Vector Spaces
Field
A non-empty set F is said to be a field if there exists two binary operations + and “.” on F such that
(i) (F, +) is an abelian group
(ii) (F \ {0}, .) is a multiplicative group and
(iii) for any a, b, c ∈ F, we have a (b + c) = ab + ac, (a + b)c = ac + bc.
Example:
(Q, +, ·) is a field
What is Q?
Q is the set of rational numbers:
Q = { p/q | p, q are integers, q ≠ 0 }.What does (Q, +, ·) mean?
This notation means we are looking at the set Q together with two operations:
(addition of rationals)
(multiplication of rationals)
So, (Q, +, ·) is not just a set, but a mathematical structure: the rationals with the usual addition and multiplication.
What does “is a field” mean?
A field is a set with two operations (+, ·) such that:(Q, +) is an abelian group (closed, associative, identity 0, additive inverses, commutative).
(Q \ {0}, ·) is an abelian group (all nonzero elements have multiplicative inverses, for example 2/3 has inverse 3/2).
Multiplication distributes over addition:
a · (b + c) = a·b + a·c for all a, b, c in Q.
So the statement (Q, +, ·) is a field means: The set of rational numbers, with usual addition and multiplication, satisfies all the field properties.
Let’s compare (Q, +, ·) and (Z, +, ·).
Integers with addition:
(Z, +) is an abelian group:
Closed: sum of two integers is integer
Associative
Identity: 0
Inverses: if a ∈ Z, then –a ∈ Z
Commutative
So this part is fine.
Integers with multiplication:
(Z, ·):
Closed, associative, commutative
Identity: 1
But the problem: not all nonzero integers have multiplicative inverses in Z.
Example: 2 ∈ Z. Its inverse should be 1/2, but 1/2 ∉ Z.
Only 1 and –1 have multiplicative inverses inside Z.
So (Z \ {0}, ·) is not a group.
Conclusion:
(Q, +, ·) is a field (every nonzero element has inverse).
(Z, +, ·) is not a field (no inverses except ±1).
That’s why we extend integers to rationals in the first place — to make division (except by 0) always possible.
Do you also want me to compress this into a very short version for quick recall?
Vector Spaces
A non-empty set V is said to be a vector space over a field F (generally, we take the field as the set of real numbers or the set of complex numbers) if it satisfies the following:
(i) (V, +) is an abelian group,
(ii) V is closed under scalar multiplication (that is, for every α ∈ F, v ∈ V we have αv ∈ V) and also this scalar multiplication satisfies the following conditions:
(a) α(v + w) = αv + αw,
(b) (α + β)v = αv + βv,
(c) α(βv) = (αβ)v, and
(d) 1·v = v for all α, β ∈ F and v, w ∈ V (here 1 is the identity of F with respect to multiplication).
Q over Q
Vectors = rationals (Q)
Scalars = rationals (Q)
Addition of rationals stays in Q, inverses exist, so (Q, +) is an abelian group.
Multiplying a rational scalar with a rational vector gives another rational.
All vector space axioms hold (distributive, associative, identity).
=> Q over Q is a vector space.
R over Q
Vectors = real numbers (R)
Scalars = rationals (Q)
Rational times real is real, addition is fine.
=> R over Q is a vector space.
C over Q
Vectors = complex numbers (C)
Scalars = rationals (Q)
Rational times complex is complex, addition is fine.
=> C over Q is a vector space.
R over R
Vectors = real numbers (R)
Scalars = real numbers (R)
Usual multiplication and addition work.
=> R over R is a vector space.
C over R
Vectors = complex numbers (C)
Scalars = real numbers (R)
Real times complex is complex, addition is fine.
=> C over R is a vector space.
C over C
Vectors = complex numbers (C)
Scalars = complex numbers (C)
Usual multiplication and addition work.
=> C over C is a vector space.
All of the above need to have, and they do have 1 as a multiplicative identity
Z over Z is not a vector space because:
Vectors = integers (Z)
Scalars = integers (Z)
Now test vector space conditions:
(Z, +) is an abelian group → ✅ True (sum of integers is integer, inverses exist, etc.)
Closed under scalar multiplication → at first seems okay, because integer × integer = integer.
But the problem is with axiom (d):
1·v = v, where 1 is the multiplicative identity of the field of scalars.
Here the scalars are Z. But Z is not a field (only ±1 have multiplicative inverses).
A vector space requires scalars to come from a field, not just any set.
So:
Since Z is not a field, Z over Z cannot be a vector space.
