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208 Chapter 4 Vector Spaces
m×n n×1
4.4.9. For A ∈ and a subspace S of , the image
A(S)= {Ax | x ∈S}
m×1
of S under A is a subspace of —recall Exercise 4.1.9. Prove
that if S∩ N (A)= 0, then dim A(S) = dim(S). Hint: Use a basis
{s 1 , s 2 ,..., s k } for S to determine a basis for A(S).
4.4.10. Explain why rank (A) − rank (B) ≤ rank (A − B).
4.4.11. If rank (A m×n )= r and rank (E m×n )= k ≤ r, explain why
r − k ≤ rank (A + E) ≤ r + k.
In words, this says that a perturbation of rank k can change the rank
by at most k.
n
4.4.12. Explain why every nonzero subspace V⊆ must possess a basis.
4.4.13. Explain why every set of m − 1 rows in the incidence matrix E of a
connected directed graph containing m nodes is linearly independent.
4.4.14. For the incidence matrix E of a directed graph, explain why
number of edges at node i when i = j,
" T #
EE =
ij −(number of edges between nodes i and j) when i = j.
4.4.15. If M and N are subsets of a space V, explain why
dim span (M∪N) = dim span (M) + dim span (N)
− dim span (M) ∩ span (N) .
4.4.16. Consider two matrices A m×n and B m×k .
(a) Explain why
rank (A | B)= rank (A)+ rank (B) − dim R (A) ∩ R (B) .
Hint: Recall Exercise 4.2.9.
(b) Now explain why
dim N (A | B) = dim N (A)+dim N (B)+dim R (A)∩R (B) .
(c) Determine dim R (C) ∩ N (C) and dim R (C)+ N (C) for
−111 −21
−103 −42
C = −103 −53 .
−103 −64
−103 −64