Expressing a 3D Vector as a Linear Combination of Three Vectors
Problem 1
Express the vector (4, 7, 3) as a linear combination of (1,1,0), (2,1,0), (0,0,1)
Step 1: Assume a linear combination
(4,7,3) = a(1,1,0) + b(2,1,0) + c(0,0,1)
Step 2: Expand
a(1,1,0) = (a,a,0)
b(2,1,0) = (2b,b,0)
c(0,0,1) = (0,0,c)
Adding:
(a + 2b, a + b, c)
Step 3: Equate components
a + 2b = 4
a + b = 7
c = 3
Step 4: Solve
Subtract equations:
(a + 2b) − (a + b) = 4 − 7
b = −3
Substitute into second equation:
a − 3 = 7
a = 10
Final Answer
(4,7,3) = 10(1,1,0) − 3(2,1,0) + 3(0,0,1)
Problem 2
Express
(5,4,6)
as a linear combination of
(1,0,1), (0,1,1), (1,1,0)
Step 1
(5,4,6) = a(1,0,1) + b(0,1,1) + c(1,1,0)
Step 2: Expand
a(1,0,1) = (a,0,a)
b(0,1,1) = (0,b,b)
c(1,1,0) = (c,c,0)
Adding:
(a + c, b + c, a + b)
Step 3: Equate components
a + c = 5
b + c = 4
a + b = 6
Step 4: Solve
a = 5 − c
b = 4 − c
Substitute into third equation:
(5 − c) + (4 − c) = 6
9 − 2c = 6
c = 1.5
a = 3.5
b = 2.5
Final Answer
(5,4,6) = 3.5(1,0,1) + 2.5(0,1,1) + 1.5(1,1,0)
Problem 3
Express
(3,2,5)
as a linear combination of
(1,1,1), (1,0,2), (0,1,1)
Step 1
(3,2,5) = a(1,1,1) + b(1,0,2) + c(0,1,1)
Step 2: Expand
a(1,1,1) = (a,a,a)
b(1,0,2) = (b,0,2b)
c(0,1,1) = (0,c,c)
Adding:
(a + b, a + c, a + 2b + c)
Step 3: Equate
a + b = 3
a + c = 2
a + 2b + c = 5
Step 4: Solve
b = 3 − a
c = 2 − a
Substitute into third equation:
a + 2(3 − a) + (2 − a) = 5
a + 6 − 2a + 2 − a = 5
8 − 2a = 5
a = 1.5
b = 1.5
c = 0.5
Final Answer
(3,2,5) = 1.5(1,1,1) + 1.5(1,0,2) + 0.5(0,1,1)
Problem 4
Express
(6,5,4)
as a linear combination of
(1,2,0), (2,1,1), (0,1,1)
Step 1
(6,5,4) = a(1,2,0) + b(2,1,1) + c(0,1,1)
Step 2: Expand
a(1,2,0) = (a,2a,0)
b(2,1,1) = (2b,b,b)
c(0,1,1) = (0,c,c)
Adding:
(a + 2b, 2a + b + c, b + c)
Step 3: Equate
a + 2b = 6
2a + b + c = 5
b + c = 4
Step 4: Solve
c = 4 − b
Substitute into second equation:
2a + b + 4 − b = 5
2a + 4 = 5
a = 0.5
Substitute into first equation:
0.5 + 2b = 6
b = 2.75
c = 1.25
Final Answer
(6,5,4) = 0.5(1,2,0) + 2.75(2,1,1) + 1.25(0,1,1)
Problem 5
Express
(2,3,4)
as a linear combination of
(1,0,1), (0,1,1), (1,1,1)
Step 1
(2,3,4) = a(1,0,1) + b(0,1,1) + c(1,1,1)
Step 2: Expand
a(1,0,1) = (a,0,a)
b(0,1,1) = (0,b,b)
c(1,1,1) = (c,c,c)
Adding:
(a + c, b + c, a + b + c)
Step 3: Equate
a + c = 2
b + c = 3
a + b + c = 4
Step 4: Solve
a = 2 − c
b = 3 − c
Substitute:
(2 − c) + (3 − c) + c = 4
5 − c = 4
c = 1
a = 1
b = 2
Final Answer
(2,3,4) = 1(1,0,1) + 2(0,1,1) + 1(1,1,1)
Problems to check for Linear Independence of Vectors
Problem 1
Check whether the vectors
V₁ = (2,5,3)
V₂ = (1,1,1)
V₃ = (4,-2,0)
are linearly independent.
Step 1: Form the Linear Combination
To check linear independence, we examine whether the equation
aV₁ + bV₂ + cV₃ = (0,0,0)
has only the trivial solution.
