CS/ECE/ME 532
Homework 2: subspaces and linear equations
1. Rank of a product. Suppose that C = AB where A ∈ R
m×k and B ∈ R
k×n
. Prove the following
inequality: rank(C) ≤ min(m, n, k). Hint: think about how we proved that rank(xyT) = 1 in class.
2. Subspace properties. Suppose S, T ⊆ R
n are subspaces.
a) Prove that the sum S + T is a subspace. Here, S + T = {s + t | s ∈ S and t ∈ T}, i.e. the set
of vectors that can be written as the sum of a vector from S and a vector from T.
b) Prove that the intersection S ∩ T is a subspace. Here, S ∩ T = {x | x ∈ S and x ∈ T}, i.e. the
set of vectors belonging to both S and T.
3. Mostly zeros. Consider the matrix
A =
1 0 0 0 0
0 0 0 0 0
0 0 0 1 0
0 0 0 0 0
a) Find a basis for range(A) and find a basis for null(A).
b) Find a vector b such that Ax = b has no solutions, or explain why no such b can exist. Repeat
the question for the case of exactly one solution, and the case of infinitely many solutions.
4. Linear equations. This problem concerns linear equations and their solutions.
a) Find all solutions to the following system of equations.
x + 3y + 6z = 1
2x + 7y + 15z = −1
b) Find all solutions to the following equation.
x + 4y + 10z = 2
c) Find all (x, y, z) that simultaneously satisfy the equations of parts (a) and (b).
d) Sketch the set of solutions to parts (a), (b), and (c) in 3D on the same axes.
5. Existence of solutions. We saw that Ax = b will have at least one solution if b ∈ range(A).
However, this property can be difficult to check! An alternate way is to compare rank(A) and
rank(
A b
). If they are the same, then Ax = b has at least one solution. If they are different, then
Ax = b has no solutions. Explain why this works. Note:
A b
∈ R
m×(n+1) is the matrix formed
by including b as an extra column of A.
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