Maximal Linearly Independent Set
Def: (Maximal linearly independent set)
Let V be a vector space over 𝔽. Let S ⊆ V.
Then S is said to be a maximal linearly independent set if -
(i) S is linearly independent
(ii) S ∪ {v} is linearly dependent for any v ∈ V \ S.
-
This means:
If you try to add any new vector v from the vector space V, but not already in S,
Then the resulting set S ∪ {v} will become linearly dependent.
Minimal Spanning Set
Def: Minimal Spanning Set:
Let V be a vector space over 𝔽. Let S ⊆ V.
Then S is said to be a minimal spanning set if
(i) S is a spanning set.
(ii) S \ {v} does not span V for any v ∈ S.
---
Setup:
Let V be a vector space over a field 𝔽, and let S ⊆ V be a subset of vectors.
What it means:
The set S is called a minimal spanning set if it satisfies two conditions:
1. S is a spanning set.
This means every vector in V can be written as a linear combination of vectors in S.
2. For any vector v ∈ S, if you remove v from S, the resulting set S \ {v} does not span V.
In other words, every vector in S is essential to span the whole space.
If you remove any vector from S, you lose the ability to span V.
Understanding Maximal Linearly Independent Set vs Minimal Spanning Set
Let’s work in the vector space ℝ².
Define the following vectors:
v₁ = (1, 0)
v₂ = (0, 1)
v₃ = (1, 1)
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Step 1: Start with a linearly independent set
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Let S = {v₁} = {(1, 0)}
- S is linearly independent ✅
- But S does NOT span ℝ² ❌ (only x-axis)
Now add v₂:
S = {v₁, v₂} = {(1, 0), (0, 1)}
- S is linearly independent ✅
- S spans ℝ² ✅
- Adding any other vector (like v₃) would make it linearly dependent ❌
=> This is a maximal linearly independent set
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Step 2: Start with a spanning set
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Let S = {v₁, v₂, v₃} = {(1, 0), (0, 1), (1, 1)}
- S spans ℝ² ✅
- But S is not minimal ❌ (v₃ = v₁ + v₂, so it's redundant)
Remove v₃:
S = {v₁, v₂} = {(1, 0), (0, 1)}
- S still spans ℝ² ✅
- Now, if we remove either v₁ or v₂, it no longer spans ℝ² ❌
=> This is a minimal spanning set
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Conclusion:
--------------------------------------------
S = {(1, 0), (0, 1)} is both:
- A maximal linearly independent set ✅
- A minimal spanning set ✅
=> Therefore, it is a basis of ℝ²
Theorem 3:
Prove that a subset of a vector space is either linearly independent or one of the vectors
can be expressed as a linear combination of preceding vectors.
Proof:
1. Let S = {v₁, v₂, ..., vₙ} be a subset of a vector space V.
- Meaning: We're looking at a finite list of vectors from V.
- Why: To analyze if they're dependent or independent.
2. If v₁, v₂, ..., vₙ are linearly independent, then there's nothing to prove.
- Meaning: If no vector depends on the others, the condition is already satisfied.
- Why: The theorem says "either independent OR dependent in a specific way."
3. Suppose v₁, v₂, ..., vₙ are linearly dependent.
- Meaning: Now consider the opposite case — they are dependent.
- Why: To show that at least one vector can be written using earlier ones.
4. Then there exist scalars αᵢ (not all zero), for i = 1 to n, such that:
α₁v₁ + α₂v₂ + ... + αₙvₙ = 0 (Equation ①)
- Meaning: This is the definition of linear dependence.
- Why: Provides a starting equation to work from.
5. Since not all αᵢ are zero, let k be the largest index such that αₖ ≠ 0.
- Meaning: Pick the last non-zero coefficient in the list.
- Why: We'll solve for vₖ and show it's dependent on earlier vectors.
6. Then: αₖ₊₁ = αₖ₊₂ = ... = αₙ = 0
- Meaning: All αᵢ after k must be zero.
- Why: Simplifies the equation to only terms up to vₖ.
7. So Equation ① becomes:
α₁v₁ + α₂v₂ + ... + αₖ₋₁vₖ₋₁ + αₖvₖ = 0
- Why: Rewrite using only non-zero terms.
8. Rearranging:
αₖvₖ = -α₁v₁ - α₂v₂ - ... - αₖ₋₁vₖ₋₁
- Why: Isolate vₖ on one side.
