Linear Algebra Problem Set Compilation

1. For the given Matrix below, find

i. A–B

1 0 4 2 3 4
A= 2 1 3 B= 𝑎 𝑏 𝑐
3 2 0 𝑑 𝑒 𝑓

1−2 0−3 4−4

A–B= −𝑎
2 1−𝑏 3−𝑐
3−𝑑 2−𝑒 0−𝑓

−1 −3 0
A–B=2−𝑎 1−𝑏 3−𝑐
3−𝑑 2−𝑒 −𝑓

ii. 𝐴2

1 0 4
A= 2 1 3
3 2 0

1 0 4 1 0 4
2
𝐴 =2 1 3 x 2 1 3
3 2 0 3 2 0

1(1) + 0(2) + 4(3) 1(0) + 0(1) + 4(2) 1(4) + 0(3) + 4(0)

2
𝐴 = 2(1) + 1(2) + 3(3) 2(0) + 1(1) + 3(2) 2(4) + 1(3) + 3(0)
3(1) + 2(2) + 0(3) 3(0) + 2(1) + 0(2) 3(4) + 2(3) + 0(0)

13 8 4
2
𝐴 = 13 7 11
7 2 18

iii. ABC

1 0 4 2 3 4 1 0 0
A= 2 1 3 B= 𝑎 𝑏 𝑐 C= 1 0 0
3 2 0 𝑑 𝑒 𝑓 0 0 1
 1 0 4 2 3 4
AB= 2 1 3 x 𝑎 𝑏 𝑐
3 2 0 𝑑 𝑒 𝑓

1(2) + 0(𝑎) + 4(𝑑) 1(3) + 0(𝑏) + 4(𝑒) 1(4) + 0(𝑐 ) + 4(𝑓)

AB= 2(2) + 1(𝑎) + 3(𝑑) 2(3) + 1(𝑏) + 3(𝑒) 2(4) + 1(𝑐 ) + 3(𝑓)
3(2) + 2(𝑎) + 0(𝑑) 3(3) + 2(𝑏) + 0(𝑒) 3(4) + 2(𝑐 ) + 0(𝑓)

2 + 4𝑑 3 + 4𝑒 4 + 4𝑓
AB= 4 + 𝑎 + 3𝑑 6 + 𝑏 + 3𝑒 8 + 𝑐 + 3𝑓
6 + 2𝑎 9 + 2𝑏 12 + 2𝑐

2 + 4𝑑 3 + 4𝑒 4 + 4𝑓 1 0 0
AB(C) = 4 + 𝑎 + 3𝑑 6 + 𝑏 + 3𝑒 8 + 𝑐 + 3𝑓 x 1 0 0
6 + 2𝑎 9 + 2𝑏 12 + 2𝑐 0 0 1

2 + 4𝑑(1) + 3 + 4𝑒(1) + 4 + 4𝑓(0) 0 4 + 4𝑓(1)

ABC= 4 + 𝑎 + 3𝑑(1) + 6 + 𝑏 + 3𝑒(1) + 0 0 8 + 𝑐 + 3𝑓(1)
6 + 2𝑎(1) + 9 + 2𝑏(1) + 0 0 12 + 2𝑐(1)

5 + 4𝑑 + 4𝑒 0 4 + 4𝑓
ABC= 10 + 𝑎 + 𝑏 + 3𝑑 + 3𝑒 0 8 + 𝑐 + 3𝑓
15 + 2𝑎 + 2𝑏 0 12 + 2𝑐

iv. 𝐵2 − 𝐶 2

2 3 4 2 3 4
2 𝑎 𝑏 𝑐 𝑎 𝑏 𝑐
𝐵 = x
𝑑 𝑒 𝑓 𝑑 𝑒 𝑓

2(2) + 3(𝑎) + 4(𝑑) 2(3) + 3(𝑏) + 4(𝑒) 2(4) + 3(𝑐 ) + 4(𝑓)

2
𝐵 = 𝑎(2) + 𝑏(𝑎) + 𝑐(𝑑) 𝑎(3) + 𝑏(𝑏) + 𝑐(𝑒) 𝑎 (4) + 𝑏(𝑐 ) + 𝑐(𝑓)
𝑑 (2) + 𝑒(𝑎) + 𝑓(𝑑) 𝑑(3) + 𝑒(𝑏) + 𝑓(𝑒) 𝑑(4) + 𝑒(𝑐 ) + 𝑓(𝑓)

4 + 3𝑎 + 4𝑑 6 + 3𝑏 + 4𝑒 8 + 3𝑐 + 4𝑓
2
𝐵 = 2𝑎 + 𝑎𝑏 + 𝑐𝑑 3𝑎 + 𝑏2 + 𝑐𝑒 4𝑎 + 𝑏𝑐 + 𝑐𝑓
2𝑑 + 𝑎𝑒 + 𝑑𝑓 3𝑑 + 𝑏𝑒 + 𝑒𝑓 4𝑑 + 𝑐𝑒 + 𝑓 2
 1 0 0 1 0 0
𝐶2 = 1 0 0 x 1 0 0
0 0 1 0 0 1

1(1) 0 0
2
𝐶 = 1(1) 0 0
0 0 1(1)

1 0 0
𝐶2 = 1 0 0
0 0 1

3 + 3𝑎 + 4𝑑 6 + 3𝑏 + 4𝑒 8 + 3𝑐 + 4𝑓
2 2
𝐵 − 𝐶 = 2𝑎 + 𝑎𝑏 + 𝑐𝑑 − 1 3𝑎 + 𝑏2 + 𝑐𝑒 4𝑎 + 𝑏𝑐 + 𝑐𝑓
2𝑑 + 𝑎𝑒 + 𝑑𝑓 3𝑑 + 𝑏𝑒 + 𝑒𝑓 4𝑑 + 𝑐𝑒 + 𝑓 2 − 1

v. 𝐵𝑇 𝐷

2 3 4 2 𝑎 𝑑 3 1
B=𝑎 𝑏 𝑐 𝐵𝑇 = 3 𝑏 𝑒 D=5 2
𝑑 𝑒 𝑓 4 𝑐 𝑓 6 2

2(3) + 𝑎 (5) + 𝑑(6) 2(1) + 𝑎(2) + 𝑑(2)

