Solutions to Assignment 8
Math 217, Fall 2002
4.6.26 If A is 6 × 4, what is the smallest possible dimension of Nul(A)?
Well, we know that rank(A)+dim(Nul(A)) = 4. It is possible that rank(A) = 4,
that is, that dim(Col(A)) = 4 because A is 6 × 4. We conclude the smallest
possible dimension of Nul(A) is zero.
4.6.20 Suppose a nonhomogeneous system of nine linear equations in ten unknowns
has a solution for all possible constants on the right sides of the equations. Is
it possible to find two nonzero solutions of the associated homogeneous system
that are not multiples of each other? Discuss.
Again, we know that rank(A) + dim(Nul(A)) = 10. If the system is consistent
for all possible constants on the right side of the equation, then the matrix of
coefficients must have a pivot in every row. As the matrix is 9 × 10, this means
that the matrix must have 9 pivots. Thus there are 9 pivot columns, and so the
rank(A) = 9. It must be the case that dim(Nul(A)) = 1, and we conclude that
a basis for the vector space of solutions to the homogeneous system of equations
contains only one element. This means that every nonzero solution is a multiple
of the one basis element. So it is not possible to find two nonzero vectors which
solve the system of equations and are not multiples of each other.
4.6.28 Justify the following equalities:
(a) dim(Row(A)) + dim(Nul(A)) where A has n columns.
Theorem 14 (pg 265) tells us that dim(Row(A)) = dim(Col(A)), and of
course, dim(Col(A)) = rank(A) by definition. Theorem 14 also tells us
that rank(A) + dim(Nul(A)) = n. So unwinding things, we have n =
rank(A) + dim(Nul(A)) = dim(Row(A)) + dim(Nul(A)) as required.
(b) dim(Col(A)) + dim(Nul(AT )) where A has m rows.
1
Note that Col(A) is isomorphic to Row(AT ), and thus their dimensions are
equal (so rank(AT ) = dim(Col(A))). We know that AT has m columns, so
by theorem 14, rank(AT ) + dim(Nul(AT )) = m. Replacing rank(AT ) with
dim(Col(A)) gives the equality desired.
Oops! I did some of the wrong problems. I’ll leave them for your viewing
pleasure (the next two).
4.7.6 If a 6 × 3 matrix A has rank 3, find dim(Nul(A)), dim(Row(A)), and rank(AT ).
We know that rank(A)+dim(Nul(A)) = 3, so if A has rank 3 then dim(Nul(A)) =
0 (this makes Nul(A) = {0}). We also saw in theorem 14, that dim(Col(A)) =
dim(Row(A)), so dim(Row(A)) = rank(A) = 3. Finally, we know that rank(AT ) =
dim(Col(AT )) = dim(Row(A)) = 3. This is because Row(A) is isomorphic to
Col(AT ) so their dimensions are equal (as we used in 4.6.28).
4.7.8 Suppose a 5 × 6 matrix A has four pivot columns. What is dim(Nul(A))? Is
Col(A) = R4 ? Why or why not?
We know that rank(A) + dim(Nul(A)) = 6, and rank(A) is the number of pivot
columns of A. If A has 4 pivots, then it must be the case that dim(Nul(A)) = 2.
Furthermore, if dim(Col(A)) = 4, we know that Col(A) has a basis of four
elements. It does turn out that Col(A) is isomorphic to R4 . It is not equal to
R4 , however, because it is a subspace of R5 (the matrix has 5 rows).
4.7.6 Let D = {d1 , d2 , d3 } and F = {f1 , f2 , f3 } be bases for a vector space V , and
suppose that f1 = 2d1 − d2 + d3 , f2 = 3d2 + d3 , and f3 = −3d1 + 2d3 .
(a) Find the change-of-coordinates matrix from F to D.
To compute
P
we just follow the formula given on page 273 (theorem
D←F
15). This is easy because we are told hold
di . The answer is

2 0
P

= −1 3
D←F
1 1
2
to write the fi in terms of the

−3
0 .
2
(b) Find [x]D for x = f1 − 2f2 + 2f3 .
Now we know that
[x]D =
So
P
D←F
[x]F .



   
2 0 −3
2 0 −3
1
−4
[x]D = −1 3 0  [x]F = −1 3 0  −2 = −7 .
1 1 2
1 1 2
2
3
−1
1
1
1
4.7.8 Let B =
,
and C =
,
be bases for R2 . Find the
8
−5
4
1
change-of-coordinates matrix from B to C and the change-of-coordinates matrix
from C to B.
We know that
P
C ←B
−1
1
=
.
8 C −5 C
−1
1
To calculate
and
we simply put the matrix
8 C
−5 C
1 1 | −1 1
4 1 | 8 −5
in row reduced echelon form (the two vectors we are
looking for willbe the right
1 1 | −1 1
two vectors in the row reduced echelon form of
). The row
4 1 | 8 −5
1 0 | 3 −2
reduced echelon form of the matrix in question is
. Thus
0 1 | −4 3
P
3 −2
=
.
C ←B
−4 3
P
We also know that
B←C
a 2 × 2 matrix:
P
B←C
=(
P
=(
)−1 . We use the formula for the inverse of
1 3 2
−1
.
) = 14
4 3
C ←B
P
C ←B
4.7.14 In P2 , find the change-of-coordinates matrix from the basis B = {1 − 3t2 , 2 +
t − 5t2 , 1 + 2t} to the standard basis C = {1, t, t2 }. The write t2 as a linear
combination of the polynomials in B.
3
As with the last problem, I need to write the basis
  elements of B in terms of the
a
standard basis C. Of course, [a + bt + ct2 ]C =  b , so the change-of-coordinates
c
matrix is


1
2 1
P
1 2 .
= 0
C ←B
−3 −5 0
So now we have that
P
C ←B
 
0
2
2

[t ]B = [t ]C = 0 .
1
The augmented matrix corresponding to

1
2
0
1
−3 −5
and in row reduced echelon form this

1 0
0 1
0 0
 
3
2

So, we see that [t ]B = −2.
1
4
this linear system is

1 0
2 0 ,
0 1
is

0 3
0 −2 .
1 1