Write your answers to multiple-choice questions in the table below.
Question 1 Question 2 Question 3 Question 4
Multiple-choice answers:
Do not write in the following table.
Total MC Question 5 Question 6 Question 7 Question 8 Total on 26
Q1.[2 points] Exactly one of the following is a subspace of the vector space M2,2(R). Which is it?
0
A=
c
a
B=
c
a
C=
c
a
D=
c
a
E=
3
b
: bc = 0, b, c ∈ R
0
b
: ad = 1, a, b, c, d ∈ R
d
b
: b − c = 0, a, b, c, d ∈ R
d
b
: a + b + c = 1, a, b, c ∈ R
0
b
: a, b, d ∈ R
d
Solution: The zero matrix does not belong to any of options
B, D and E. So those are not subspaces.
0 1
0 0
Option A is not closed under addition: matrices
and
are both in option A but their sum
0 0
1 0
0 1
is not. Option C is a subspace since it is closed under addition and scalar multiplication, and
1 0
0 0
belongs to it.
0 0
Solution: C
Write your answer to multiple-choice questions in table on page 2.
Q2.[2 points] Which of the following statements are TRUE?
     
   
0
0 
0 
 2
 1
I. span 2 , 1 = span 4 , −1 , 0 .
 6
 3
0
0 
0 
   
 
0 
 1
 1 
II. span 2 , 1 = span 2 .
 3
 3 
0 
III. The function f (x) = 5 is a linear combination of the functions g(x) = cos2 (x) and h(x) = sin2 (x).
     
1
0
1
IV. Every vector in R3 is a linear combination of the vectors 0 , 1 , 1.
0
0
0
V. For vectors u, v in an arbitrary vector space V , span{u, v} = span{u + v, −v}.
A. Only I and IV
B. Only III and V
C. Only II, III, and IV
D. Only II, IV, and V
E. Only I, III, and V
F. All of the statements are true.
   
   
     
   
0
0 
0 
0
0
1
 2
 1
 2

















4 , −1
2 , 1
4 , −1 , 0
⊆ span
. Thus, span
Solution: E Solution: 2 , 1 ∈ span





0
6
0
3
0
0
6
0
3
(by Big Theorem).
   
     
    
     
0 
0
0 
2
0
0
0 
 1
 2
 1



















2 , 1
2 , 1
4 , −1 , 0
⊆ span
Similarly, 4 , −1 , 0 ∈ span
. Thus, span





0
0
3
0 
6
6
0
0
3
0
(by Big Theorem).
Thus, these two spans are equal.
 
   
 