9. Divide both sides by αₖ:
vₖ = (-α₁/αₖ)v₁ + (-α₂/αₖ)v₂ + ... + (-αₖ₋₁/αₖ)vₖ₋₁
- Meaning: vₖ is a linear combination of previous vectors.
- Why: This proves the theorem in the dependent case.
10. Note: (-αᵢ/αₖ) ∈ 𝔽, for 1 ≤ i ≤ k−1
- Meaning: These are valid field elements (scalars).
- Why: Ensures the expression is valid in the vector space.
Conclusion:
Hence, vₖ is a linear combination of {v₁, ..., vₖ₋₁}, so the theorem is proved.
Theorem 4:
In a vector space V, a maximal linearly independent set forms a basis.
Proof:
Let S = {v₁, v₂, ..., vₙ} be a maximal linearly independent set in V.
To prove: S forms a basis for V.
Step 1:
We already know that S is linearly independent.
So, we only need to show that S spans V.
Step 2:
Take any vector v ∈ V.
Assume there exists a linear relation involving v:
α₁v₁ + α₂v₂ + ... + αₙvₙ + αv = 0 (Equation ⊗)
We now consider two cases:
---------------------------------------------
Case (i): α = 0
Then Equation ⊗ becomes:
α₁v₁ + α₂v₂ + ... + αₙvₙ = 0
But S is linearly independent,
so this implies:
α₁ = α₂ = ... = αₙ = 0
Therefore:
All scalars are zero ⇒ S ∪ {v} is linearly independent
This contradicts the assumption that S is maximal, since we were able to add v to S and still maintain independence.
So, our assumption α = 0 must be false.
---------------------------------------------
The only possibility is: α ≠ 0
Now, go back to Equation ⊗:
α₁v₁ + α₂v₂ + ... + αₙvₙ + αv = 0
Rearrange:
αv = -α₁v₁ - α₂v₂ - ... - αₙvₙ
Divide both sides by α:
v = (-α₁/α)v₁ + (-α₂/α)v₂ + ... + (-αₙ/α)vₙ
This means:
v is a linear combination of vectors in S
So, any v ∈ V can be expressed as a combination of vectors in S
⇒ S spans V
Conclusion:
S is linearly independent (given),
and S spans V (just proved)
∴ S is a basis for V.
Theorem 5:
In a vector space V, a minimal spanning set of vectors forms a basis.
Proof Breakdown (Step-by-step)
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Step 1: (Top of image 1)
"Let S = {v1, v2, ..., vn} be a minimal spanning set for V."
Why?
We’re starting by assuming we have a spanning set S for V that is minimal, meaning no vector in S can be removed without losing the ability to span V.
---
Step 2:
"Then S spans V. i.e., L(S) = V."
Why?
This just restates what it means to be a spanning set — the linear span of S covers all of V.
---
Step 3:
"To prove that S is a basis for V, it is enough to show that S is linearly independent."
Why?
A basis must both:
- Span the vector space V
- Be linearly independent
We already know S spans V, so we only need to check linear independence.
---
Step 4:
"Assume that S is linearly dependent."
Why?
We use proof by contradiction — assume the opposite of what we want, and show it leads to a contradiction.
---
Step 5:
"Then there exists vj (for some j, 1 ≤ j ≤ n) which is a linear combination of its preceding ones. [by Theorem 3]"
Why?
If S is dependent, at least one vector can be written as a linear combination of the others.
---
Step 6:
"i.e., vj = α1 v1 + α2 v2 + ... + α(j-1) v(j-1), where the αi's are scalars."
Why?
This explicitly shows the linear dependence: vj is not needed to span the space since it can be written using other vectors.
---
What this means:
If we can write vj using the other vectors, we could remove it from S and still span V, contradicting the assumption that S is minimal.
---
Now we move to the second image:
---
Step 7:
"Let x ∈ L(S) = V."
Why?
We want to show that removing vj doesn’t prevent us from expressing any x in V — which would contradict minimality.
---
Step 8:
"Then x = β1 v1 + β2 v2 + ... + βn vn, for some scalars βi."
Why?
Since S spans V, any vector x in V can be written using vectors from S.
---
Step 9:
Replace vj in the expression using its linear combination from earlier.
That means rewrite:
x = β1 v1 + ... + β(j-1) v(j-1) + βj vj + β(j+1) v(j+1) + ... + βn vn
by plugging in:
vj = α1 v1 + α2 v2 + ... + α(j-1) v(j-1)
Why?
We're showing that even without vj, we can still write any vector in V.
