MATH 240 – FALL 2014

H O M EWO R K
A S S I GN M E NT # 4
CORRECTION

Algebra I

14/10/2014
UNIVERSITY OF TORONTO

Professor: Dror Bar-Natan

Correction Homework Assignment #4

Exercise 8 page 41-42: Let 𝑆 = {(1,1,0), (1,0,1), (0,1,1)} be a subset of the vector space 𝐹 3

(a) Prove that if 𝐹 = ℝ, then 𝑆 is linearly independent

Let 𝑆 = {(1,1,0), (1,0,1), (0,1,1)} be a subset of the vector space ℝ3

Suppose 𝑎1 , 𝑎2 , 𝑎3 ∈ ℝ such that

(0,0,0) = 𝑎1 (1,1,0) + 𝑎2 (1,0,1) + 𝑎3 (0,1,1)
We obtain the following system of equation:

𝑎1 + 𝑎2 = 0 𝑒1 2𝑎1 = 0 𝑒1 + 𝑒2 − 𝑒3
𝑎1 + 𝑎3 = 0 𝑒2 → 𝑎1 + 𝑎3 = 0 𝑒2
𝑎2 + 𝑎3 = 0 𝑒3 𝑎2 + 𝑎3 = 0 𝑒3

𝑎1 = 0 1/2 ⋅ 𝑒1 𝑎1 = 0 𝑒1
𝑎1 + 𝑎3 = 0 𝑒2 → 𝑎3 = 0 𝑒2 − 𝑒1
𝑎2 + 𝑎3 = 0 𝑒3 𝑎2 + 𝑎3 = 0 𝑒3

𝑎1 = 0 𝑒1
𝑎3 = 0 𝑒2
𝑎2 = 0 𝑒3 − 𝑒2

Therefore the only solution for this system is 𝑎1 = 𝑎2 = 𝑎3 = 0. Si 𝑆 is linearly independent.

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(b) Prove that if 𝐹 has characteristic 2, then 𝑆 is linearly dependent

Let 𝑆 = {(1,1,0), (1,0,1), (0,1,1)} be a subset of the vector space 𝐹 3 where 𝐹 is a field with
characteristic 2.

Since 𝐹 is a field with characteristic 2 we have that: 1 + 1 = 0. Therefore if we consider the

linear combination of the vectors of 𝑆 where 𝑎1 , 𝑎2 , 𝑎3 ∈ 𝐹 and 𝑎1 = 𝑎2 = 𝑎3 = 1, we get:

𝑎1 (1,1,0) + 𝑎2 (1,0,1) + 𝑎3 (0,1,1) = 1 ⋅ (1,1,0) + 1 ⋅ (1,0,1) + 1 ⋅ (0,1,1)

= (1,1,0) + (1,0,1) + (0,1,1) = (1 + 1,1 + 1,1 + 1) = (0,0,0)

Therefore, in a field with characteristic 2, 𝑎1 = 𝑎2 = 𝑎3 = 1 is a finite linear combination of

vectors of 𝑆 whose coefficients are not all equals to 0. Hence, 𝑆 is linearly dependent.
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Exercise 9 page 41: Let 𝑢 and 𝑣 be distinct vectors in a vector space 𝑉. Show that {𝑢, 𝑣} is linearly
dependent if and only if 𝑢 or 𝑣 is a multiple of the other.

Let 𝑢 and 𝑣 be distinct vectors in a vector space 𝑉 over a field 𝐹 .

We show first that 𝑢 or 𝑣 is a multiple of the other ⟹ {𝑢, 𝑣} linearly dependent.
- Suppose 𝑢 is a multiple of 𝑣. Then there exists 𝑞 ∈ 𝐹 such that 𝑢 = 𝑣𝑞. From this, we
have that: 𝑢 − 𝑣𝑞 = 0𝑉 . Thus, there exists a non trivial representation of the zero vector
as a linear combination of distinct vectors of {𝑢, 𝑣} since we have 𝑎1 = 1.
- Suppose 𝑣 is a multiple of 𝑢. Then there exists 𝑞 ∈ 𝐹 such that 𝑣 = 𝑢𝑞. From this, we
have that: 𝑣 − 𝑢𝑞 = 0𝑉 . Thus, there exists a non trivial representation of the zero vector
as a linear combination of distinct vectors of {𝑢, 𝑣} since we have 𝑎1 = 1.
Hence 𝑢 or 𝑣 is a multiple of the other ⟹ {𝑢, 𝑣} linearly dependent.

