Complementary Mathematics: Algebra

Fall Semester 2020/2021

Teaching assistant: Olga Bychkova
Date: 12/10/2020

Exercise Session #1
Suggested Solutions
Linear vector spaces. Linear dependence and independence of vectors. Basis.
Arithmetic operations with vectors.

Definition 1. A vector space (over R) consists of a set V along with two operations ‘+’
and ‘·’ such that
~ ∈ V , then their vector sum ~v + w
(1) if ~v , w ~ ∈ V and
• ~v + w ~ =w ~ + ~v
• (~v + w)~ + ~u = ~v + (w ~ + ~u), where ~u ∈ V
• there is a zero vector ~0 ∈ V such that ~v + ~0 = ~v for all ~v ∈ V
• each ~v ∈ V has an additive inverse w ~ ∈ V such that w~ + ~v = ~0
(2) if r and s are scalars (r, s ∈ R) and ~v , w ~ ∈ V , then each scalar multiple r · ~v ∈ V and
• (r + s) · ~v = r · ~v + s · ~v
• r · (~v + w)~ = r · ~v + r · w ~
• (r · s) · ~v = r · (s · ~v )
• 1 · ~v = ~v
Problem 1. Show that the set R2 is a vector space if the operations ‘+’ and ‘·’ have their
usual meaning.

Solution: We shall check all of the conditions in the definition.

There are five conditions in item (1). First, for closure of addition, note that for any v1 ,
v2 , w1 , w2 ∈ R the result of the sum
     
v1 w1 v1 + w1
+ =
v2 w2 v2 + w2

is a column array with two real entries, and so is in R2 .

Second, to show that addition of vectors commutes, take all entries to be real numbers
and compute the following:
           
v1 w1 v1 + w 1 w 1 + v1 w1 v
+ = = = + 1 .
v2 w2 v2 + w 2 w 2 + v2 w2 v2

The second equality follows from the fact that the components of the vectors are real
numbers, and the addition of real numbers is commutative.
The third condition, associativity of vector addition, is similar:
           
v1 w1 u1 v1 + w1 u1 (v1 + w1 ) + u1
+ + = + = =
v2 w2 u2 v2 + w2 u2 (v2 + w2 ) + u2

1
            
v1 + (w1 + u1 ) v1 w 1 + u1 v w1 u1
= = + = 1 + + .
v2 + (w2 + u2 ) v2 w 2 + u2 v2 w2 u2
For the fourth, we must produce a zero element, i.e. the vector of zeroes:
     
v1 0 v
+ = 1 .
v2 0 v2

Fifth, to produce an additive inverse, note that for any v1 , v2 ∈ R we have

     
−v1 v 0
+ 1 = .
−v2 v2 0

So the first vector is the desired additive inverse of the second.

The checks for the five conditions in item (2) are similar. First, for closure under scalar
multiplication, where r, v1 , v2 ∈ R,
   
v rv1
r· 1 =
v2 rv2

is a column array with two real entries, and so is in R2 .

The following checks the second condition:
             
v1 (r + s)v1 rv1 + sv1 rv1 sv1 v1 v
(r + s) · = = = + =r· +s· 1 .
v2 (r + s)v2 rv2 + sv2 rv2 sv2 v2 v2

For the third condition, the scalar multiplication distributes from the left over vector
addition:        
v1 w1 v1 + w1 r(v1 + w1 )
r· + =r· = =
v2 w2 v2 + w2 r(v2 + w2 )
         
rv1 + rw1 rv1 rw1 v1 w1
= = + =r· +r· .
rv2 + rw2 rv2 rw2 v2 w2
The fourth condition follows from
          
v1 (rs)v1 r(sv1 ) sv1 v
(rs) · = = =r· =r· s· 1 .
v2 (rs)v2 r(sv2 ) sv2 v2

And fifth condition is:      

v1 1 · v1 v
1· = = 1 .
v2 1 · v2 v2
In a similar way, each Rn is a vector space with the usual operations of vector addition
and scalar multiplication.
Definition 2. For any vector space, a subspace is a subset that is itself a vector space,
under the inherited operations.
Definition 3. Vectors x1 , ..., xn ∈ V are said to be linearly dependent if there are scalars
a1 , ..., an , not all zero, such that a1 x1 + · · · + an xn = 0.
In other words, x1 , ..., xn are linearly dependent if some xi is expressible as a linear
combination of the remaining vectors.