---
Step 10:
Simplify the expression:
x = (β1 + βj α1) v1 + (β2 + βj α2) v2 + ... + (β(j-1) + βj α(j-1)) v(j-1) + β(j+1) v(j+1) + ... + βn vn
Why?
Now x is expressed without using vj! So the set S \ {vj} still spans V, contradicting the assumption that S was minimal.
---
Final Steps:
"⇒ x can be written as a linear combination of elements of S \ {vj}."
Why?
This is the contradiction — we said S was minimal, but we just removed vj and still spanned V.
---
Conclusion:
"Therefore our assumption is wrong. So, S is linearly independent and hence forms a basis for V."
Why?
We've shown that assuming dependence leads to a contradiction. So S must be linearly independent, and since it also spans V, it is a basis.
---
Summary of Logic:
1. We start with a minimal spanning set S.
2. To show it's a basis, we must prove it's linearly independent.
3. Assume it's not (i.e., it's dependent).
4. Show this would allow us to remove a vector and still span V.
5. But this contradicts the minimality of S.
6. Therefore, S must be linearly independent.
7. So S is a basis.
---
L
Theorem 6
If the vector space V has a basis B with n elements, then:
(i) Any set with (n + 1) vectors is linearly dependent.
(ii) Any set with (n - 1) vectors does not span V.
Proof:
(i) follows from Theorem 4.
(ii) follows from Theorem 5.
Theorem 7:
Prove that an orthogonal set of non-zero vectors is linearly independent.
🧠 Goal:
We want to prove that if we have vectors that are:
- Orthogonal (each pair is perpendicular),
- Non-zero,
Then the set must be linearly independent.
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🟢 Step 1: Define the set
Let A = {v₁, v₂, ..., vₙ} be an orthogonal set of non-zero vectors.
🧠 Why?
This is the setup — we're proving something about such a set A.
------------------------------------------------------------
🔴 Step 2: Assume a linear combination equals zero
Suppose:
α₁v₁ + α₂v₂ + ... + αₙvₙ = 0 (1)
🧠 Why?
To test for linear independence, we assume that a linear combination of the vectors adds up to the zero vector.
We'll try to prove that all αᵢ must be 0.
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🔵 Step 3: Take dot product with vₖ
Take dot product of both sides of (1) with one of the vectors vₖ:
⟨α₁v₁ + α₂v₂ + ... + αₙvₙ, vₖ⟩ = ⟨0, vₖ⟩
⇒ α₁⟨v₁, vₖ⟩ + α₂⟨v₂, vₖ⟩ + ... + αₙ⟨vₙ, vₖ⟩ = 0 (2)
🧠 Why?
The dot product is linear, so it distributes over sums and scalars.
We apply it to both sides to isolate terms involving vₖ.
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🟣 Step 4: Use orthogonality
Since A is orthogonal:
⟨vᵢ, vₖ⟩ = 0 if i ≠ k
⟨vₖ, vₖ⟩ = ‖vₖ‖²
So all terms in (2) are zero except the one where i = k.
Thus:
αₖ⟨vₖ, vₖ⟩ = αₖ‖vₖ‖² = 0 (3)
🧠 Why?
Because only the "self-dot-product" survives due to orthogonality.
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🟢 Step 5: Use that vectors are non-zero
Since vₖ is a non-zero vector, we know:
‖vₖ‖² ≠ 0
So from (3):
αₖ‖vₖ‖² = 0 ⇒ αₖ = 0
🧠 Why?
The product is zero, and the magnitude is non-zero, so the scalar must be zero.
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🔵 Step 6: Repeat for all k
This argument holds for all k = 1, 2, ..., n.
So:
α₁ = α₂ = ... = αₙ = 0
🧠 Why?
We showed that each αₖ must be zero individually.
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✅ Conclusion:
Since the only solution to equation (1) is the trivial solution, the set A is linearly independent.
∴ A is linearly independent. ∎
------------------------------------------------------------₁ = α₂ = ... = αₙ = 0
🧠 Why?
We showed that each α_k must be zero individually.
------------------------------------------------------------
✅ Conclusion:
Since the only solution to equation (1) is the trivial solution, the set A is linearly independent.
∴ A is linearly independent. ∎
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Theorem 8:
Prove that an orthonormal set of non-zero vectors is linearly independent.
Same as the previous proof
the only difference is <αₖ,vₖ> = 1
so for αₖ*1=0
αₖ must be 0
hence α₁ = α₂ = ... = αₙ = 0