𝑇
𝐵 𝐷= 3(3) + 𝑏(5) + 𝑒(6) 3(1) + 𝑏(2) + 𝑒(2)
4(3) + 𝑐(5) + 𝑓(6) 4(1) + 𝑐(2) + 𝑓(2)

6 + 5𝑎 + 6𝑑 2 + 2𝑎 + 2𝑑
𝐵𝑇 𝐷= 9 + 5𝑏 + 6𝑒 3 + 2𝑏 + 2𝑒
12 + 5𝑐 + 6𝑓 4 + 2𝑐 + 2𝑓

2. Find the Determinant and the inverse of the given matrix if it exist
𝟏 𝟏 𝟏 𝟏
𝟏 𝟑 𝟏 𝟐
𝟏 𝟐 −𝟏 𝟏
𝟓 𝟗 𝟏 𝟔
 3 1 2 1 1 2 1 3 2 1 3 1
Det= 1 [2 − 1 1 ] −1 [1 − 1 1 ] +1 [1 2 1 ] −1 [ 1 2 −1]
9 1 6 5 1 6 5 9 6 5 9 1
Det = −2 − 4 − 2 + 8
=0
NOTE: Since the determinant is 0 hence the inverse DOES NOT EXIST.

3. Solve the following by way of reducing the system to echelon form.

𝟏 𝟏 𝟐 −𝟏
𝟏 −𝟐 𝟏 −𝟓
𝟑 𝟏 𝟏 𝟑

1 1 2 −1 1 1 2 −1
1 −2 1 −5 Eliminate 1st Column 0 −3 −1 −4
3 1 1 3 0 2 −5 6

5 7
1 1 2 1 0 −3
−1 3
nd 1 4 1 4
Divide 2 row by −3 0 1 3 3
Eliminate 2nd Column 0 1 3 3
0 −2 −5 6 0 0 −
13 26
3 3

5 7
1 0 −3 1 0 0 1
3
13
Divide 3rd row by − 0 1
1 4 Eliminate Column 3 0 1 0 2
3
3 3 0 0 1 −2
0 0 1 −2

𝑎 = 1 𝑏 = 2 𝑐 = −2
Show that the following is a vector space.

𝒂𝒃
4. The set of all 2x2 matrices of the form ( ) with the standard operations is a
𝒄𝟎
vector space.
𝑎 𝑏 𝑎 𝑏 2𝑎 2𝑏
ANSWER: Since + = , it shows that it is closed under addition.
𝑐 0 𝑐 0 2𝑐 0
 𝑎 𝑏 𝑎𝑘 𝑏𝑘
Furthermore 𝑘 = is also closed under scalar multiplication. Hence, it is a vector
𝑐 0 𝑐𝑘 0
space.

5. The set of all 2x2 diagonal matrices with the standard operation.
1 0 1 0 2 0
ANSWER: Since + = is closed under addition hence, it is not a vector space.
0 1 0 1 0 2

6. Suppose that S={ 𝒗𝟏 , 𝒗𝟐 , 𝒗𝟑 }is linearly independent set of vectors in a vector

space. Show that T= { 𝒘𝟏 , 𝒘𝟐 , 𝒘𝟑 } is also linearly independent where 𝒘𝟏 = 𝒗𝟏 +
𝒗𝟐 + 𝒗𝟑 , 𝒘𝟐 = 𝒗𝟏 + 𝒗𝟐 , 𝒘𝟑 = 𝒗𝟑
ANSWER: Note that for a set of vector (𝑣1 , 𝑣2 , 𝑣3 ) in 𝑅𝑛 is said to be linearly independent if the
vector equation
𝑥1 𝑣1 + 𝑥2 𝑣2 + 𝑥3 𝑣3 … 𝑥𝑝 𝑣𝑝 = 0 such that 𝑣1 = 𝑣2 = 𝑣3 = 0 has only trivial solutions.

Now, since S is linearly independent it is safe to say that 𝑣1 + 𝑣2 + 𝑣3 = 0

𝑎1 = 0 𝑎2 + 𝑎3 = 0 𝑎1 + 𝑎2 + 𝑎3 = 0
Suppose 𝑎1 𝑤1 + 𝑎2 𝑤2 + 𝑎3 𝑤3 = 𝑎1 (𝑣1 + 𝑣2 + 𝑣3 ) + 𝑎2 (𝑣1 + 𝑣2 ) + 𝑎3 (𝑣3 ) = 0
We need to show that {𝑤1 + 𝑤2 + 𝑤3 } = (0,0,0) substituting we have
{𝑣1 + 𝑣2 + 𝑣3 , 𝑣1 + 𝑣2 , 𝑣3 } = ( 0 , 0 ,0 ) forming the equation
𝑣1 + 𝑣2 + 𝑣3 = 0 𝑒𝑞. 1
𝑣2 + 𝑣3 = 0 𝑒𝑞. 2
𝑣3 = 0 𝑒𝑞. 3

7. Let A be a n x n matrix. Show that each of the following is symmetric.

a) 𝐴𝐴𝑇
To determine 𝐴𝐴𝑇 to be symmetric, we need to show that it is equal to its transpose.
𝑎 𝑏 𝑎 𝑐
Let A= 𝐴𝑇 =
𝑐 𝑑 𝑏 𝑑

2 2
𝐴𝐴𝑇 = 𝑎 +𝑏 𝑎𝑐 + 𝑏𝑑
𝑎𝑐 + 𝑏𝑑 𝑐 2 + 𝑑2
2 2
(𝐴𝐴𝑇 )𝑇 = 𝑎 +𝑏 𝑎𝑐 + 𝑏𝑑
𝑎𝑐 + 𝑏𝑑 𝑐 2 + 𝑑2
 Using the properties, we can write it as

𝐴𝐴𝑇 = (𝐴𝐴𝑇 )𝑇 = (𝐴𝑇 )𝑇 (𝐴𝑇 ) = 𝐴𝐴𝑇

b) 𝐴 + 𝐴𝑇

To determine 𝐴 + 𝐴𝑡 to be symmetric, we need to show that it is equal to its transpose.