 
0
0 
 1 
 1
 1 
Since 1 ̸∈ span 2 , we conclude that span 2 , 1 ̸= span 2 .
 3 
 3
 3 
0
0 
Since 5 = 5 cos2 (x)+5 sin2 (x), the function f (x) = 5 is a linear combination of the functions g(x) = cos2 (x)
and h(x) = sin2 (x).
Vectors in R3 in 
which
 coordinate is non-zero cannot be expressed as a linear combination of the
 the
 third
1
1
0
three vectors 0 , 1 , 1.
0
0
0
Since u = 1(u+v)+1(−v) and v = 0(u+v)+(−1)(−v), it follows by the Big Theorem that span{u, v} ⊆
span{u + v, −v}.
Since u + v = 1u + 1v and −v = 0u + (−1)v, it follows by the Big Theorem that span{u + v, −v} ⊆
span{u, v}.
Thus, span{u, v} = span{u + v, −v}.
Write your answer to multiple-choice questions in table on page 2.
Q3.[2 points] Let P2 = {p(x) = a + bx + cx2 : a, b, c ∈ R} be the vector space of polynomials of degree
at most 2.
Consider the following three subsets of P2 :
S = {1 + x, x + x2 , 1 + x + x2 }
T = {1 + x, x + x2 , 1 + 2x + x2 }
U = {0, 1, x, x2 }
Which of the following statements are true?
I. S, T , and U are each linearly dependent.
II. S and U are linearly independent but T is linearly dependent.
III. T and U are linearly independent but S is linearly dependent.
IV. S is linearly independent, but T and U are each linearly dependent.
V. U spans P2 .
A. Only I.
B. Only II.
C. Only III.
D. Only IV.
E. Only I and V.
F. Only II and V.
G. Only III and V.
H. Only IV and V.
Solution: H Solution: For S, we have
a(1 + x) + b(x + x2 ) + c(1 + x + x2 ) = 0 =⇒ (b + c)x2 + (a + b + c)x + (a + c) = 0
=⇒ b + c = 0, a + b + c = 0 and a + c = 0
So b = −c, and so a + b + c = a + (−c) + c = 0 =⇒ a = 0.
Finally, from a + c = 0, we have c = −a, so c = 0.
The only solution to the dependence equation is the trivial solution, so S is LI.
For T , e
a(1 + x) + b(x + x2 ) + c(1 + 2x + x2 ) = 0 =⇒ (b + c)x2 + (a + b + 2c)x + (a + c) = 0
=⇒ b + c = 0, a + b + 2c = 0 and a + c = 0
So b = −c and a = −2c − b = −2c − (−c) = −c and a = −c.
c can take on any real value to give a solution, so we find that non-trivial solutions to the dependence
equation exist!
For instance 1(1 + x) + 1(x + x2 ) + (−1)(1 + 2x + x2 ) = 0.
Thus, T is LD.
U contains the zero polynomial, so U is LD.
Every vector in P2 can be written as a linear combination of the vectors in U since ax2 + bx + c =
1(0) + c(1) + b(x) + a(x2 ).
Thus, P2 = span(U ).
Write your answer to multiple-choice questions in table on page 2.
Q4.[2 points] Which of the following statements are TRUE?
I. The set W = {(x, y, z) ∈ R3 | x + y + z = 0} is a subspace of R3 .
II. The set of all 2 × 2 matrices satisfying,
a b
a c
=
c d
b d
forms a vector space under standard matrix addition and scalar multiplication.
III. The set of all polynomials of degree exactly 3 forms a vector space under standard polynomial
addition and scalar multiplication.
IV. The set V = {(x, y) ∈ R2 | x2 + y 2 = 0} is a subspace of R2 .
A. Only I and II
B. Only I and IV
C. Only II and III
D. Only I, II, and IV
E. Only I, III, and IV
F. All of the statements are true
Solution: D Solution:
a
U=
c
a
=
b
b
a b
a c
∈ M2,2 (R) :
=
d
c d
b d
b
: a, b, d ∈ R
d
1 0
0 1
0 0
= span
,
,
0 0
1 0
0 1
Thus, U is a subspace of M2,2 (R), hence a vector space.
The set of all polynomials of degree exactly 3 does not contain the zero polynomial Z(x) = 0 (degree ̸= 3).
Thus, this set does not form a vector space.
V = {(0, 0)} is a subspace of R2 (the trivial space).
W is a plane passing through the origin hence is a subspace of R3 .
Write your answer to multiple-choice questions in table on page 2.
Q5. [4 points] Let V = {f : R → R} be the set of all real-valued functions with domain R, with
following non-standard operations:
• **Addition**: For any f, g ∈ V , define (f ⊕ g)(x) = f (x) + g(x) + 1
for all x ∈ R
• **Scalar Multiplication**: For any f ∈ V and a ∈ R, define (a ⊙ f )(x) = af (x) + a for all x ∈ R.
(a) For f (x) = sin(x) − 5 and g(x) = 2ex + x2 , what is f ⊕ g and what is 3 ⊙ f ?
(f ⊕ g)(x) = Solution: (f ⊕ g)(x) = sin(x) − 5 + 2ex + x2 + 1 = sin(x) + 2ex − 4
(3 ⊙ f )(x) =
Solution: (kf )(x) = 3(sin(x) − 5) + 3 = 3 sin(x) − 12
(b) Show that V has a zero vector (a distinguished function Z(x) satisfying Z ⊕ f = f for all vectors
f ∈ V ).
Solution: (b) We have
Z ⊕ f = f ⇐⇒ Z(x) + f (x) + 1 = f (x) for all x ∈ R
⇐⇒ Z(x) = f (x) − f (x) − 1 for all x ∈ R
⇐⇒ Z(x) = −1 for all x ∈ R
Thus, the constant function Z(x) = −1 is the zero vector of V .
(c) Show that for each f in V there exists a negative vector (−f ) ∈ V which satisfies f ⊕ (−f ) = Z
for the function Z(x) you found in part (b).
Solution: (c) We have
f ⊕ (−f ) = Z ⇐⇒ f (x) + (−f )(x) + 1 = Z(x) for all x ∈ R
⇐⇒ (−f )(x) = Z(x) − f (x) − 1 for all x ∈ R
⇐⇒ (−f )(x) = −1 − f (x) − 1 for all x ∈ R
⇐⇒ (−f )(x) = −f (x) − 2 for all x ∈ R
Thus, for each f ∈ V , the negative vector (−f ) is the function (−f )(x) = −f (x) − 2.
(d) Give a counterexample (with concrete vectors f, g ∈ V and scalar k ∈ R) to show that the arithmetic
property k ⊙ (f ⊕ g) = (k ⊙ f ) ⊕ (k ⊙ g) does not hold for all vectors in V .
Solution: (d) Here is one counterexample (but there are many other possible counterexamples):
Let f (x) = sin(x), g(x) = x and k = 5.
Then
(5 ⊙ (f ⊕ g))(x) = 5(f ⊕ g)(x) + 5
= 5(sin(x) + x + 1) + 5
= 5 sin(x) + 5x + 10
((5 ⊙ f ) ⊕ (5 ⊙ g))(x) = (5 sin(x) + 5) ⊕ (5x + 5)
= (5 sin(x) + 5) + (5x + 5) + 1
= 5 sin(x) + 5x + 11
Since 5 ⊙ (f ⊕ g) ̸= (5 ⊙ f ) ⊕ (5 ⊙ g) for these functions, this arithmetic property does not hold for
all vectors in V .
Q6.[3 points] Let S = {v ∈ R3 | v · (1, −2, 3) = 0}, with the usual operations of R3.
Use the subspace test to show that S is a subspace of R3 .
Solution: First, we show S is closed under addition. Let u, v ∈ S, then by definition of S, u.(1, −2, 3) = 0
and v.(1, −2, 3) = 0. Now, (u + v).(1, −2, 3) = u.(1, −2, 3) + v.(1, −2, 3) = 0. Hence, by definition of S,
(u + v) ∈ S.
Second, we show S is closed under scalar multiplication. Let k ∈ R and v ∈ S. So,
(kv).(1, −2, 3) = k(v.(1, −2, 3)) = k.0 = 0. By definition of S, kv ∈ S.
Finally, we verify 0 ∈ S. Since 0.(1, −2, 3) = (0, 0, 0).(1, −2, 3) = 0. By definition of S, we conclude that
0 ∈ S.
Q7. [5 points] Let M2,2(R) denote the vector space of 2 × 2 matrices with real entries and consider the
following subspace:
U=
a
b−a
a + 2b
0
: a, b ∈ R .
(a) Find a spanning set for U .
Solution: (a)
a
b−a
U=
: a, b ∈ R
a + 2b
0
1 −1
0 1
+b
: a, b ∈ R
= a
1 0
2 0
1 −1
0 1
= span
,
1 0
2 0
(this question continues on the next page)
(previous question continued)
Recall: U =
(b) Find a matrix A ∈ M2,2 (R) such that A ̸∈ U .
a
b−a
a + 2b
0
: a, b ∈ R .
Justify your answer.
Solution: There are many possible answers.
0 0
For instance,
̸∈ U since its (2, 2)-entry does not equal 0.
0 1
1 2
(c) Show that B =
∈ U.
7 0
Justify your answer.
Solution:
1 2
1 2
a
b−a
∈ U ⇐⇒
=
for some a, b ∈ R.
7 0
7 0
a + 2b
0

