Math 102, Winter 2009,
Homework 7
(1) Find the standard matrix of the linear transformation T : R3 −→ R3
obtained by reflection through the plane −x + z = 0 followed by a rotation
about the positive x-axes by 60◦ .
Solution : Let S denotes the reflection and R denotes the rotation. The
standard matrix of reflection S through the plane −x + z = 0 can be obtained
by using the formula
S(u) = u − 2projn u,
where n = (−1, 0, 1) is the normal vector of the plane and u ∈ R3 .
Since,
S(e1 ) = (1, 0, 0) − 2(
(1, 0, 0) · (−1, 0, 1)
)(−1, 0, 1) = (0, 0, 1)
2
(0, 1, 0) · (−1, 0, 1)
)(−1, 0, 1) = (0, 1, 0)
2
(0, 0, 1) · (−1, 0, 1)
S(e3 ) = (0, 0, 1) − 2(
)(−1, 0, 1) = (1, 0, 0),
2
the standard matrix of S is:


0 0 1
[S] = [S(e1 ) S(e2 ) S(e3 )] =  0 1 0  .
1 0 0
S(e2 ) = (0, 1, 0) − 2(
The rotation R by π/3 about the x-axes has the standard matrix


 
1
0
0
1
0
0
√
[R] =  0 cos π/3 − sin π/3  =  0 √1/2 − 3/2  .
0 sin π/3 cos π/3
0
3/2
1/2
Since T = R ◦ S, the required matrix is

 


0
0
0
1
0
0
0
0
1
√
√
[T ] = [R◦S] = [R][S] =  0 √1/2 − 3/2   0 1 0  =  − 3/2 √1/2 0  .
1 0 0
0
3/2
1/2
1/2
3/2 0
(2) Determine if the given linear transformation is one-to-one and/or onto.
Justify your answer.
(a) T : R4 −→ R3 , T (x, y, z, w) = (x − 4y + 8z + w, 2y − z + 3w, 5w).
(b) T : R3 −→ R4 , T (x, y, z) = (2x + 3y, 3x + y + z, x − z, y − z).
Solution :


1 −4 8 1
(a) We first observe that the standard matrix [T ] of T is [T ] =  0 2 −1 3  .
0 0
0 5
The matrix [T ] is already in an echelon form, and we see that each row
of [T ] has a pivot position, therefore T is onto. Also, not every column
of T has a pivot position, that is, T is not one-to-one.
(b) We have

2
 3
[T ] = 
 1
0


3 0

1 1 
 ∼ ··· ∼ 

0 −1 
1 −1
1
0
0
0

0 −1
1 4 
.
0 −10 
0 0
Thus, each column of [T ] has a pivot position, hence T is one-to-one. On
the other hand, not every row of [T ] has a pivot position, that is, T is
not onto.
(3) ~u = (1, 2), ~v = (3, 5) and w
~ = (7, 5) be vectors in R2 .
(a) Express w
~ as a linear combination of ~u and ~v , that is, find scalars α and
β such that w
~ = α~u + β~v .
(b) Let T : R2 −→ R2 be the linear transformation such that T (~u) = (−4, 3)
and T (~v ) = (7, 8). Find T (w).
~
Solution :
(a) The vector equation α~u + β~v = w,
~ that is, α(1, 2) + β(3, 5) = (7, 5), is
equivalent to the linear system
α + 3β = 7
.
2α + 5β = 5
Solving the system we have α = −20, β = 9. Thus, w
~ = (−20)~u + 9~v .
(b) Since T is a linear transformation,
T (w)
~ = T (α~u+β~v ) = αT (~u)+βT (~v ) = (−20)(−4, 3)+9(7, 8) = (143, 12).
(4) Determine whether w lies in Span{u, v}.
(a) u = (0, −2, 2), v = (1, 3, −1), w = (3, 1, 5).
(b) u = (2, 1, −3, 5), v = (−1, 2, 2, −3), w = (2, 5, 8, 3).
Solution :
(a)
w ∈ Span{u, v} ⇐⇒ x1 u + x2 v = w,
x1 , x2 ∈ R.
The last vector equation is equivalent
to the linear

 system whose aug0
1 | 3
mented matrix is: [ u v | w] =  −2 3 | 1  . An echelon form of
2 −1 | 5