Substitute the vectors:
(2,5,3)a + (1,1,1)b + (4,-2,0)c = (0,0,0)
Step 2: Expand the Vectors
a(2,5,3) = (2a,5a,3a)
b(1,1,1) = (b,b,b)
c(4,-2,0) = (4c,-2c,0)
Adding the vectors:
(2a + b + 4c , 5a + b − 2c , 3a + b) = (0,0,0)
Step 3: Form the System of Equations
Equating corresponding components:
2a + b + 4c = 0
5a + b − 2c = 0
3a + b = 0
Step 4: Solve the Equations
From the third equation:
3a + b = 0
b = −3a
Substitute into the first equation:
2a + (−3a) + 4c = 0
−a + 4c = 0
a = 4c
Substitute into the second equation:
5(4c) + (−3(4c)) − 2c = 0
20c − 12c − 2c = 0
6c = 0
c = 0
Now compute the remaining variables:
a = 4c = 0
b = −3a = 0
Step 5: Conclusion
The only solution is
a = 0, b = 0, c = 0
Since the trivial solution is the only solution, the vectors are linearly independent.
Final Result
The vectors
(2,5,3), (1,1,1), and (4,-2,0)
are linearly independent.
Problem:
Find a condition on 
Solution:
Condition for a Vector to be in the Span of Two Vectors in R³
We are given the vectors u = (1, -3, 2) and v = (2, -1, 1) and a general vector w = (a, b, c)
We must find the condition on a, b, c such that w is a linear combination of u and v.
If this condition is satisfied, then
w ∈ span(u, v)
Step 1: Express w as a Linear Combination
For w to belong to the span of u and v, there must exist scalars α and β such that
w = αu + βv
Substitute the vectors:
(a, b, c) = α(1, -3, 2) + β(2, -1, 1)
Step 2: Expand the Right Side
α(1, -3, 2) = (α, -3α, 2α)
β(2, -1, 1) = (2β, -β, β)
Adding the vectors:
(α + 2β , -3α – β , 2α + β)
Thus
(a, b, c) = (α + 2β , -3α – β , 2α + β)
Step 3: Equate Corresponding Components
a = α + 2β
b = -3α – β
c = 2α + β
This gives a system of three equations.
Step 4: Solve the System
From the first equation:
α = a – 2β
Substitute into the third equation:
c = 2(a – 2β) + β
c = 2a – 4β + β
c = 2a – 3β
Solve for β:
β = (2a – c) / 3
Now substitute β into the second equation:
b = -3(a – 2β) – β
b = -3a + 6β – β
b = -3a + 5β
Substitute β:
b = -3a + 5(2a – c)/3
Multiply by 3:
3b = -9a + 10a – 5c
3b = a – 5c
Step 5: Final Condition
Rearranging:
a – 3b – 5c = 0
Final Result
The vector
w = (a, b, c)
is a linear combination of u and v if and only if
a – 3b – 5c = 0
Interpretation
This condition describes a plane passing through the origin in R³.
Therefore,
span(u, v)
represents a two-dimensional subspace (a plane) in R³, and any vector satisfying
a – 3b – 5c = 0
lies in that plane.
Problem
Find the basis and dimension of the subspace spanned by the vectors 
Solution
Finding the Basis and Dimension of a Subspace in V₃(R)
We are given the vectors
v₁ = (2,4,2)
v₂ = (1,-1,0)
v₃ = (1,2,1)
v₄ = (0,3,1)
We must find the basis and dimension of the subspace spanned by these vectors.
Step 1: Concept
The span of a set of vectors is the set of all linear combinations of those vectors.
To find the basis, we need to identify the linearly independent vectors among them.
A convenient method is to form a matrix using the vectors as rows (or columns) and perform Gaussian elimination (row reduction).
The non-zero rows in row-echelon form form a basis.
Step 2: Form the Matrix
Place the vectors as rows of a matrix.
| 2 4 2 |
| 1 -1 0 |
| 1 2 1 |
| 0 3 1 |
Step 3: Row Reduction (Gaussian Elimination)
First interchange row 1 and row 2 for convenience.
| 1 -1 0 |
| 2 4 2 |
| 1 2 1 |
| 0 3 1 |
Eliminate first column
R₂ → R₂ − 2R₁
| 1 -1 0 |
| 0 6 2 |
| 1 2 1 |
| 0 3 1 |
R₃ → R₃ − R₁
| 1 -1 0 |
| 0 6 2 |
| 0 3 1 |
| 0 3 1 |
Eliminate second column
R₃ → R₃ − (1/2)R₂
| 1 -1 0 |
| 0 6 2 |
| 0 0 0 |
| 0 3 1 |
R₄ → R₄ − (1/2)R₂
| 1 -1 0 |
| 0 6 2 |
| 0 0 0 |
| 0 0 0 |
Step 4: Identify Independent Rows
The resulting matrix has two non-zero rows:
(1, -1, 0)
(0, 6, 2)
These rows are linearly independent.
