Math 222 - Selected Homework Solutions for
Assignment 6
Instructor - Al Boggess
Fall 1998
Section 3.5
11 There is a mistake in this problem. It should read, the transition
matrix from F to E rather than the other way around. The transition
matrix from F = [v1 ; : : : ; vn ] to E = [u1 ; : : : ; un ] is U ,1 V . On the
other hand, the reduced row echelon form of (U jV ) involves executing
row operations until U becomes the identity matrix. This is equivalent
to multiplying both left and right hand sides by U ,1 , i.e.
U ,1 U jU ,1 V = I jU ,1V
Thus the transition matrix from F to E is the result (i.e the right
hand side) of the reduced row echelon form of (U jV ).
Section 3.6
7 We are given that A is m n with m > n and that N (A) = 0.
a) Since N (A) = 0, the column vectors of A are linearly independent
(Theorem 3.6.2). There are n column vectors and n < m, so the
column vectors cannot span Rm .
b) If b is not in the column space of A, there will be no solutions to
Ax = b. If b is in the column space of A, there will be exactly one
solution to Ax = b (again, by Theorem 3.6.2).
8 We are given that A is a 6 5 matrix. If the dimension of N (A) = 2,
then the rank of A is 3 (using Theorem 3.6.4 since 3 + 2 = 5). If the
rank of B is 4, then the dimension of N (B ) is 1.
1
10 If Ax = b, then the row echelon form of the augmented matrix (Ajb)
has the same number of nonzero rows as the row echelon form of A.
Otherwise, there would be a row of zeros for the row echelon form for
A with something nonzero on the right side which would lead to an
inconsistency (no solution). Since the number of nonzero rows equals
the rank, we conclude that the rank of A and the rank of Ajb are the
same.
Conversely, if the rank of Ajb equals the rank of A then their row
echelon forms have the same number of rows. In particular, the row
echelon form of Ajb does NOT have a row of zeros on the left with
something nonzero on the right. Therefore, the system Ax = b can be
solved, which means the system Ax = b is consistent.
11 We are given that A is an m n matrix.
a) We are to show that if B is a nonsingular m m matrix then BA
and A have the same nullspace. If (BA)x = B (Ax) = 0 then Ax lies
in the nullspace of B . Since B is nonsingular, its nullspace consists
only of f0g. Thus Ax must be zero. This implies x belongs to the
nullspace of A. We conclude that N (BA) N (A).
Conversely, if Ax = 0, then (BA)(x) = B (Ax) = 0 and so x belongs
to N (BA). Therefore N (A) N (BA). In the previous paragraph, we
showed N (BA) N (A), so N (BA) = N (A).
b) We are given that C is nonsingular. So C t is nonsingular. By part
a), the nullity of C t At = (AC )t is the same as the nullity of At . Thus
the rank of (AC )t is the same as the rank of At (Theorem 3.6.4). By
Theorem 3.6.5, the rank of a matrix equals the rank of its transpose.
So we conclude that the rank of A equals the rank of AC .
15 We are given that x0 is a particular solution to Ax = b.
a) Suppose y is any solution to Ax = b, we must show that y = x0 + z
where z 2 N (A). Consider the dierence z = y , x0 . Apply A to this
vector and we obtain
A(z ) = A(y , x0 ) = A(y) , A(x0 ) = b , b = 0
Therefore z belongs to N (A) and y = z + x0 .
Conversely, if y = x0 + z with z 2 N (A), then
Ay = A(x0 ) + A(z ) = b + 0 = b
2
b) If N (A) = 0, then the vector z in part a) must be zero. Therefore
y = x0 is the only solution to Ax = b.
3