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Correction Homework Assignment #4

Now let us show that {𝑢, 𝑣} linearly dependent ⟹ 𝑢 or 𝑣 is a multiple of the other

Suppose {𝑢, 𝑣} linearly dependent. Then there exists 𝑎, 𝑏 ∈ 𝐹 such that 𝑎, 𝑏 are not both equal
to zero such that 𝑎𝑢 + 𝑏𝑣 = 0𝑉
- Suppose 𝑎 ≠ 0. Then 𝑎𝑢 + 𝑏𝑣 = 0𝑉 ⟹ 𝑎𝑢 = −𝑏𝑣. Moreover, since 𝑎 ≠ 0, ∃𝑎−1 ∈
𝐹 such that 𝑎 ⋅ 𝑎−1 = 1. Thus, we get: 𝑎𝑢 = −𝑏𝑣 ⟹ 𝑎−1 𝑎𝑢 = 𝑎−1 (−𝑏𝑣)
⟹ 1 ⋅ 𝑢 = 𝑎−1 (−𝑏𝑣) ⟹ 𝑢 = (−𝑏𝑎−1 )𝑣. Therefore 𝑢 is a multiple of 𝑣.
- Suppose 𝑏 ≠ 0. Then 𝑎𝑢 + 𝑏𝑣 = 0𝑉 ⟹ 𝑏𝑣 = −𝑎𝑢. Moreover, since 𝑏 ≠ 0, ∃𝑏 −1 ∈
𝐹 such that 𝑏 ⋅ 𝑏 −1 = 1. Thus, we get: 𝑏𝑣 = −𝑎𝑢 ⟹ 𝑏 −1 𝑏𝑣 = 𝑏 −1 (−𝑎𝑢)
⟹ 1 ⋅ 𝑣 = 𝑏 −1 (−𝑎𝑢) ⟹ 𝑣 = (−𝑎𝑏 −1 )𝑢. Therefore 𝑣 is a multiple of 𝑢.
Thus {𝑢, 𝑣} linearly dependent ⟹ 𝑢 or 𝑣 is a multiple of the other.
Hence 𝑢 or 𝑣 is a multiple of the other ⟺ {𝑢, 𝑣} linearly dependent.
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Exercise 11 page 42: Let 𝑆 = {𝑢1 , 𝑢2 , … , 𝑢𝑛 } be a linearly independent subset of a vector space 𝑉 over a
field ℤ2 . How many vectors are there in 𝑆𝑝𝑎𝑛(𝑆)? Justify your answer.

Let 𝑆 = {𝑢1 , 𝑢2 , … , 𝑢𝑛 } be a linearly independent subset of a vector space 𝑉 over the field ℤ2 .
Let denote by 𝑛 the number of elements of 𝑆 which we denote by |𝑆|. If 𝑛 = 0, then 𝑆 = ∅
(possible since the empty set is linearly independent) then 𝑆𝑝𝑎𝑛(𝑆) = {0}. Therefore,
|𝑆𝑝𝑎𝑛(𝑆)| = 1

Suppose now that 𝑛 ≥ 1. We have 𝑆 = {(1)} because {(0)} is linearly dependent and any
nonempty subset of a vector space that contain 0 is linearly dependent.
Moreover we have (1) = 1 ⋅ (1) and (0) = 0 ⋅ (1). Thus 𝑆𝑝𝑎𝑛(𝑆) = ℤ2 and it has 2 elements.

So the number of elements of the span of a linearly independent subset of a vector space 𝑉 over
a field ℤ2 is 1 if the number of elements the subset is 0 and 2 is the number of element of the
subset is greater than 1.

General case for a linearly independent subset of a vector space 𝑉 over a field ℤ𝑘2 with 𝑘 a positive integer

Let 𝑆 = {𝑢1 , 𝑢2 , … , 𝑢𝑛 } be a linearly independent subset of a vector space 𝑉 over the field ℤ𝑘2 .