𝑎 𝑏 𝑎 𝑐
Let A= 𝐴𝑇 =
𝑐 𝑑 𝑏 𝑑

𝑎 𝑏 𝑎 𝑐
𝐴 + 𝐴𝑇 = +
𝑐 𝑑 𝑏 𝑑

2𝑎 𝑏+𝑐
=
𝑏+𝑐 2𝑑

2𝑎 𝑏+𝑐
(𝐴 + 𝐴𝑇 )𝑇 =
𝑏+𝑐 2𝑑
Using the properties, we can write it as
𝐴 + 𝐴𝑇 = (𝐴 + 𝐴𝑇 )𝑇

= (𝐴𝑇 )𝑇 + 𝐴𝑇
= 𝐴 + 𝐴𝑇

8. Determine whether the set S= {1, 𝒙𝟐 , 𝒙𝟐 + 𝟐} spans 𝑷𝟐.

To determine if S spans 𝑃2. , we need to show that S= {1, x, 𝑥 2 }
𝑆 = 1(1) + (1)𝑥 2 + 1(𝑥 2 + 2)
= 1 + 𝑥2 + 𝑥2 + 2
= 3 + 2𝑥 2
Thus, the set of S= {1, 𝑥 2 , 𝑥 2 + 2} does not span 𝑃2.
9. Explain why 𝑺 = {(𝟏, 𝟑, 𝟎), (𝟒, 𝟏, 𝟐), (−𝟐, 𝟓, −𝟐)} is not a basis for 𝑹𝟑 .To test for
linearly independence, we form the vector equation. 𝒄𝟏 𝒗𝟏 + 𝒄𝟐 𝒗𝟐 + 𝒄𝟑 𝒗𝟑 = 𝟎,
expanding this, we have
𝑐1 (1,3,0) + 𝑐2 (4,1,2) + 𝑐3 (−2,5, −2) = (0,0,0)
{𝑐1 + 4𝑐2 − 2𝑐3 , 3𝑐1 + 𝑐2 + 5𝑐3 , 2𝑐2 − 2𝑐3 } = (0,0,0)

𝑐1 + 4𝑐2 − 2𝑐3 , = 0
3𝑐1 + 𝑐2 + 5𝑐3 = 0
2𝑐2 − 2𝑐3 = 0
Augmented Matrix
1 4 −2 0
3 1 5 0
0 2 −2 0
Using Gauss-Jordan Elimination
1 0 2 0
0 1 −1 0
0 0 0 0
This implies that the system has nontrivial solutions. Hence, it is not linearly independent and
therefore is not a basis.
𝟏 𝟎𝟏 𝟎
10. Explain why 𝑺 = { , } is not a basis for 𝑴𝟐𝟐
𝟎 𝟏𝟎 𝟏
To test for linearly independence we have
𝑐1 𝑣1 𝑐2 𝑣2 = 0 expanding the equation we have,
1 0 1 0 0 0
𝑐1 + 𝑐2 = which produce the system
0 1 0 1 0 0

𝑐1 + 𝑐2 = 0
𝑐1 + 𝑐2 = 0
Augmented Matrix
1 1 0
0 0 0
0 0 0
1 1 0
Using Gauss-Jordan Elimination
1 1 0
0 0 0
0 0 0
0 0 0
This implies that the system has nontrivial solution
11. What is a Vector Space?
A vector space, also known as a linear space, is a fundamental concept in linear
algebra. It provides a mathematical framework for understanding and working with
vectors and their operations. A vector space is defined by a set of vectors and a set
of rules that govern how these vectors can be combined and manipulated.

Formally, a vector space V over a field F consists of the following components:

1. A set of vectors: This is a collection of elements, each of which is called a vector.

These vectors can be represented as arrays of numbers or symbols.
2. A field of scalars: This is a set of numbers from a field, usually the real numbers
(denoted as ℝ) or complex numbers (denoted as ℂ), with two main operations:
addition and multiplication.
3. Vector addition: For any two vectors u and v in the vector space V, there exists a
unique vector u + v in V, called their sum. This operation satisfies properties like
associativity and commutativity.
4. Scalar multiplication: For any scalar (a number from the field) c and any vector u
in V, there exists a unique vector cu in V, called the scalar multiple of u by c. This
operation also satisfies properties like distributivity and compatibility with field
multiplication.
5. Compatibility of operations: The scalar multiplication and vector addition
operations must be compatible, meaning that for any scalar c and vectors u, v in
V, the distributive property holds: c(u + v) = cu + cv.
6. Zero vector: There exists a special vector, denoted as 0, which acts as the additive
identity. For any vector u, u + 0 = u.
7. Additive inverses: For every vector u in V, there exists a vector -u in V such that u
+ (-u) = 0.

These properties ensure that vector spaces maintain consistency and behave
analogously to the intuitive notions of vectors and operations, allowing for the study of
various mathematical and physical concepts in a structured manner. Common
examples of vector spaces include Euclidean spaces (n-dimensional spaces with real
or complex coordinates), function spaces, and spaces of matrices.

12. Let𝑺 = {(𝟏, 𝟐, −𝟏), (𝟑, 𝟏, 𝟎), (𝟎, −𝟓, 𝟑)} ⊂ ℝ𝟑 . Consider ℝ𝟑 as a vector space
over ℝ.

(a) Is S linearly independent?

For S to be linearly independent, the vector equation should have only the trivial solutions
where all the coefficients are zero. Hence,

1 3 0 0
𝑎 [ 2 ] + 𝑏 [1] + 𝑐 [−5] = [0]
−1 0 3 0

Equivalently, we can solve the matrix equation

1 3 0 𝑎 0
[2 1 −5] [𝑏] = [0]
−1 0 3 𝑐 0
Forming the augmented matrix,

1 3 0 0
[2 1 −5 0]
−1 0 3 0

Using Gauss-Jordan Elimination, we get

1 0 −3 0
[0 1 1 0 ]
0 0 0 0
This implies that the system has a non-trivial solution hence, it is not linearly independent.