 1=a
2=b−a
⇐⇒
 7 = a + 2b
a=1
⇐⇒
b=3
a
b−a
1 2
Using a = 1 and b = 3, we see that
=
= B ∈ U.
a + 2b
0
7 0
(d) Is W =
x 2x
7x 0
: x∈R
a subspace of U ?
Justify your answer.
1 2
1 2
Solution: Notice that W = span
and, from part (c),
∈ U . By the Big Theorem,
7 0
7 0
1 2
span
is a subspace of U .
7 0
Alternative (longer) solution: since U is a vector space, we can show that W is a subset of U and
apply the Subspace Test:
x 2x
x
3x − x
Let
be an arbitrary element of W . Then W =
∈ U (by using a = x
7x 0
x + 2(3x)
0
and b = 3x.
Thus, W ⊆ U .
x 2x
y 2y
Let X =
,Y =
∈ W and let k ∈ R. Then
7x 0
7y 0
x + y 2x + 2y
(x + y) 2(x + y)
X +Y =
=
∈ W.
7x + 7y 0 + 0
7(x + y)
0
Thus, W is closed under addition.
x 2x
kx k(2x)
kx 2(kx)
Moreover, kX = k
=
=
∈ W.
7x 0
k(7x) k(0)
7(kx)
0
Thus, W is closed under scalar multiplication.
0 0
0 2(0)
Finally, 0 =
=
∈ W.
0 0
7(0) 0
By the Subspace Test, W is a subspace of U .
Q8. [6 points]
a) Let V be a vector space and u, v be two vectors in V . Given that {u, v} is linearly independent,
prove that {u + v, v} is linearly independent.
Solution: a) Let V be a vector space and u, v ∈ V .
Assume {u, v} is linearly independent.
Then
a(u + v) + bv = 0 =⇒ au + (a + b)v = 0
=⇒ a = 0 and a + b = 0
=⇒ a = 0 and b = 0.
since {u, v} is LI
Since the only solution to the dependence equation a(u + v) + bv = 0 is the trivial solution
(a = b = 0), we conclude that {u + v, v} is linearly independent.
b) Let V be a vector space and u, v, w be three vectors in V . Given that {u, v} is linearly dependent,
prove that {u, v, w} is linearly dependent.
Solution: b) Let V be a vector space and u, v ∈ V .
Assume {u, v} is linearly dependent.
Then there exists a non-trivial solution to the dependence equation au+bv = 0 (so a ̸= 0 or b ̸= 0).
Assume without loss of generality that a ̸= 0.
Then au + bv + 0w = 0 is a non-trivial solution showing that {u, v, w} is LD.
c) Suppose that u, v are non-zero vectors in a vector space V . Prove that if {u, v} is linearly dependent
then they are multiples of one another, i.e. u = cv for some scalar c.
Solution: c) Suppose that u, v are non-zero vectors in a vector space V . Assume {u, v} is linearly
dependent.
Since {u, v} is linearly dependent, there exist some scalars, not all zero, a, b ∈ R such that
au + bv = 0.
If a ̸= 0, then we have u = − ab v.
If b ̸= 0, then we have v = − ab u. In either case, the vectors u
and v are multiples of one another.
(end of the exam!!)