We claim that the number of vectors in 𝑆𝑝𝑎𝑛(𝑆) = 2𝑛 . We will show it by induction over
|𝑆| = 𝑛.

If 𝑛 = 0, then 𝑆 = ∅ (possible since the empty set is linearly independent) then 𝑆𝑝𝑎𝑛(𝑆) =
{0𝑉 }. Therefore, |𝑆𝑝𝑎𝑛(𝑆)| = 1 = 20

Base case: We now suppose that 𝑛 = 1. So 𝑆 = {𝑢1 } is non empty. If 𝑣 ∈ 𝑠𝑝𝑎𝑛(𝑆), there exists
𝑎 ∈ ℤ2 such that
𝑣 = 𝑎𝑢1
Since ℤ2 has only two elements we have that 𝑎 = 1 or 𝑎 = 0. Therefore, 𝑠𝑝𝑎𝑛(𝑆) = {𝑢1 , 0𝑉 }
and |𝑆𝑝𝑎𝑛(𝑆)| = 2 = 21

Suppose that |𝑆𝑝𝑎𝑛(𝑆)| = 2𝑛 for any linearly independent subset of a vector space 𝑉 over ℤ𝑘2
such that |𝑆| = 𝑛 ≥ 1. Let us show that it is true for 𝑛 + 1.

Suppose that 𝑆 ∪ {𝑢𝑛+1 } is linearly independent. Moreover 𝑆 ⊂ 𝑆 ∪ {𝑢𝑛+1 } ⟹ 𝑆𝑝𝑎𝑛(𝑆) ⊂

𝑆𝑝𝑎𝑛(𝑆 ∪ {𝑢𝑛+1 }) ⟹ |𝑆𝑝𝑎𝑛(𝑆)| = 2𝑛 ≤ |𝑆𝑝𝑎𝑛(𝑆 ∪ {𝑢𝑛+1 })| since 𝑆 is finite (and thus so is

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Correction Homework Assignment #4

its span). We want to know how much new vectors does 𝑢𝑛+1 brings to the span of this new set.
Suppose that 𝑣 ∈ 𝑆𝑝𝑎𝑛(𝑆 ∪ 𝑢𝑛+1 ). Then
𝑛

𝑣 = ∑ 𝑎𝑖 𝑢𝑖 + 𝑎𝑛+1 𝑢𝑛+1
𝑖=1
For every new vector of 𝑆𝑝𝑎𝑛(𝑆 ∪ 𝑢𝑛+1 ) we must have that 𝑎𝑛+1 = 1 otherwise 𝑣 was already
in 𝑆𝑝𝑎𝑛(𝑆). So therefore we have that:
𝑛

𝑣 = ∑ 𝑎𝑖 𝑢𝑖 + 𝑢𝑛+1
𝑖=1
We claim that any vector of this form is not already in 𝑆𝑝𝑎𝑛(𝑆). In fact if it were in the 𝑆𝑝𝑎𝑛(𝑆)
we would have that:
𝑛

𝑣 = ∑ 𝑏𝑖 𝑢𝑖
𝑖=1

This implies that (𝑣 + 𝑣 = 0 because 1 + 1 = 0 + 0 = 0):

𝑛 𝑛 𝑛

𝑣 + 𝑣 = 0 = ∑ 𝑏𝑖 𝑢𝑖 + ∑ 𝑎𝑖 𝑢𝑖 + 𝑢𝑛+1 = ∑(𝑎𝑖 + 𝑏𝑖 )𝑢𝑖 + 𝑢𝑛+1

𝑖=1 𝑖=1 𝑖=1
This would implies that there exists a non trivial linear combination of 0 with distinct vectors of
𝑆 ∪ 𝑢𝑛+1 since 𝑎𝑛+1 = 1. Thus it contradicts the fact that 𝑆 ∪ 𝑢𝑛+1 is linearly independent.
Thus every new vectors in 𝑆 ∪ 𝑢𝑛+1 is of the form:
𝑛