(b) Find a subset of S that is a basis for span(S).

By looking at the row echelon form of the given matrix, it shows that vectors in column 1 and 2
provide a basis for the set of vectors.
1 0 −3 0
[0 1 1 0]
0 0 0 0

Since the 1st and 2nd column has the leading term we get the vectors in its original matrix which
will be the subset of S that provides a basis and that is
1 3 0 0
[ 2 1 −5 0]
−1 0 3 0
Furthermore, the vectors (1,2, −1), (3,1,0) are subset of S that is a basis of span (S).

13. Suppose V is a vector space and W1 and W2 are subspaces of V. Let U be the
set of all
vectors in V that can be written as a sum of a vector in W1 and a vector in W2.
That is,
𝑼 = {𝒘𝟏 + 𝒘𝟐 |𝒘𝟏 ∈ 𝑾𝟏 ; 𝒘𝟐 ∈ 𝑾𝟐 }
Show that U is a subspace of V.
14. Let 𝑽 = {𝟎 } consist of a single vector 0 and define 0 + 0 = 0 and c0 = 0 for each
scalar c in F. Prove that V is a vector space over F.

This can be proved by showing that V satisfies the following properties of a vector space.
Now, since V contains the zero vector 0, and for any scalar c in F, c0=0. Therefore, the set V is
closed under addition and scalar multiplication.
For Commutativity of addition. Obviously if 𝑥 + 𝑦 does equal to 𝑦 + 𝑥 because both are 0 which
is the only vector in the space.
Therefore, V is a vector space over F.

15. Let V denote the set of ordered pairs of real numbers. If (a1, a2) and (b1, b2) are elements
of V and c ∈ R, define (𝒂𝟏 , 𝒂𝟐 ) + (𝒃𝟏 , 𝒃𝟐 ) = (𝒂𝟏 + 𝒃𝟏 , 𝒂𝟐 𝒃𝟐 ) and 𝒄(𝒂𝟏 , 𝒂𝟐 ) =
(𝒄𝒂𝟏 , 𝒂𝟐 ). Is V a vector space over R with these operations? Justify your answer.

First, we have to check that all properties holds for us to say that V is a vector space over R with
the given operation.
However, checking for Distributive property under scalar multiplication where,
(𝑐 + 𝑑 )(𝑎1 , 𝑎2 ) = (𝑐𝑎1 + 𝑑𝑎1 , 𝑎2 )
Where 𝑐, 𝑑 𝜖 ℝ
But, notice that,
𝑐 (𝑎1 , 𝑎2 ) + 𝑑 (𝑎1 , 𝑎2 ) = (𝑐𝑎1 + 𝑑𝑎1 , 𝑎22 )
Thus, it fails to hold this property. Therefore V is not a vector space over ℝ under these operation.

16. Let 𝑽 = {(𝒂𝟏 , 𝒂𝟐 ): 𝒂𝟏 , 𝒂𝟐 ∈ 𝑭}, where F is a field. Define addition of elements of V

coordinatewise, and for c ∈ F and (𝒂𝟏 , 𝒂𝟐 ) ∈ 𝑽, define 𝒄(𝒂𝟏 , 𝒂𝟐 ) = (𝒂𝟏 , 𝟎). Is V a vector
space over F with these operations? Justify your answer.

First, we have to check that all properties holds for us to say that V is a vector space over F with
the given operation.
However, checking for an identity scalar element where for any scalar c, then
𝑐 (𝑎1 , 𝑎2 ) = (𝑎, 0) = (𝑎1 , 𝑎2 )
This applies only if 𝑎2 = 0. Thus, not every element 𝑥 𝜖 V has the property that the scalar identity
takes it to itself.
Therefore there does not exist an identity scalar element.Hence, V is not a vector space over F
with these operation.

17. Let 𝑽 = {(𝒂𝟏 , 𝒂𝟐 ): 𝒂𝟏 , 𝒂𝟐 ∈ 𝑹}. For (𝒂𝟏 , 𝒂𝟐 ), (𝒃𝟏 , 𝒃𝟐 ) ∈ 𝑽 and c ∈ R, define (𝒂𝟏 , 𝒂𝟐 ) +
(𝒃𝟏 , 𝒃𝟐 ) = (𝒂𝟏 + 𝟐𝒃𝟏 , 𝒂𝟐 + 𝟑𝒃𝟐 ) and 𝒄(𝒂𝟏 , 𝒂𝟐 ) = (𝒄𝒂𝟏 , 𝒄𝒂𝟐 ). Is V a vector space over
R with these operations? Justify your answer.
18. Let V and W be vector spaces over a field F. Let 𝒁 = {(𝒗, 𝒘): 𝒗 ∈ 𝑽 𝒂𝒏𝒅 𝒘 ∈ 𝑾}.
Prove that Z is a vector space over F with the operations (𝒗𝟏 , 𝒘𝟏 ) + (𝒗𝟐 , 𝒘𝟐 ) = (𝒗𝟏 +
𝒗𝟐 , 𝒘𝟏 + 𝒘𝟐 ) 𝒂𝒏𝒅 𝒄(𝒗𝟏 , 𝒘𝟏 ) = (𝒄𝒗𝟏 , 𝒄𝒘𝟏 ).