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x1 , ..., xn are called linearly independent if they are not linearly dependent. Hence x1 , ...,
xn are linearly independent if and only if the equation
a1 x 1 + · · · + an x n = 0
has only the trivial solution a1 = 0, ..., an = 0.
Problem 2. Determine whether the following vectors are linearly dependent or indepen-
dent:
(a) u = (1, 2, 3, 0), v = (2, 0, 1, 1), w = (1, 1, 1, 1).
(b) u = (2, 1, −1), v = (3, −4, −2), w = (7, −2, −4).

Solution:
(a) au + bv + cw = a · (1, 2, 3, 0) + b · (2, 0, 1, 1) + c · (1, 1, 1, 1) =
= (a + 2b + c, 2a + c, 3a + b + c, b + c) = (0, 0, 0, 0).


 a + 2b + c = 0
2a + c = 0


 3a +b+c=0
b+c=0


The last two equations imply a = 0. Substituting it into the second equation, we
obtain c = 0. Then from the last equation, we get b = 0. Therefore, vectors u, v, and
w are linearly independent.
(b) au + bv + cw = a · (2, 1, −1) + b · (3, −4, −2) + c · (7, −2, −4) =
= (2a + 3b + 7c, a − 4b − 2c, −a − 2b − 4c) = (0, 0, 0).

 2a + 3b + 7c = 0
a − 4b − 2c = 0
−a − 2b − 4c = 0


The sum of the last two equations implies −6b − 6c = 0 =⇒ b = −c. Substituting
it into the second equation, we obtain a + 2c = 0 =⇒ a = −2c. Then from the first
equation, we get an identity (−4c − 3c + 7c = 0). Therefore, vectors u, v, and w are
linearly dependent. In particular, w = 2u + v (c = −1 =⇒ a = 2, b = 1).
Problem 3. Determine whether x is a linear combination of vectors u, v, and w:
(a) x = (2, 4, 6), u = (1, 0, 0), v = (0, 1, 0), w = (0, 0, 1).
(b) x = (1, 3, 6), u = (1, 1, 2), v = (2, 1, −1), w = (1, 2, 1).
(c) x = (1, 1, 1), u = (1, 1, −1), v = (2, 1, −2), w = (1, 2, −1).

Solution:
(a) x = (2, 4, 6) = 2 · (1, 0, 0) + 4 · (0, 1, 0) + 6 · (0, 0, 1) = 2u + 4v + 6w. Therefore, x is a
linear combination of vectors u, v, and w.
(b) x = au + bv + cw = a · (1, 1, 2) + b · (2, 1, −1) + c · (1, 2, 1) =
= (a + 2b + c, a + b + 2c, 2a − b + c) = (1, 3, 6).

 a + 2b + c = 1
a + b + 2c = 3
2a − b + c = 6


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 The difference of the first two equations implies b−c = −2 =⇒ b = c−2. Substituting
it into the third equation, we obtain a = 2. The sum of the last two equations implies
3a + 3c = 9 =⇒ c = 3 − a = 1, b = c − 2 = −1. Therefore, x is a linear combination
of vectors u, v, and w: x = 2u − v + w.
(c) x = au + bv + cw = a · (1, 1, −1) + b · (2, 1, −2) + c · (1, 2, −1) =
= (a + 2b + c, a + b + 2c, −a − 2b − c) = (1, 1, 1).

 a + 2b + c = 1
a + b + 2c = 1
−a − 2b − c = 1


The first and third equations contradict to each other (they imply 1 = −1). Therefore,
the system of equations doesn’t have a solution, and x isn’t a linear combination of
vectors u, v, and w.
Definition 4. Vectors x1 , ..., xn ∈ S are said to form a basis of subspace S if
(a) x1 , ..., xn are linearly independent;
(b) every vector in S is a linear combination of x1 , ..., xn .
Problem 4. Show that the vectors a1 = (2, 4) and a2 = (1, 1) form a basis for R2 .

Solution: First, we show that the vectors are linearly independent:

xa1 + ya2 = x · (2, 4) + y · (1, 1) = (2x + y, 4x + y) = (0, 0).