𝑣 = ∑ 𝑎𝑖 𝑢𝑖 + 𝑢𝑛+1
𝑖=1
Since there are 2𝑛 vectors of the form ∑𝑛𝑖=1 𝑎𝑖 𝑢𝑖 since |𝑆𝑝𝑎𝑛(𝑆)| = 2𝑛 by our induction
hypothesis. Therefore, there are 2𝑛 vectors of the form ∑𝑛𝑖=1 𝑎𝑖 𝑢𝑖 + 𝑢𝑛+1 . Since these vectors
does not belongs to 𝑆𝑝𝑎𝑛(𝑆) we have that:
𝑆𝑝𝑎𝑛(𝑆 ∪ 𝑢𝑛+1 ) = 2𝑛 + 2𝑛 = 2𝑛+1

Hence, we proved that |𝑆𝑝𝑎𝑛(𝑆)| = 2𝑛 for any linearly independent subset of a vector space 𝑉
over ℤ𝑘2 such that |𝑆| = 𝑛 ≥ 1.
Actually as we proved it before, this is also true for 𝑛 = 0.

Exercise 4 page 54: Do the polynomials 𝑥 3 − 2𝑥 2 + 1,4𝑥 2 − 𝑥 + 3, and 3𝑥 − 2 generate 𝑃3 (ℝ)?

Justify your answer.

Let us consider the subset 𝑆 = {𝑥 3 − 2𝑥 2 + 1,4𝑥 2 − 𝑥 + 3, 3𝑥 − 2} of the vector space 𝑃3 (𝑅).

Moreover, we have that dim(𝑃3 (ℝ)) = 4. By the corollary 2 of the replacement theorem, we
have that any finite generating set of a vector space of finite dimension 𝑛 should at least contains
𝑛 elements. Here we have that |𝑆| = 3 < dim(𝑃3 (ℝ)) = 4. Therefore, 𝑆 does not generates
𝑃3 (ℝ).
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Exercise 5 page 54: Is {(1,4, −6), (1,5,8), (2,1,1), (0,1,0)} a linearly independent subset of ℝ3 ?
Justify your answer.

Let us consider the subset 𝑆 = {(1,4, −6), (1,5,8), (2,1,1), (0,1,0)} of the vector space ℝ3 .

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Correction Homework Assignment #4

Moreover, we have that dim(ℝ3 ) = 3.

Let 𝑛 be the dimension of a finite dimensional vector space. By the corollary 2 of replacement
theorem, we know that if 𝐵 is a basis the vector space then |𝐵| = 𝑛. Moreover a basis generates
the vector space by definition. Besides, by the replacement theorem, we know that if 𝐿 is a
linearly independent subset of the vector space that contains 𝑚 elements and |𝐺| is a generating
subset of the vector space, then |𝐿| ≤ |𝐺|. So in particular for the basis we have |𝐿| ≤ 𝑛 = |𝐵|.

In this case, |𝑆| = 4 > dim(ℝ3 ) = 3 so 𝑆 is not linearly independent.

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Exercise 9 page 55: The vectors 𝑢1 = (1,1,1,1), 𝑢2 = (0,1,1,1), 𝑢3 = (0,0,1,1) and 𝑢4 =

(0,0,0,1) for a basis for 𝐹 4 . Find the unique representation of an arbitrary vector (𝑎1 , 𝑎2 , 𝑎3 , 𝑎4 ) in 𝐹 4 as a
linear combination of 𝑢1 , 𝑢2 , 𝑢3 and 𝑢4 .

Since 𝐵 = {𝑢1 , 𝑢2 , 𝑢3 , 𝑢4 } is a basis on 𝐹4 and that (𝑎1 , 𝑎2 , 𝑎3 , 𝑎4 ) ∈ 𝐹 4 , then ∃! 𝑏1 , 𝑏2 , 𝑏3 , 𝑏4

such that :

(𝑎1 , 𝑎2 , 𝑎3 , 𝑎4 ) = 𝑏1 𝑢1 + 𝑏2 𝑢2 + 𝑏3 𝑢3 + 𝑏4 𝑢4
This implies the following system of equation:

𝑏1 = 𝑎1 𝑒1 𝑏1 = 𝑎1 𝑒1
𝑏1 + 𝑏2 = 𝑎2 𝑒2 𝑏1 + 𝑏2 = 𝑎2 𝑒2
𝑏1 + 𝑏2 + 𝑏3 = 𝑎3 𝑒3 → 𝑏1 + 𝑏2 + 𝑏3 = 𝑎3 𝑒3
𝑏1 + 𝑏2 + 𝑏3 + 𝑏4 = 𝑎4 𝑒4 𝑏4 = 𝑎4 − 𝑎3 𝑒4 − 𝑒3