19. Prove that (𝒂𝑨 + 𝒃𝑩)𝒕 = 𝒂𝑨𝒕 + 𝒃𝑩𝒕 for any A,B ∈ Mm×n(F) and any a, b ∈ F.

20. Prove that diagonal matrices are symmetric matrices.

To prove that a matrix is symmetric is to find its transpose. When we say transpose of a matrix,
we mean that the rows becomes the columns and the columns becomes the rows in a given
matrix. If the matrix and its transpose is identical, then we can say that it is symmetric.
A Diagonal matrix on the other hand is a square matrix in which every element except the
principal diagonal elements is zero.
Let A be an 𝑛 𝑥 𝑛 matrix whose (𝑖, 𝑗) entry is 𝑎𝑖𝑗 . Then, since A is diagonal.
𝑖 ≠ 𝑗 implies 𝑎𝑖𝑗 = 0
To show that 𝐴𝑇 = 𝐴. We need to show that the (𝑖, 𝑗) entry of 𝐴𝑇 is the same as the (𝑖, 𝑗) entry
of A. Consider two cases:
Case 1: If 𝑖 ≠ 𝑗 then
(𝑖, 𝑗) entry of 𝐴𝑇 = (𝑗, 𝑖 ) entry of A= 0 = (𝑖, 𝑗) entry of A

Case 2: If 𝑖 = 𝑗 𝑡ℎ𝑒𝑛 𝑐𝑙𝑒𝑎𝑟𝑙𝑦,

(𝑖, 𝑖) entry of 𝐴𝑇 = 𝑎𝑖𝑖 = (𝑖, 𝑖) entry of A
Therefore, the (𝑖, 𝑗) entry of 𝐴 𝑎𝑛𝑑 𝐴𝑇 transpose coincide. Hence it is symmetric.

21. Prove that a subset W of a vector space V is a subspace of V if and only if 0 ∈ W and ax
+ y ∈ W whenever a ∈ F and x, y ∈ W.

22. Show that the set V= {(𝒙, 𝒚)𝝐 ℝ𝟐 / 𝒙𝒚 ≥ 𝟎} is not a vector space of ℝ𝟐 .

Solution:
For V to be a vector space, it is required that V must satisfy all the axioms on which vector
addition and scalar multiplication are defined. Furthermore, V must be closed under addition,
that is for any 𝑥 and 𝑦 in V, 𝑥 + 𝑦 𝜖 𝑉. So, we let (−1,0) and (0, −1) 𝜖 𝑉
Now, (−1,0) + (0, −1) = (−1 + 0, 0 + 1) = (−1,1)
But, −1 x 1 = −1 < 0 ⇒ (−1,1) ∉ 𝑉

Therefore, V is not a vector space in ℝ2

23. Let V= {𝒙/𝒙 𝝐ℝ, 𝒙 > 𝟎} defined by addition and scalar multiplication as follows.
𝒙 + 𝒚 = 𝒙 + 𝒚 where 𝒙, 𝒚 𝝐 𝑽 and 𝒓𝒙 = 𝒓. 𝒙 where 𝒓𝝐ℝ 𝒂𝒏𝒅 𝒙 𝝐 𝑽.
Show that V is not a vector space.
Solution:
To show that the structure is not a vector space, all we have to show is that at least
one of the axioms fails to hold. Since, if 𝑟 < 0, 𝑟𝑥 = 𝑟. 𝑥 < 0 and 𝑟𝑥 ∉ 𝑉.
24. Determine whether 𝑺 = {𝟏 + 𝒙 , 𝒙 + 𝒙𝟐 , 𝟏 + 𝒙𝟐 } is linearly independent in 𝑷𝟐.
Solution:
For the set S to be linearly independent in 𝑷𝟐. If the vector equation has only the trivial
solutions .Now, consider 𝑐1 (1 + 𝑥)+𝑐2 (𝑥 + 𝑥 2 ) + 𝑐3 (1 + 𝑥 2 ) = 𝑒 = 0 + 0𝑥 + 0𝑥 2
By collecting terms on the left side , this equation can be rewritten as
(𝑐1 +𝑐3 ) + (𝑐1 + 𝑐2 )𝑥 + (𝑐2 + 𝑐3 )𝑥 2 = 0 + 0𝑥 + 0𝑥 2 = 𝑒
From Algebra we know that a polynomial is identically zero only when all the
coefficients are zero. So we have ,
𝑐1 + 𝑐3 = 0
𝑐1 + 𝑐2 =0
𝑐2 + 𝑐3 = 0
Which has only the trivial solutions. Therefore S is linearly independent.
25. Let 𝑷𝟑 be the set of all polynomials of degree at most 3 and let
𝑼 = {𝒇𝝐 𝑷𝟑 ∶ 𝒇(𝟐) = 𝒇(𝟏) = 𝟐𝒇(𝟎)}
Show that U is a subspace 𝑷𝟑 and find a basis for it.
Solutions:
To say that f(x) = 𝑎𝑥 3 + 𝑏𝑥 2 + 𝑐𝑥 + 𝑑 satisfies the first condition is to say that
𝑓 (2) = 𝑓 (1) ⇒ 8𝑎 + 4𝑏 + 2𝑐 + 𝑑 = 𝑎 + 𝑏 + 𝑐 + 𝑑 ⇒ 𝑐 = −7𝑎 − 3𝑏
To say that 𝑓 (𝑥 ) = 𝑎𝑥 3 + 𝑏𝑥 2 + 𝑐𝑥 + 𝑑 satisfies the second condition is to say that
𝑓(1) = 2𝑓 (0) ⇒ 𝑎 + 𝑏 + 𝑐 + 𝑑 = 2𝑑 ⇒ 𝑑 = 𝑎 + 𝑏 + 𝑐
It easily follows that 𝑈 is a subspace because U can be expressed in the form
𝑈 = { 𝑎𝑥 3 + 𝑏𝑥 2 − 7𝑎𝑥 − 3𝑏𝑥 − 6𝑎 − 2𝑏: 𝑎, 𝑏 𝜖 ℝ}
= 𝑆𝑝𝑎𝑛 { 𝑥 3 − 7𝑥 − 6, 𝑥 2 − 3𝑥 − 2}
To show that the last two polynomials are linearly independent, we note that
𝑎(𝑥 3 − 7𝑥 − 6) + 𝑏(𝑥 2 − 3𝑥 − 2) = 𝑎𝑥 3 + 𝑏𝑥 2 − (7𝑎𝑥 − 3𝑏)𝑥 − (6𝑎 + 2𝑏).
If this linear combination happens to be zero, then the coefficient of 𝑥 3 , 𝑥 2 must both be
zero and so 𝑎 = 𝑏 = 0. We conclude that 𝑥 3 − 7𝑥 − 6 𝑎𝑛𝑑 𝑥 2 − 3𝑥 − 2 form a basis of
𝑈.
𝒙𝟏
26. Given the subset 𝑺𝟏 = {𝒙𝟐 𝝐 ℝ𝟑 𝒘𝒉𝒆𝒓𝒆 𝒙𝟏 ≥ 𝟎} show why it is not a subset of the vector
𝒙𝟑
space.
 Solution:
By definition, for us to say that a subset is a subspace, the following conditions should be
satisfied:
1. The subset 𝑆1 contains the zero vector of V.
2. If 𝑢, 𝑣 𝜖 𝑆1 , 𝑡ℎ𝑒𝑛 𝑢 + 𝑣 𝜖 𝑆1 .
3. If 𝑢 𝜖 𝑆1 𝑎𝑛𝑑 𝑎 𝜖 𝐾, 𝑡ℎ𝑒𝑛 𝑎𝑢 𝜖 𝑆1
Thus, to prove a subset 𝑆1 is not a subspace, we just need to find a counterexample of any
of the three conditions

