r of them, dim R (A)= r = rank (A).
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Similarly, the entire set of rows in A spans R A , but the set of all rows

is not a basis when dependencies exist. Recall from (4.2.7) that if U = C r×n
0
is any row echelon form that is row equivalent to A, then the rows of C span
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R A . Since rank (C)= r, (4.3.5) insures that the rows of C are linearly
T
independent. Consequently, the rows in C are a basis for R A , and, since
T
there are r of them, dim R A = r = rank (A). Older texts referred to

dim R A T as the row rank of A, while dim R (A) was called the column rank
of A, and it was a major task to prove that the row rank always agrees with the
column rank. Notice that this is a consequence of the discussion above where it
T
was observed that dim R A = r = dim R (A).
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Turning to the nullspaces, let’s first examine N A . We know from
(4.2.12) that if P is a nonsingular matrix such that PA = U is in row echelon
T
form, then the last m − r rows in P span N A . Because the set of rows
in a nonsingular matrix is a linearly independent set, and because any subset