2 −1 | 5

this matrix is 0 2 | 6  . Therefore, the system is consistent ( it
0 0 | 0
has the unique solution, x1 = 4, x2 = 3) and w is in Span{u, v}.
(b) As in the part (a):
w ∈ Span{u, v} ⇐⇒ x1 u + x2 v = w,
x1 , x2 ∈ R.
The last vector equation is equivalent
to the linear

 system whose aug2 −1 | 2
 1
2 | 5 

mented matrix is: [ u v | w] = 
 −3 2 | 8  . An echelon form of
5 −3 | 3


1 2 | 5
 0 −5 | −8 

this matrix is 
 0 8 | 23  . Therefore, the system is inconsistent
0 0 | 9
and w ∈
/ Span{u, v}.
(5) Under what condition(s) does b = (b1 , b2 , b3 ) lie in the span of
S = {(1, 2, −1), (6, 4, 2)}?
Does w = (9, 2, 7) lie in Span{S}? What about w0 = (4, −1, 8)? Explain your
answer.
Solution : Let u = (1, 2, −1), v = (6, 4, 2). Then, as in the previous
question,
b ∈ Span{u, v} ⇐⇒ x1 u + x2 v = b,
x1 , x2 ∈ R.
The last vector equation
 is equivalent
1 6 | b1

2 4 | b2
matrix is: [ u v | b] =
−1 2 | b3

 
1 6 | b1
1



2 4 | b2 = 20
[ u v | b] =
−1 2 | b3
0
to
 the linear system whose augmented
 . Using row operations,
 

6 |
b1
1 6 |
b1
.
−8 | b2 − 2b1  =  20 −8 |
b2 − 2b1
8 | b3 + b1
0 0 | −b1 + b2 + b3
Thus, b = (b1 , b2 , b3 ) lies in the span of S if and only if
−b1 + b2 + b3 = 0, · · · (∗)
which is an equation of a plane in R3 through the origin. The vector w =
(9, 2, 7) satisfies the condition (∗), hence lies in Span{S}, while the vector
w0 = (4, −1, 8) does not satisfies the condition (∗), hence, does not lie in
Span{S}.
(6) Determine if the following vectors span R3 or not:
{(2, −1, 1), (1, 0, −1), (1, 2, 1), (2, 2, 0)}
.
Solution : The set {(2, −1, 1), (1, 0, −1), (1, 2, 1), (2, 2, 0)} spans R3 means
that for any vector b = (b1 , b2 , b3 ) in R3 , the vector equation x1 (2, −1, 1) +
x2 (1, 0, −1) + x3 (1, 2, 1) + x4 (2, 2, 0) = (b1 , b2 , b3 ) has a solution.This vector
 to the linear system whose augmented matrix is
 equation is equivalent
2
1 1 2 | b1
 −1 0 2 2 | b2  . An echelon form of this matrix is
 1 −1 1 0 | b3

1 −1 1 0 |
b3
 0 −1 3 2 |
 . Since no values of b1 , b2 , b3 can produce
b2 + b3
0 0 8 8 | b1 + 3b2 + b3
a row of the form (0 0 0 0 | ∗), (where ∗ a non-zero number), the system is
consistent. Hence, the vectors span R3 .
(7) Let v1 = (0, 3, 1, −1), v2 = (6, 0, 5, 1), v3 = (4, −7, 1, c), where c is a
constant.
(a) Find value(s) of c so that {v1 , v2 , v3 } is linearly dependent in R4 .
(b) For each value of c found in (a), find a nontrivial linear dependence relation
among v1 , v2 and v3 .
Solution :
(a) The set {v1 , v2 , v3 } is linearly dependent if and only if the vector equation
x1 v1 +x2 v2 +x3 v3 = 0 has a nontrivial solution, that is, if and only if the
linear system whose augmented matrix is [v1 v2 v
3 | 0] has a nontrivial

0 6 4 | 0
 3 0 −7 | 0 

solution. An echelon form of [v1 v2 v3 | 0] = 
 1 5 1 | 0 
−1 1 c | 0


1 5
1
| 0
 0 3
2
| 0 

is 
 0 0 c − 3 | 0  . Hence, for c = 3, there is a free variable and
0 0
0
| 0
{v1 , v2 , v3 } is linearly dependent.

