Mathematics 2270
2004
1. a) Find
 the
1 2
A = 4 5
7 8

1 0

B= 0 1
0 0

1 0
C = 0 1
0 2
Solutions
PRACTICE EXAM I-Solutions
Fall
rank
of the following matrices:
3
6
10

3 1
2 −1
0 0

1 0
1 −1
0 1


1 2 3
4 5 6  R2 − 4R1 → R1
7 8 10 R3 − 7R1 → R3

1
0
0

1
0
0

1
0
0

1
0
0

2
3
−1
−3 −6 
× R2
3
−6 −11

2
3
R1 − 2R2 → R2
1
2 
−6 −11 R3 + 6R2 → R3

0 −1 R1 + R3 → R1
1 2  R2 − 2R3 → R2
0 1

0 0
1 0
0 1
Then Rank A = 3.
B is already in row reduced form, then Rank B = 2.


1 0 1 0
0 1 1 −1
0 2 0 1 R3 − 2R2 → R3

1
0
0

1
0
0

1
0
0

0 1
0
1 1 −1
−1
× R3
0 −2 3
2
0 1
0
R1 − R 3 → R 1
1 1 −1  R2 − R3 → R2
0 1 −3/2

0 0 3/2
1 0 1/2 
0 1 −3/2
Then Rank C = 3.
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 we found that Rank A = 3 and A is a 3x3 square matrix, then A is invertible,
~
meaning
 that
 the linear system A~x = b has always unique solution.
0

B~y = 1
1
Solution:
 
0
The system B~y = 1 has augmented matrix
1


1 0 3 1 |0
0 1 2 −1 | 1
0 0 0 0 |1
The last row then means that this system is inconsistent, no solution.
C~z = d~
Solution:
As then rank of C is equal to the number of rows, the system is consistent. Moreover
as Rank C = # of columns−1 there is one free variable, then the system has infinitely
many solutions.
1
1
2
2
√
.
2. Let T : R → R be the projection on the line parallel to ~u = 2
1
a) Find the matrix representing T .
Solution:
Applying the formula we find
1/2 1/2
A=
1/2 1/2
b) Show that ker T = L where L is the line perpendicular to ~u. Hint: use the formula
T (~x) = (~x.~u)~u
Solution:
Let ~x ∈ R2 , then we have
~x ∈ ker T ⇐⇒ T (~x) = ~0 ⇐⇒ (~x.~u)~u = ~0
But as ~u 6= 0, then last equation is equivalent to ~x.~u = 0 which is equivalent to ~x
being perpendicular to ~u then to L. So we showed that
~x ∈ ker T ⇐⇒ ~x ⊥ L
which means that ker T = L.
c) Is T one-to-one?
Solution: No because ker T = L 6= {~0}.
3. Let T : R2 → R3 and U : R3 → R2 be two linear maps defined by


 
2x1 + x2
y1
x1
y
1
=  −x2 
T
U y2  =
x2
y1 + y2 + y3
−x1
y3
a) Write down the matrices A and B representing T and U respectively.
Solution: We have


2
1
1 0 0
A =  0 −1
B=
1 1 1
−1 0
b) Calculate the map U ◦ T and the matrix C representing it.
Solution: We have