Step 5: Basis of the Subspace
A basis for the span can therefore be taken as
{ (1,-1,0), (0,6,2) }
Since any scalar multiple of a basis vector can also be used, we may simplify the second vector:
(0,6,2) = 2(0,3,1)
Thus a cleaner basis is
{ (1,-1,0), (0,3,1) }
Step 6: Dimension
The dimension of a subspace is the number of vectors in its basis.
Here the basis contains 2 vectors.
Therefore Dimension = 2
Final Result
Basis of the subspace: { (1,-1,0), (0,3,1) }
Dimension of the subspace: 2
Interpretation
Although four vectors were given, they are not all linearly independent.
They span a two-dimensional subspace (a plane) in R³.
Thus the subspace spanned by the given vectors is a plane passing through the origin in V₃(R).
Problem
Show that the functions 
Solution
Orthogonality of Functions in Pₙ
Given the functions
f(x) = 3x − 2
g(x) = x
We need to
- Find the inner product ⟨f,g⟩ over the interval [0,1]
- Check whether the functions are orthogonal.
Step 1: Definition of Inner Product
For polynomials in (P_n), the inner product over the interval [0,1] is defined as
⟨f,g⟩ = ∫₀¹ f(x)g(x) dx
Two functions are orthogonal if
⟨f,g⟩ = 0
Step 2: Compute the Product f(x)g(x)
f(x)g(x) = (3x − 2)(x)
Expand:
f(x)g(x) = 3x² − 2x
Step 3: Compute the Inner Product
⟨f,g⟩ = ∫₀¹ (3x² − 2x) dx
Split the integral:
⟨f,g⟩ = ∫₀¹ 3x² dx − ∫₀¹ 2x dx
Step 4: Evaluate the Integrals
First integral
∫₀¹ 3x² dx
= 3 ∫₀¹ x² dx
= 3 [x³/3]₀¹
= 3(1/3 − 0)
= 1
Second integral
∫₀¹ 2x dx
= [x²]₀¹
= 1 − 0
= 1
Step 5: Compute the Inner Product
⟨f,g⟩ = 1 − 1
⟨f,g⟩ = 0
Step 6: Check Orthogonality
Since
⟨f,g⟩ = 0
the functions are orthogonal.
Final Result
Inner Product:
⟨f,g⟩ = 0
Conclusion:
The functions f(x) = 3x − 2 and g(x) = x are orthogonal in (P_n) over the interval [0,1].
Problem
Consider 
Solution
Finding the Inner Product of Two Functions
Given the functions
f(t) = 3t − 5
g(t) = t²
Find the inner product ⟨f,g⟩ over the interval [0,1].
Step 1: Definition of Inner Product
For functions in (P_n), the inner product over the interval [0,1] is defined as
⟨f,g⟩ = ∫₀¹ f(t)g(t) dt
Step 2: Compute the Product f(t)g(t)
f(t)g(t) = (3t − 5)(t²)
Expand:
f(t)g(t) = 3t³ − 5t²
Step 3: Substitute into the Inner Product
⟨f,g⟩ = ∫₀¹ (3t³ − 5t²) dt
Separate the integrals:
⟨f,g⟩ = ∫₀¹ 3t³ dt − ∫₀¹ 5t² dt
Step 4: Evaluate the Integrals
First integral
∫₀¹ 3t³ dt
= 3 ∫₀¹ t³ dt
= 3 [t⁴/4]₀¹
= 3(1/4 − 0)
= 3/4
Second integral
∫₀¹ 5t² dt
= 5 ∫₀¹ t² dt
= 5 [t³/3]₀¹
= 5(1/3)
= 5/3
Step 5: Compute the Inner Product
⟨f,g⟩ = 3/4 − 5/3
Take LCM = 12
3/4 = 9/12
5/3 = 20/12
⟨f,g⟩ = 9/12 − 20/12
⟨f,g⟩ = −11/12
Final Result
⟨f,g⟩ = −11/12
Therefore, the inner product of the functions f(t) = 3t − 5 and g(t) = t² over the interval [0,1] is
−11/12.
Dr. Selvaraj Vadivelu 