1
Let us consider 𝑥 = 0 Since , 𝑥1 = 1 ≥ 0, the vector 𝑥 𝜖 𝑆1
0
Then consider the scalar product of 𝑥 and the scalar −1. Then we have,
−1
(−1). 𝑥 = [ 0 ]
0
Notice the first entry is −1, hence – 𝑥 is not in 𝑆1 . Thus 𝑆1 does not satisfy condition 3 and
it is not a subspace of ℝ3
27. Consider the following basis for ℝ𝟐 :
𝟏𝟑
𝑬={ , }
𝟐𝟓
−𝟐
Find the coordinates for the vector [ ]in terms of the basis E.
𝟒
Solution:
We need to find numbers 𝑐1 and 𝑐2 such that
1 3 −2
𝑐1 [ ] + 𝑐2 [ ] = [ ]
2 5 4
Thus, we need to solve the following system of line,ar equations:
𝑐1 + 3𝑐2 = −2
2𝑐1 + 5𝑐2 = 4
Solving, we get 𝑐1 = 22 and 𝑐2 = −8. Thus, the coordinates in terms of basis E are
22
[ ]
−8
28. Let V and W be vector spaces over a field F. Let Z= {(𝒗, 𝒘): 𝒗 ∈ 𝑽 and w∈ 𝑾}.
Prove that Z is a vector space over F with operations (𝒗𝟏 , 𝒘𝟏 ) + (𝒗𝟐 , 𝒘𝟐 ) = (𝒗𝟏 + 𝒗𝟐 , 𝒘𝟏 +
𝒘𝟐 ) and c (𝒗𝟏 , 𝒘𝟏 ) = (𝒄𝒗𝟏 , 𝒄𝒘𝟏 ).
Proof: Generally, the first component of a vector in Z inherits vector space properties from V,
while the second component of a vector in Z inherits vector space properties from W.
Note that since V is a vector space and W is a vector space, for all (𝑣1 , 𝑢1 ), (𝑣2 , 𝑢2 ) ∈
𝑉, (𝑣1 , 𝑢1 ) + (𝑣2 , 𝑢2 ) = (𝑣1 + 𝑣2 , 𝑢1 + 𝑢2 ) = (𝑣2 , 𝑢2 ) + (𝑣1 , 𝑢1 ) = (𝑣2 + 𝑣1 , 𝑢2 + 𝑢1 )
For the first component we have 𝑣1 + 𝑣2 = 𝑣2 + 𝑣1 . For the second component we have,
For all (𝑥1 , 𝑤1 ), (𝑥2 , 𝑤2 ) ∈ 𝑊, (𝑥1 , 𝑤1 ) + (𝑥2 , 𝑤2 ) = (𝑥1 + 𝑥2 , 𝑤1 + 𝑤2 ) = (𝑥2 , 𝑤2 ) + (𝑥1 , 𝑤1 )
= (𝑥2 + 𝑥1 , 𝑤2 + 𝑤1 ) Hence, we get 𝑤1 + 𝑤2 = 𝑤2 + 𝑤1 .
By definition, (𝑣1 , 𝑤1 ) + (𝑣2 , 𝑤2 ) = (𝑣1 + 𝑣2 , 𝑤1 + 𝑤2 ) = (𝑣2 + 𝑣1 , 𝑤2 + 𝑤1 ) = (𝑣2 , 𝑤2 ) +
(𝑣1 , 𝑤1 ) where additive commutativity holds for Z.
Also, note that since V and W are vector spaces then for all (𝑣1 , 𝑢1 ), (𝑣2 , 𝑢2 ), (𝑣3 , 𝑢3 ) ∈
𝑉,((𝑣1 , 𝑢1 ) + (𝑣2 𝑢2 )) + (𝑣3 , 𝑢3 ) = (𝑣1 , 𝑢1 ) + ((𝑣2 , 𝑢2 ) + (𝑣3 , 𝑢3 )) for the first component, we
have (𝑣1 + 𝑣2 ) + 𝑣3 = 𝑣1 + (𝑣2 + 𝑣3 ). For the second component we get, (𝑤1 + 𝑤2 ) + 𝑤3 =
𝑤1 + (𝑤2 + 𝑤3 ). ((𝑣1 + 𝑣2 ) + 𝑣3, (𝑤1 + 𝑤2 ) + 𝑤3 ) = (𝑣1 + (𝑣2 + 𝑣3 ), 𝑤1 + (𝑤2 + 𝑤3 )). By
definition, ((𝑣1 , 𝑤1 ) + (𝑣2 , 𝑤2 )) + (𝑣3 , 𝑤3 ) = (𝑣1 , 𝑤1 ) + ((𝑣2 , 𝑤2 ) + (𝑣3 , 𝑤3 )) where
(𝑣1 , 𝑤1 ), (𝑣2 , 𝑤2 ), (𝑣3 , 𝑤3 ) ∈ 𝑍 by additive associativity holds for Z.
And also since V, W are vector spaces that,is for every zero vector for V and zero vector for W.
The zero vector for Z can be formed by taking the first component of the zero vector of V and the
second component of the zero vector of W. Check that, (0𝑣 , 0𝑤 )is the zero vector in Z . (𝑣, 𝑤) +
(0𝑣 0𝑤 ) = (𝑣 + 0𝑣 , 𝑤 + 0𝑤 ) = (𝑣, 𝑤).
Since, for all (𝑣, 𝑢) ∈ 𝑉, for every (-v,-u) such that (𝑣, 𝑢) + (−𝑣, −𝑢) = 0𝑣 ⇒ for all v such that
𝑣 + (−𝑣) = 0. For all (𝑥, 𝑤) ∈ 𝑊 for every (−𝑥, −𝑤) such that (𝑥, 𝑤) + (−𝑥, −𝑤) = 0𝑤 ⇒
For all w, every – 𝑤 such that 𝑤 + (−𝑤) = 0 ⇒ For all (𝑣, 𝑤) ∈ 𝑍, for every
(−𝑣, −𝑤)𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡 9𝑣, 𝑤) + (−𝑣, −𝑤) = 0𝑧
For all (v,u) ∈ 𝑉, 1(𝑣, 𝑢) = (𝑣, 𝑢) ⇒ 1. 𝑣 = 𝑣 and for all (𝑥, 𝑤) ∈ 𝑊, 1(𝑥, 𝑤) = (𝑥, 𝑤) ⇒ 1. 𝑤 =
𝑤. By definition, 1(𝑣, 𝑤) = (1. 𝑣, 1. 𝑤)𝑤ℎ𝑒𝑟𝑒 𝑐 = 1 𝑏𝑢𝑡 (1. 𝑣, 1. 𝑤) = (𝑣, 𝑤) ⇒ 1(𝑣, 𝑤) =
(𝑣, 𝑤)
For all a,b ∈ 𝐹, for all (𝑣, 𝑢) ∈ 𝑉, a(b(v,u))=(ab)(v,u) and for all (𝑥, 𝑤) ∈ 𝑊, 𝑎(𝑏(𝑥, 𝑤) such that
𝑎(𝑏𝑣) = (𝑎𝑏)𝑣 and 𝑎(𝑏𝑤) = (𝑎𝑏)𝑤 ⇒ 𝑓𝑜𝑟 𝑎𝑙𝑙 (𝑣, 𝑤) ∈ 𝑍, 𝑎(𝑏(𝑣, 𝑤)) =
𝑎(𝑏𝑣, 𝑏𝑤) 𝑤ℎ𝑒𝑟𝑒 𝑐 = 𝑏 = (𝑎(𝑏𝑣), 𝑎 (𝑏𝑤))𝑤ℎ𝑒𝑟𝑒 𝑐 = 𝑎 = ((𝑎𝑏)𝑣, (𝑎𝑏)𝑤)𝑏𝑦 (∗) =
(𝑎𝑏)(𝑣, 𝑤)𝑤ℎ𝑒𝑟𝑒 𝑐 = 𝑎𝑏

