Download Homework assignment, Feb. 18, 2004. Solutions

Homework assignment, Feb. 18, 2004. Solutions
1. True or false:
a) The rank of a matrix equal to the number of its non-zero columns. False
b) The m × n zero matrix is the only m × n matrix having rank 0. True, rank 0 means
no pivots, which can happen only if all entries are 0.
c) Elementary row operations preserve rank. True
d) Elementary column operations do not necessarily preserve rank. False
e) The rank of a matrix is equal to the maximum number of linearly independent columns
in the matrix. True
f) The rank of a matrix is equal to the maximum number of linearly independent rows
in the matrix. True
g) The rank of an n × n matrix is at most n. True
h) An n × n matrix having rank n is invertible. True
2. A 54 × 37 matrix has rank 31. What are dimensions of all 4 fundamental subspaces?
dim Ran A = dim Ran AT = 31
(by the definition of rank and Rank Theorem), and
dim Ker A = 37 − 31 = 6,
dim Ker AT = 54 − 31 = 23.
3. Compute rank and find bases of all four fundamental subspaces for the matrices
⎛
⎞
⎛
⎞
1 2 3 1 1
1 1 0
⎜ 1 4 0 1 2 ⎟
⎜
⎟
⎝ 0 1 1 ⎠,
⎝ 0 2 −3 0 1 ⎠
1 1 0
1 0 0 0 0
4. Prove that if A : X → Y and V is a subspace of X then dim AV ≤ rank A. (AV here
means the subspace V transformed by the transformation A, i.e. any vector in AV can be
represented as Av, v ∈ V ). Deduce from here that rank(AB) ≤ rank A.
Remark: Here one can use the fact that if V ⊂ W then dim V ≤ dim W . Do you understand
why is it true?
Proof.
Proposition. If subspaces V and W satisfy V ⊂ W , then dim V ≤ dim W
Proof. Let dim V = r and let v1 , v2 , . . . , vr be a basis in V . Vectors v1 , v2 , . . . , vr are in W ,
and they are linearly independent. Therefore, r ≤ dim W
Since V ⊂ X we have after the transformation A that AV ⊂ AX =: Ran A. By the
Proposition dim AV ≤ dim AX = dim Ran A = rank A.
To prove the statement about ranks, let us denote V := Ran B. Then Ran AB = AV ,
and so rank AB = dim AV ≤ rank A.
5. Prove that if A : X → Y and V is a subspace of X then dim AV ≤ dim V . Deduce from
here that rank(AB) ≤ rank B.
Proof. Let dim V = r and let v1 , v2 , . . . , vr be a basis in V . Then the system of vectors
Av1 , Av2 , . . . , Avr isa generating system for AV . Indeed, every vector v ∈ V can be
αk vk , so any vector of form Av, v ∈ V can be represented as a
represented as v = rk=1
linear combination Av = k αk Avk .
Since any generating set in V must have at least dim V vectors (see Prop. 4.3), we have
that r ≥ dim AV , i.e. dim AV ≤ r = dim V .
To prove the statement about ranks, let us again denote V := Ran B, so rank B = dim V .
Then Ran(AB) = AV , and so rank AB = dim AV ≤ dim V = rank B.
6. Prove that if the product AB of two n × n matrices is invertible, then both A and B are
invertible. Even if you know about determinants, do not use them, we did not cover them
yet. Hint: use previous 2 problems.
Proof. Since AB is an n × n matrix, it is invertible if and only if rank(AB) = n (pivots in
every row and every column). By previous problems rank A ≥ n and rank B ≥ n. But A
and B are n × n matrices, so their rank cannot be more than n. Therefore, both A and B
have rank n, so they have pivots in every row and column, so they both are invertible.
You also can show that if C = AB is invertible, then A is right invertible, and B is left
invertible (was in one of the previous assignments). Since A and B are square matrices, we
can conclude by counting pivots, that they are invertible.