𝑏1 = 𝑎1 𝑒1 𝑏1 = 𝑎1 𝑒1
𝑏1 + 𝑏2 = 𝑎2 𝑒2 𝑏2 = 𝑎2 − 𝑎1 𝑒2 − 𝑒1
→
𝑏3 = 𝑎3 − 𝑎2 𝑒3 − 𝑒2 𝑏3 = 𝑎3 − 𝑎2 𝑒3
𝑏4 = 𝑎4 − 𝑎3 𝑒4 𝑏4 = 𝑎4 − 𝑎3 𝑒4

Therefore, the unique representation of an arbitrary vector (𝑎1 , 𝑎2 , 𝑎3 , 𝑎4 ) ∈ 𝐹 4 is

(𝑎1 , 𝑎2 − 𝑎1 , 𝑎3 − 𝑎2 , 𝑎4 − 𝑎3 ) as a linear combination of 𝑢1 , 𝑢2 , 𝑢3 and 𝑢4 .

Exercise 12 page 55: Let 𝑢, 𝑣 and 𝑤 be distinct vectors of a vector space 𝑉. Show that if {𝑢, 𝑣, 𝑤} is a
basis for 𝑉 then {𝑢 + 𝑣 + 𝑤, 𝑣 + 𝑤, 𝑤} is also a basis for 𝑉

Let 𝑢, 𝑣 and 𝑤 be distinct vectors of a vector space 𝑉. Suppose that 𝐵 = {𝑢, 𝑣, 𝑤} is a basis for
𝑉.

First notice that 𝑢 + 𝑣 + 𝑤, 𝑣 + 𝑤, 𝑤 are distinct vectors since:

- 𝑢 + 𝑣 + 𝑤 = 𝑣 + 𝑤 ⟹ 𝑢 = 0 ⟹ 𝐵 = {0, 𝑣, 𝑤} ⟹ 𝐵 is linearly dependent. This
contradicts the fact that 𝐵 is a basis. ⟹ 𝑢 + 𝑣 + 𝑤 ≠ 𝑣 + 𝑤
- 𝑣 + 𝑤 = 𝑤 ⟹ 𝑣 = 0 ⟹ 𝐵 = {𝑢, 0, 𝑤} ⟹ 𝐵 is linearly dependent. This contradicts
the fact that 𝐵 is a basis ⟹ 𝑣 + 𝑤 ≠ 𝑤
- 𝑢 + 𝑣 + 𝑤 = 𝑤 ⟹ 𝑢 + 𝑣 = 0 ⟹ There exists a non trivial linear combination of 0
with vectors of 𝐵 ⟹ 𝐵 is linearly dependent. This contradicts the fact that 𝐵 is a basis.
⟹ 𝑢+𝑣+𝑤 ≠𝑤

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Correction Homework Assignment #4

Therefore 𝑢 + 𝑣 + 𝑤, 𝑣 + 𝑤, 𝑤 are 3 distinct elements and 𝐵′ = {𝑢 + 𝑣 + 𝑤, 𝑣 + 𝑤, 𝑤} has 3

elements.

Moreover let 𝑎, 𝑏, 𝑐 ∈ 𝐹 such that:

0 = 𝑎(𝑢 + 𝑣 + 𝑤) + 𝑏(𝑣 + 𝑤) + 𝑐𝑤 = 𝑎𝑢 + (𝑎 + 𝑏)𝑣 + (𝑎 + 𝑏 + 𝑐)𝑤
Since 𝐵 is a basis we have that the only linear combination of vectors of 𝐵 that gives the zero
vector is the linear combination where all the coefficients equal zero. So by identification we have
that

𝑎=0 𝑎=0
𝑎+𝑏 =0 → 𝑏=0
𝑎+𝑏+𝑐 = 0 𝑐=0

Therefore, the only combination of vectors of 𝐵′ that gives the zero vector is the combination
where all coefficient are equal to 0. Therefore 𝐵′ is linearly independent.
Since 𝐵′ is linearly and has exactly 3 elements, 𝐵′ is a basis of 𝑉 by corollary 2 of the
replacement theorem.