Next, we have for all 𝑐 ∈ 𝐹, 𝑓𝑜𝑟 𝑎𝑙𝑙 (𝑣1 , 𝑢1 ), (𝑣2 , 𝑢2 ) ∈ 𝑉, 𝑐((𝑣1 , 𝑢1 ) + (𝑣2 , 𝑢2 )) = 𝑐 (𝑣1 , 𝑢1 ) +
𝑐 (𝑣2 , 𝑢2 ) ⇒, 𝑐 (𝑣1 + 𝑣2 ) = 𝑐𝑣1 + 𝑐𝑣2 and also for all (𝑥1 , 𝑤1 ), (𝑥2 , 𝑤2 ) ∈ 𝑊, 𝑐((𝑥1 , 𝑤1 ) +
(𝑥2 , 𝑤2 ) ⇒ 𝑐(𝑤1 + 𝑤2 ) = 𝑐𝑤1 + 𝑐𝑤2 𝑐𝑜𝑟𝑟𝑒𝑠𝑝𝑜𝑛𝑑𝑠 𝑡𝑜 , 𝑓𝑜𝑟 𝑎𝑙𝑙 (𝑣1 , 𝑤1 ), (𝑣2 , 𝑤2 ) ∈ 𝑍,

Thus, 𝑐((𝑣1 , 𝑤1 ) + (𝑣2 , 𝑤2 )) = 𝑐 (𝑣1 , 𝑣2 ) + 𝑐 (𝑤1 , 𝑤2 ) = (𝑐𝑣1 , 𝑐𝑣2 , 𝑐𝑤1 , 𝑐𝑤2 ) = 𝑐(𝑣1 , 𝑤1 ) +
(𝑣2 , 𝑤2 )
Lastly, for all 𝑎, 𝑏 ∈ F, for all (v,u) ∈ V, (𝑎 + 𝑏)(𝑣, 𝑢) = 𝑎(𝑣, 𝑢) + 𝑏(𝑣, 𝑢) = (𝑎𝑣 + 𝑏𝑣, 𝑎𝑢 +
𝑏𝑢) corresponds to (𝑎 + 𝑏)𝑣 = 𝑎𝑣 + 𝑏𝑣 for all (𝑥, 𝑤) ∈ 𝑊, (𝑎 + 𝑏)(𝑥, 𝑤) = 𝑎(𝑥, 𝑤) +
𝑏(𝑥, 𝑤) = (𝑎𝑥 + 𝑏𝑥, 𝑎𝑤 + 𝑏𝑤) corresponds to (𝑎 + 𝑏)𝑤 = 𝑎𝑤 + 𝑏𝑤 corresponds to for all
(𝑣, 𝑤) ∈ 𝑍, (𝑎 + 𝑏)(𝑣, 𝑤) = (𝑎𝑣 + 𝑏𝑣, 𝑎𝑤 + 𝑏𝑤) = 𝑎(𝑣, 𝑤) + 𝑏(𝑣, 𝑤)
29. Let H, K be subspaces of vector space V. Show 𝑯 ∩ 𝑲 is a subspace of V.
To prove this, we have to show that the following properties holds:
First, 0 ∈ 𝐻 ∩ 𝐾. Since H and K are subspaces of the vector space . hence, 0 ∈ 𝐻 and 0 ∈ 𝐾.
Therefore, the zero vector is in the intersection of 𝐻 and 𝐾, 0 ∈ 𝐻 ∩ 𝐾.
Second, If we have any vector say, 𝑢, 𝑣 ∈ 𝐻 ∩ 𝐾 𝑡ℎ𝑒𝑛 𝑢 + 𝑣 ∈ 𝐻 ∩ 𝐾. Suppose 𝑢, 𝑣 ∈ 𝐻 ∩ 𝐾
so we know that 𝑢 and 𝑣 are in 𝐻 and at the same time 𝑢, 𝑣 are also in 𝐾 by definition of
intersection and since we know that 𝑢 and 𝑣 are in 𝐻 and 𝐾 then 𝑢 + 𝑣 ∈ 𝐻 and 𝑢 + 𝑣 ∈ 𝐾.
Therefore, 𝑢 + 𝑣 ∈ 𝐻 ∩ 𝐾.
Third, If we have some vector 𝑢 ∈ 𝐻 ∩ 𝐾, then we have any scalar 𝑐 such that 𝑐𝑢 ∈ 𝐻 ∩ 𝐾.
Assume that we have 𝑢 ∈ 𝐻 ∩ 𝐾,that means 𝑢 is in 𝐻 and 𝑢 is in 𝐾 since it is an intersection and
since 𝐻 is a subspace then 𝑐𝑢 ∈ 𝐻 and 𝑐𝑢 ∈ 𝐾 then therefore 𝑐𝑢 ∈ 𝐻 ∩ 𝐾.
Hence, 𝐻 ∩ 𝐾 is a subspace of 𝑉.
30. Verify whether the polynomials 𝒙𝟑 − 𝟓𝒙𝟐 − 𝟐𝒙 + 𝟑 , 𝒙𝟑 − 𝟏 , 𝒙𝟑 + 𝟐𝒙 + 𝟒 are linearly
independent.
Solutions:
We may construct a matrix with coefficients of 𝑥 3 , 𝑥 2 , 𝑥 and constant terms.
1 1 1
[−5 0 0]
−2 0 2
3 −1 4
To find the rank of A let us reduce it to row echelon form by applying elementary
