Mathematics 2270
EXAM I-Solutions
Fall 2004
1. a) Find
 the rank
 of the following matrices:
1 1 0
A = 2 3 4 
1 0 1
Solution:
By rowreduction you find that rank A = 3

1 0 5 6
B = 0 1 0 −1
0 0 0 0
Solution: B is already in row reduced form, then rank B = 2.
b) Give the number of solutions for the following linear systems (you do not need to
solve them! You can use part a)); explain:
A~x = ~b
Solution: As rank A = 3 it is invertible, which means that the system has always
uniquesolution.

0
B~y = 1
0
Solution: The augmented matrix of the system is


1 0 5 6 |0
0 1 0 −1 | 1
0 0 0 0 |0
This system is consistent and has rank B = 2 it has two free variables, then it has
infinite number of solutions.
1
1
2
2
√
2. Let T : R → R be the projection on the line parallel to ~u = 5
.
−2
a) Find the matrix representing T .
Solution: We name the matrix A; using the formula we find
1/5 −2/5
A=
−2/5 4/5
b) Show that T is not one-to-one.
Solution: We prove that by showing that ker T 6= {0}. It is sufficient then to find
a non-zero vector in ker T . Moreover as
know that T (x) = (x.u)u we will choose
we
2
. Then
it perpendicular to u. So we take x =
1
1
x.u = √ (2 × 1 + (−2) × 1) = 0
5
and then T (x) = 0 which means that x ∈ ker T ; and as x 6= 0 then ker T 6= {0} so T
is not one-to-one.
3. Let T : R2 → R3 and U : R3 → R2 be two linear maps represented by the matrices A
and B respectively, where


1 −1
0
2
0
A= 2
0  and B =
1 −1 1
−1 2
a) Compute the matrix product C = BA.
Solution: We compute the product:
C = BA =

1 −1
0 2 0 
4 0

2
0
=
1 −1 1
−2 1
−1 2

b) Deduce from a) that U ◦ T is a bijection (invertible).
Solution: C = BA represents U ◦ T . Moreover C is invertible because
det C = 4 + 2 = 6 6= 0
so U ◦ T is a bijection.
c) Show that U is onto.
Solution: Let z be a vector in R2 , we look for an antecedent of z by U (meaning we
look for a vector y ∈ R3 such that U (y) = z). As U ◦ T is a bijection we know that it
exists a (unique) vector x ∈ R3 such that U ◦ T (x) = z. If we define y = T (x) ∈ R3
we have
z = U ◦ T (x) = U (T (x)) = U (y)
Which proves that U is onto.
4. Define the following matrices:
A=
1 −1 0
2 0 1


2 −1
and B =  2 −1
−4 2
Let T : R3 → R2 and U : R2 → R3 be the linear maps represented by A and B
respectively.
a) Compute the matrix product C = AB. Deduce the map T ◦ U .
Solution: We compute
0 0
C = AB =
0 0
As C represents T ◦U , then T ◦U = 0 (the zero map, that is for all x ∈ R2 , T ◦U (x) = 0.
b) Prove that the 2 vectors
 
 
−1
2
v1 =  2  and v2 = −1
2
−4
are in the kernel of T .
Solution: for i = 1, 2 we have vi = U (ei ), then
T (vi ) = T (U (ei )) = T ◦ U (ei ) = 0 because T ◦ U = 0
Then v1 , v2 ∈ ker T .
c) [Extra credit]: General case: prove that for 2 linear maps f : Rm → Rp and
g : Rp → Rn such that g ◦ f = 0 (that is the zero map, for all ~x ∈ Rm , g ◦ f (~x) = ~0)
we have
Im f ⊂ ker g
Let y ∈ Im f . Then it exists x ∈ Rm such that f (x) = y. Then we
Solution:
compute
g(y) = g(f (x)) = g ◦ f (x) = 0 because g ◦ f is the zero map
So that y ∈ ker g. So we proved Im f ⊂ ker g.
5. a) Show that
 
(
 x1
5x1 − x2 + 3x3 = 0
3
V = x2  ∈ R |

x1 + 6x3 = 0
x3
is a linear subspace of R3 .
Solution: We define the following linear map



T : R 3 → R2
 
x1
5x
−
x
+
3x
1
2
3
x2  7→
x1 + 6x3
x3
then we have ker T 
=
V , which 
proves that V is 
a linear subspace.

 2t + u
b) Show that W = −t + 3u ∈ R3 | t, u ∈ R is a vector subspace of R3 .


5t − u
Solution: We have

   

 

2
1

 
 2t + u
W = −t + 3u ∈ R3 | t, u ∈ R = t −1 + u  3  ∈ R3 | t, u ∈ R

 

5
−1
5t − u
   
1 
 2



−1 , 3 
= span


−1
5
which proves that W
is alinear subspace.
x1
∈ R2 | x2 = x21 is not a linear subspace of R3 .
c) Show that X =
x2
Solution:
We will prove that X is not closed
under scaling. We remark that
1
1
2
∈
/ X because 2 6= 22 = 4. Then
∈ X because it satisfies 1 = 1 . But then 2
1
1
X is not closed under scaling, then X is not a linear subspace.