2x + y = 0
4x + y = 0
The difference of the two equations implies 2x = 0 =⇒ x = 0. Substituting it into any
eqution, we obtain y = 0.
Second, we show that every vector a3 = (u, v) ∈ R2 is a linear combination of a1 and a2 :

a3 = xa1 + ya2 = x · (2, 4) + y · (1, 1) = (2x + y, 4x + y) = (u, v).


2x + y = u
4x + y = v
The difference of the two equations implies 2x = v − u =⇒ x = (v−u)/2. Substituting it
into the first equation, we obtain y = u − 2x = u − (v − u) = 2u − v.
Problem 5. Show that the vectors a1 = (1, 0, −1), a2 = (1, 2, 1), and a3 = (0, −3, 2) form
a basis for R3 .

Solution: First, we show that the vectors are linearly independent:

xa1 +ya2 +za3 = x·(1, 0, −1)+y·(1, 2, 1)+z·(0, −3, 2) = (x+y, 2y−3z, −x+y+2z) = (0, 0, 0).

 x+y =0
2y − 3z = 0
−x + y + 2z = 0


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From the first equation, we obtain x = −y. Substituting it into the third equation, we get
2y + 2z = 0. Together with the second equation it implies 5z = 0 =⇒ z = 0. Then from
the second equation, we obtain 2y = 0 =⇒ y = 0 =⇒ x = 0.
Second, we show that every vector a4 = (u, v, w) ∈ R3 is a linear combination of a1 , a2 ,
and a3 :

a4 = xa1 +ya2 +za3 = x·(1, 0, −1)+y·(1, 2, 1)+z·(0, −3, 2) = (x+y, 2y−3z, −x+y+2z) = (u, v, w).

 x+y =u
2y − 3z = v
−x + y + 2z = w


From the first equation, we obtain x = u − y. Substituting it into the third equation, we
get 2y − u + 2z = w. Together with the second equation it implies 5z − u = w − v =⇒
z = (w−v+u)/5. Then from the second equation, we obtain 2y = v +3z = v + 3/5(w −v +u) =
1/5(3w + 2v + 3u) =⇒ y = 1/10(3w + 2v + 3u) =⇒ x = u − y = u − 1/10(3w + 2v + 3u) =
1/10(7u − 3w − 2v).

Problem 6. Calculate the following:

(a) x + y, where x = (1, 3, 5) and y = (2, 4, 6).
(b) u − v, where u = (7, 9, 11) and v = (8, 9, 10).
(c) c · v, where c = 15 and v = (1, 3, 6).
(d) 2u + (−3)w, where u = (−1, −7) and w = (3, 1).
(e) 1/3 · w + 3u, where u = (−2, 5) and w = (3, 1).

Solution:
(a) x + y = (1, 3, 5) + (2, 4, 6) = (1 + 2, 3 + 4, 5 + 6) = (3, 7, 11).
(b) u − v = (7, 9, 11) − (8, 9, 10) = (7 − 8, 9 − 9, 11 − 10) = (−1, 0, 1).
(c) c · v = 15 · (1, 3, 6) = (15 · 1, 15 · 3, 15 · 6) = (15, 45, 90).
(d) 2u + (−3)w = 2 · (−1, −7) − 3 · (3, 1) = (2 · (−1) − 3 · 3, 2 · (−7) − 3 · 1) = (−11, −17).
(e) 1/3 · w + 3u = 1/3 · (3, 1) + 3 · (−2, 5) = (1/3 · 3 + 3 · (−2), 1/3 · 1 + 3 · 5) = (−5, 151/3).

Problem 7. Compute the dot product for each of the following:

(a) u = (5, −8), v = (1, 2).
(b) x = (−6, 8), y = (5, 12).
(c) a = (0, 3, −7), b = (2, 3, 1).
(d) u = (9, 2, 7), w = (4, 8, 10).

Solution:
(a) u · v = (5, −8) · (1, 2) = 5 · 1 + (−8) · 2 = −11.
(b) x · y = (−6, 8) · (5, 12) = −6 · 5 + 8 · 12 = 66.
(c) a · b = (0, 3, −7) · (2, 3, 1) = 0 · 2 + 3 · 3 + (−7) · 1 = 2.
(d) u · w = (9, 2, 7) · (4, 8, 10) = 9 · 4 + 2 · 8 + 7 · 10 = 122.