transformations.
Applying 𝑅2 → 𝑅2 + 5𝑅1 , 𝑅3 → 𝑅3 + 2𝑅1 , 𝑎𝑛𝑑 𝑅4 → 𝑅4 − 3𝑅1
1 1 1
[0 5 5]
0 2 4
0 −4 1
1
Applying 𝑅2 → (5 )𝑅2

1 1 1
[0 1 1]
0 2 4
0 −4 1
Applying 𝑅3 → 𝑅3 − 2𝑅2 , 𝑅4 → 𝑅4 + 4𝑅2
1 1 1
[0 1 1]
0 0 2
0 0 5
5
Applying 𝑅4 → 𝑅4 − (2)𝑅3

1 1 1
[0 1 1]
0 0 2
0 0 0
Rank of A=3=number of column vectors. So the given vectors are linearly independent
31. Show that the given subset of vectors of 𝑹𝟑 forms a basis for 𝑹𝟑 .
{(1,2,1) , (2,1,0) , (1,-1,2)}
Solutions:
We let S= {(1,2,1) , (2,1,0) , (1,-1,2)}
We know that any set of 𝑛 linearly independent vectors forms the basis of n-dimensional vector
space. Now , dim 𝑅3 = 3,
We just need to prove that vectors in S are linearly independent.
1 2 1
Let [2 1 −1]
1 0 2
We reduce this matrix to row echelon form to check the rank of A. Applying 𝑅2 → 𝑅2 + (−2)𝑅1
and 𝑅3 + (−1)𝑅1 we get
1 2 1
[0 −3 −3]
0 −2 1
1
Applying 𝑅2 → (− 3) 𝑅2 and 𝑅3 → 𝑅3 + 2𝑅2 , we get

1 2 1
[0 1 1]
0 0 3
Clearly, rank of A=3=number of vectors. Thus, the given vectors are linearly independent.
⇒S forms the basis of 𝑅3
32. Show that the set of vectors {(2 , 1) and (-1,-1)} is a spanning set in 𝑹𝟐 .

It means that you need to find a linear combination of this which could represent any vector
in 𝑅2 in equation form we have,

𝑎 (2,1) + 𝑏 (−1, −1) = (𝑥, 𝑦) where 𝑎 and 𝑏 ∈ ℝ

Their linear combination can represent any vector in 𝑅2 doing scalar multiplication we get

(2𝑎, 𝑎) + (−𝑏, −𝑏) = (𝑥, 𝑦)

Then we do properties of vector addition

(2𝑎 − 𝑏, 𝑎 − 𝑏) = (𝑥 , 𝑦)

Now if this linear combination has to be equals to (𝑥, 𝑦) that means,

2𝑎 − 𝑏 = 𝑥 → 𝑒𝑞. 1 in which their corresponding components should be the same

𝑎 − 𝑏 = 𝑦 → 𝑒𝑞. 2

Equation1 minus equation 2 is,

𝑎 =𝑥−𝑦
𝑏 =𝑎+𝑦
=𝑥−𝑦+𝑦
=𝑥
𝑏=𝑥

Such that,
(𝑥, 𝑦) = (𝑥 − 𝑦)(2,1) + 𝑥(−1, −1)

Since we can write any vector in 𝑅2 as a linear combination of the given set. We say that it is
a spanning set.

33.