1 0 −7/3 |
 0 1 2/3 |
(b) For c = 3, the reduced row echelon form of [v1 v2 v3 | 0] is 
 0 0
0
|
0 0
0
|
7
2
Therefore, x1 = 3 x3 , x2 = − 3 , where x3 is free. For example, if x3 = 3,
then x1 = 7 and x2 = −2. Thus, a nontrivial linear dependence relation
among v1 , v2 , v3 , is 7v1 − 2v2 + 3v3 = 0.
(8) Suppose {v1 , v2 } are linearly independent vectors in R4 . Given w1 =
v1 − 3v2 , w2 = v1 + v2 and w3 = −2v1 + v2 , show that {w1 , w2 , w3 } is
linearly dependent set.
Solution : The set {w1 , w2 , w3 } is linearly dependent set if and only if
the vector equation x1 w1 + x2 w2 + x3 w3 = 0 has a nontrivial solution , that
is, if and only if the linear system whose augmented matrix is [w1 w2 w3 | 0]
has a nontrivial solution. We have
x1 w1 +x2 w2 +x3 w3 = 0 ⇐⇒ x1 (v1 −3v2 )+x2 (v1 +v2 )+x3 (−2v1 +v2 ) = 0 ⇐⇒
(x1 + x2 − 2x3 )v1 + (−3x1 + x2 + x3 )v2 = 0.
Since the set {v1 , v2 } is linearly independent, only trivial linear combination
of v1 , v2 is equal 0, that is
x1
+ x2 − 2x3 = 0
−3x1 + x2 + x3 = 0
This homogeneous system must have a nontrivial solution ( more unknown
than equations). Therefore, the set {w1 , w2 , w3 } is linearly dependent set.
Alternative Solution : We can show that {w1 , w2 , w3 } is linearly dependent set, by expressing one of w1 , w2 , w3 as a linear combination of the

0
0 
.
0 
0
other two. Since w2 = v1 + v2 and w3 = −2v1 + v2 , by subtracting them,
we get, w2 − w3 = 3v1 , that is, v1 = 31 w2 − 13 w3 . Plugging this into the first
equation, we get, w2 = 13 w2 − 13 w3 + v2 . Solving for v2 gives v2 = 23 w2 + 13 w3 .
Finally,
1
1
2
1
5
4
w1 = v1 − 3v2 = w2 − w3 − 3( w2 + w3 ) = (− )w2 − w3 .
3
3
3
3
3
3

1
4
5
6
9
 3 −2 1
4 −1 

(9) Let A = 
 −1 0 −1 −2 −1  . Find the following:
2
3
5
7
8

1. a basis for N ul(A), the nul space of A,
2. a basis for Col(A), the column space of A,
3. a basis for Row(A), the row space of A,
4. rank(A),
5. nullity(A)
Solution :

1
 0
The reduced echelon form of A is B = 
 0
0
0
1
0
0
1
1
0
0
2
1
0
0

1
2 
.
0 
0
1. The augmented matrix [A | 0] of the linear system Ax = 0, where x =
[x1 , x2 , x3 , x4 , x5 ]T and 0 = [0, 0, 0, 0]T has the reduced row echelon form
[B | 0]. Thus , x1 and x2 are leading variables and x3 , x4 , x5 are free
variables. Therefore, x is in N ul(A) means that

 







x1
−x3 − 2x4 − x5
−1
−2
−1
 x2   −x3 − x4 − 2x5 
 −1 
 −1 
 −2 

 







 = x3  1 +x4  0 +x5  0  .
=
x
x
x=
3
3





 



 x4  

 0 
 1 
 0 
x4
x5
x5
0
0
1

 
 

−1
−2
−1






  −1   −2 

−1


 
 







Hence,  1  ,  0  ,  0  is a basis of N ul(A).




  1   0 


 0



0
0
1
2. The pivot columns of B, that is, columns of B which contain a leading
one , are columns 1and
 1 and 2 of A form a basis of
 columns
 2. Thus,
4 
1




  −2 
3

 
Col(A). Therefore, 
 −1  ,  0  ia a basis of Col(A).





2
3
3. Nonzero rows of B, that is {(1, 0, 1, 2, 1), (0, 1, 1, 1, 2)}, form a basis of
Row(B) = Row(A).
4. Rank(A) = dim(Col(A)) = 2.
5. N ullity(A) = dim(N ul(A)) = 3.
(10) Let v1 = (1, 2, 1), v2 = (2, 9, 0), v3 = (3, 3, 4).
(a) Show that B = {v1 , v2 , v3 } form a basis of R3 .
(b) Find the coordinate vector (w)B of w relative to B if w = (5, −1, 9).
Solution :


