Page 211 - Matrix Analysis & Applied Linear Algebra

206 Chapter 4 Vector Spaces

Since B Y is an independent set, it follows that all of the γ k ’s (as well as all
t m
δ i ’s) are zero, and (4.4.20) reduces to α i z i + β j x j = 0. But B X is

i=1 j=1
also an independent set, so the only way this can hold is for all of the α i ’s as
well as all of the β j ’s to be zero. Therefore, the only possible solution for the
α ’s, β ’s, and γ ’s in the homogeneous equation (4.4.20) is the trivial solution,
and thus B is linearly independent. Since B is an independent spanning set, it
is a basis for X + Y and, consequently,

dim (X + Y)= t+m+n =(t+m)+(t+n)−t = dim X +dim Y−dim (X∩ Y) .

Example 4.4.8
Problem: Show that rank (A + B) ≤ rank (A)+ rank (B).
Solution: Observe that

R (A + B) ⊆ R (A)+ R (B)
because if b ∈ R (A + B), then there is a vector x such that

b =(A + B)x = Ax + Bx ∈ R (A)+ R (B).
Recall from (4.4.5) that if M and N are vector spaces such that M⊆N, then
dim M≤ dim N. Use this together with formula (4.4.19) for the dimension of a
sum to conclude that

rank (A + B) = dim R (A + B) ≤ dim R (A)+ R (B)

= dim R (A) + dim R (B) − dim R (A) ∩ R (B)
≤ dim R (A) + dim R (B)= rank (A)+ rank (B).

Exercises for section 4.4

4.4.1. Find the dimensions of the four fundamental subspaces associated with
 
1223
A =  2413  .
3614

4.4.2. Find a basis for each of the four fundamental subspaces associated with
 
12021
A =  36196  .
24175