2x1 + x2
x1
2x1 + x2
2x1 + x2




−x2
=U
U ◦T
=
=
x2
2x1 + x2 + (−x2 ) + (−x1 )
x1
−x1
Then its matrix is
C=
2 1
1 0
c) Show that U ◦ T is a bijection.
Solution: To prove that U ◦ T is a bijection we will prove that its matrix C is
invertible. To do that we reduce it:
1 0
1 0
2 1
R1 ↔ R 2
0 1
2 1 R2 − 2R1 → R2
1 0
Which proves that C is invertible, then that U ◦ T is a bijection.
d) Show that for any ~z ∈ R2 it exists ~y ∈ R3 such that U (~y ) = ~z.
Solution: We know that U ◦ T is bijective. So for any ~z ∈ R2 there exists a (unique)
vector ~x ∈ R2 such that U ◦ T (~x) = ~z. But then, defining ~y = T (~x) ∈ R3 we have
U (~y ) = U (T (~x)) = U ◦ T (~x) = ~z
e) Show that ker T = {0}.
Solution: Let ~x ∈ ker T . We will prove that ~x must be the null vector. Indeed we
have T (~x) = ~0 but then we also have
U (T (~x)) = U (~0) = ~0 =⇒ U ◦ T (~x) = ~0 =⇒ ~x ∈ ker(U ◦ T )
but we know that U ◦ T is bijective then we have ker(U ◦ T ) = {~0}, then we must
have ~x = ~0.
We have then proved that ~0 is the only element of ker T , which means that ker T = {~0}.
4. Let T : Rm → Rp and U : Rp → Rn be two linear maps and vectors ~v1 , . . . , ~vm ∈ Rp
and w
~ 1, . . . , w
~ m ∈ Rn such that
∀i ∈ {1, . . . , m}, T (~ei ) = ~vi and U (~vi ) = w
~i
Give Im(U ◦ T ) as a span of vectors.
Solution: We know that
Im(U ◦T ) = span{U ◦T (~e1 ), . . . U ◦T (~em )} = span{U (T (~e1 )), . . . , U (T (~em ))} = span{w
~ 1, . . . , w
~ m}
because for all i ∈ {1, . . . , m} we have U (T (~ei )) = U (~vi ) = w
~ i.
 

 x1

5. a) Show that V = x2  ∈ R3 | x1 − x2 + 3x3 = 0 is a linear subspace of R3 .


x3
Solution: We can use 2 methods:
First one: we check all properties of the definition of a linear subspace.
First ~0 ∈ Vbecause
0−
0 +

3 × 0 = 0.
x1
y1
Take ~x = x2  , ~y = y2  ∈ R3 , and assume that ~x, ~y ∈ V . Then let’s prove that
x3
y3
~x + ~
y ∈ V . We have


x1 + y1
~x + ~y = x2 + y2 
x3 + y3
We see if ~x + ~y is in V , by checking that its coordinates satisfy the equation defining
V:
(x1 + y1 ) − (x2 + y2 ) + 3(x3 + y3 ) = (x1 − x2 + 3x3 ) + (y1 − y2 + 3y3 ) = 0
Then ~x + ~y ∈ V .


λx1
Take now λ ∈ R, we check that λ~x = λx2  ∈ V :
λx3
λx1 − (λx2 ) + 3(λx3 ) = λ(x1 − x2 + 3x3 ) = 0
then λ~x ∈ V . Then we have proved that V is a linear subspace.
The second method is to identify V as a “known” linear subspace. Here we will
identify V = ker T , where T is the following linear map:
T : R3 → R
 
x1
x2  7→ x1 − x2 + 3x3
x3
It is easy to see that T is a linear map. Moreover it is clear that V = ker T , then
as we know that the kernel of a linear map is a linear subspace, then V is a linear
subspace.
2t + u
2
∈ R | t, u ∈ R is a vector subspace of R2 .
b) Show that W =
−t + 3u
Solution: Here we just remark that W is a span of 2 vectors:
2
1
W = span
,
−1
3
 

 x1

3


x2 ∈ R | x1 − x2 + 3x3 = 1 is not a linear subspace of R3
c) Show that X =


x3
Solution: Here we remark that ~0 ∈
/ X, because 0 − 0 + 3 × 0 = 0 6= 1. Then X is not
a linear subspace.
 
   
0
2
1
6. a) Tell if the vectors 1, 1 and 1 are linearly independant or not.
1
1
1
2 can be expressed as a linear combination of the
b) Show that any
vector
~
x
∈
R
1
−1
x1
vectors ~u =
and ~v =
. Moreover if ~x =
, and if ~x = λ~u + µ~v , find
1
2
x2
λ, µ in function of x1 , x2 .