1 2 3
(a) Let A = [v1 v2 v3 ] =  2 9 3  . Then det(A) = −1 6= 0. Thus,
1 0 4
vectors v1 , v2 , v3 are linearly independent. Since dim(R3 ) = 3, by the
dimension argument (every three linearly independent vectors in a threedimensional vector space form a basis, see Theorem 5.4.5, Section 5.4 of
the text ), B = {v1 , v2 , v3 } form a basis for R3 .
(b) The coordinate vector (w)B of w relative to B is a vector (x1 , x2 , x3 )
such that x1 v1 + x2 v2 + x3 v3 = w. This vector equation is equivalent
system whose augmented matrix is [v1 v2 v3 |w] =
 to the linear 
1 2 3 | −5
 2 9 3 | −1  . The system must have the unique solution ( since
1 0 4 | 9
B is a basis of R3 ). The solution is x1 = 1, x2 = −1, x3 = 2.
Therefore,(w)B = (1, −1, 2).
(11) Let S = {v1 , v2 , v3 , v4 , v5 }, where
v1 = (1, −1, 5, 2), v2 = (−2, 3, 1, 0), v3 = (4, −5, 9, 4), v4 = (0, 4, 2, −3), v5 = (−7, 18, 2, −8).
(a) Find a subset B of S that form a basis for the subspace W = Span{v1 , v2 , v3 , v4 , v5 }.
(b) For each vector in S that is not in B find the coordinate vector relative to
B.
Solution :

1 −2 4
0 −7
 −1 3 −5 4 18 
.
(a) Let A = [v1 v2 v3 v4 v5 ] = 
 5
1
9
2
2 
2
0
4 −3 −8
Then W = Span{v1 , v2 , v3 ,v4 , v5 } = Col(A). The reduced echelon
1 0 2 0 −1
 0 1 −1 0 3 
0
0
0
0
0

form of A is the matrix B = 
 0 0 0 1 2  = [v1 v2 v3 v4 v5 ].
0 0 0 0 0
We see that the basis for Col(B) is {v10 , v20 , v40 }. Thus, a basis B for
W = Span{v1 , v2 , v3 , v4 , v5 } = Col(A) is B = {v1 , v2 , v4 }.

(b) By inspection of columns of B we see that
0
v3 = 2v10 − v20
v50 = −v10 + 3v20 + 2v40
.
Since row operations preserve linear dependence relations we have
v3 = 2v1 − v2
.
v5 = −v1 + 3v2 + 2v4
Therefore, coordinate vectors of v3 and v5 relative to the basis B are:
(v3 )B = (2, −1, 0), (v5 )B = (−1, 3, 2).
(12) Find a vector v in R4 so that {(1, 0, 0, 1), (0, 1, 1, 0), (2, 1, 1, 1), v} is a
basis of R4 .
Solution : Vectors v1 = (1, 0, 0, 1), v2 = (0, 1, 1, 0), v3 = (2, 1, 1, 1) are
linearly independent. Thus, the set {v1 , v2 , v3 } can be extended to a basis of
R4 .




1 0 2 | 1 0 0 0
1 0 0 | −1 0 0
2
 0 1 1 | 0 1 0 0 

1 

 ∼ · · · ∼  0 1 0 | −1 0 1

 0 1 1 | 0 0 1 0 
 0 0 1 | 1 0 0 −1  .
1 0 1 | 0 0 0 1
0 0 0 | 0 1 −1 0
So,we can take v = e2 = (0, 1, 0, 0), hence, {v1 , v2 , v3 , e2 }, is a basis for R4 .
Note that this is not the unique answer.
(13) Suppose that A is 4×5 matrix and b is a column vector in R4 . Given the
following data, determine whether each of the 6 systems of equations Ax = b
has (a) no solution (b) one solution (c) infinitely many solutions (d) the data
do not give enough information to tell (e) the data are impossible.
rank(A) rank([A | b])
(i)
4
5
(ii)
4
4
(iii)
4
3
.
(iv)
3
4
(v)
3
5
(vi)
3
3
Briefly explain your answers. Here [A | b] denotes the augmented matrix of
Ax = b.
Solution :
rank(A)
(i)
4
(ii) 4
(iii) 4
(iv) 3
(v) 3
(vi) 3
rank([A | b])
5
impossible, maximal rank of [A|b] is 4
4
infinitely many solutions, one free variable
3
impossible, rank(A) ≤ rank([A|b])
.
4
no solution, rank(A) =
6 rank([A|b])
5
impossible
3
infinitely many solutions, 2 free variables