Vector Spaces II

Quidquid latine dictum sit, altum sonatur, - Whatever is said in Latin sounds profound.

We will continue having lots of useless documentation to save our asses, tons of pointless meeting, and fruitless and endless discussions until we find out why no work is getting done and the code is unmaintainable. Meanwhile, I will keep using unrelated words and single letters in naming functions and variables, recycling variables and making as many of them static as I possibly can, documenting the obvious, but never ever writing a variable declaration or explaining what some section is attempting to accomplish, and avoiding testing like the plague, Apocalypse, Anawim, #justtothepoint.

Recall that a field is a commutative ring with unity such that each nonzero element has a multiplicative inverse, e.g., every finite integral domain, ℤ_p (p prime), ℚ, ℝ, and ℂ. A field has characteristic zero or characteristic p with p prime and F[x] is a field.

Definition. Let F be a field and V a set of two operations:

• +: VxV → V, ∀v, w ∈ V, v + w ∈ V
• ·: FxV → V, ∀v ∈ V, a ∈ F, av ∈ V. This operation is called scalar multiplication

We say V is a vector space over F if (V, +) is an Abelian group under addition, and ∀a, b ∈ F, u, v ∈V the following conditions hold:

1. a(v + u) = av + au, (a + b)v = av + bv
2. a(bv) = (ab)v
3. 1v = v.

Definition. Suppose V is a vector space over a field F and let S be a set of vectors from V, S = {v₁, v₂, ···, v_n}. We say S is linearly dependent if there are elements from F, not all zero, α₁, α₂, ···, α_n, such that they satisfy the following equation α₁v₁ + α₂v₂ + ··· + α_nv_n = 0 or alternatively, ∃v₁, v₂, ···, v_n ∈ V, α₁, α₂, ···, α_n∈ F such that α₁v₁ + α₂v₂ + ··· + α_nv_n = 0 We say S is linearly independent if α₁v₁ + α₂v₂ + ··· + α_nv_n = 0 ⇒ α₁ = α₂ = ··· = α_n = 0.

Examples:

• ℝ², S = {$(\begin{smallmatrix}1\\ 2\end{smallmatrix}), (\begin{smallmatrix}3\\ -4\end{smallmatrix})$}

x$(\begin{smallmatrix}1\\ 2\end{smallmatrix}) + y(\begin{smallmatrix}3\\ -4\end{smallmatrix})=(\begin{smallmatrix}0\\ 0\end{smallmatrix}) ⇒ (\begin{smallmatrix}x+3y\\ 2x-4y\end{smallmatrix})=(\begin{smallmatrix}0\\ 0\end{smallmatrix})$ ⇒ x + 3y = 0, 2x - 4y = 0 ⇒ x = y = 0.

• ℝ³, S = {(1, 0, 0), (1, 0, 1), (1, 1, 1)} are linearly independent because a(1, 0, 0) + b(1, 0, 1) + c(1, 1, 1) = (a + b + c, c, b + c) = (0, 0, 0) ⇒ a = b = c = 0.

• ℝ³, S = {(1, 2, 1), (0, 1, 2), (1, 1, 0)} are linearly independent because a(1, 2, 1) + b(0, 1, 2) + c(1, 1, 0) = (a + c, 2a + b + c, a + 2b) = (0, 0, 0) ⇒ a = b = c = 0.

• ℝ³, S = {(1, 1, 0), (0, 1, 1), (2, 3, 1)} are linearly dependent, a = -2, b = -1, c = 1.

$-2(\begin{smallmatrix}1\\ 1\\ 0\end{smallmatrix}) -1(\begin{smallmatrix}0\\ 1\\ 1\end{smallmatrix}) + (\begin{smallmatrix}2\\ 3\\ 1\end{smallmatrix})=(\begin{smallmatrix}0\\ 0\\ 0\end{smallmatrix})$

Proposition. Suppose V is a vector space over a field F and let S be a set of vectors from V, S = {v₁, v₂, ···, v_n}. {v₁, v₂, ···, v_n} is linearly dependent if and only if one of the v_i can be written as a linear combination of the others.

Proof.

⇒) Suppose {v₁, v₂, ···, v_n} is linearly dependent ⇒ ∃α₁, α₂, ···, α_n ∈ F not all zero, so ∃α_i ≠ 0 for some 1 ≤ i ≤ n, such that α₁v₁ + α₂v₂ + ··· + α_nv_n = 0 ⇒ α_iv_i = -(α₁v₁ + α₂v₂ + ··· α_i-1v_i-1 + α_i+1v_i+1 + ··· + α_nv_n) ⇒ [F field, α_i ≠ 0, it has a multiplicative inverse] v_i = (-α₁α_i^-1)v₁ (-α₂α_i^-1)v₂ + ··· (-α_i-1α_i^-1)v_i-1 (-α_i+1α_i^-1)v_i+1 + ··· (-α_nα_i^-1)v_n ∎

⇐) Suppose v_i = β₁v₁ +β₂v₂ + ··· β_i-1v_i-1 +β_i+1v_i+1 + ··· +β_nv_n ⇒ β₁v₁ +β₂v₂ + ··· β_i-1v_i-1 -v_i +β_i+1v_i+1 + ··· +β_nv_n = 0, and one of the coefficient, namely -1 ≠ 0 ∎

0, 1 and -1 are necessary elements in every field. 0 and 1 are the addition and multiplication identities respectively, and -1 is the additive inverse of 1 in F. However, those elements can be the same, e.g., -1 = 1 in ℤ₂.

Definition. A collection of vectors B ⊆ V is called a basis for V if B is linearly independent and span{v₁, v₂, ···, v_n} = {α₁v₁ + α₂v₂ + ··· + α_nv_n| α_i ∈ F} = V.

Proposition. B is a basis of V if and only if every vector v ∈ V can be written uniquely as a linear combination of elements of the basis.

Proof.

⇒) Let’s take v ∈ V = span{v₁, v₂, ···, v_n} = {α₁v₁ + α₂v₂ + ··· + α_nv_n| α_i ∈ F} = V, and suppose v = α₁v₁ + α₂v₂ + ··· + α_nv_n = β₁v₁ + β₂v₂ + ··· + β_nv_n where α_i, β_i ∈ F, v_i ∈ B. If there were different elements of V in both expressions, we can complete them by giving them zero-weights (α_l = 0) ⇒ (α₁-β₁)v₁ + (α₂-β₂)v₂ + ··· + (α_n-β_n)v_n = 0 ⇒ [B is basis for V, so B is linearly independent] α_i-β_i = 0, 1 ≤ i ≤ n ⇒ α_i = β_i = 0, 1 ≤ i ≤ n∎

⇐) Every vector v ∈ V can be written uniquely as a linear combination of elements of the basis ⇒ V = span{v₁, v₂, ···, v_n} = {α₁v₁ + α₂v₂ + ··· + α_nv_n| α_i ∈ F}

Let’s prove that B is linearly independent.

α₁v₁ + α₂v₂ + ··· + α_nv_n = 0 = 0v₁ + 0v₂ + ··· +0v_n ⇒ [By assumption, every vector v ∈ V can be written uniquely as a linear combination of elements of the basis] β_i = 0, 1 ≤ i ≤ n∎

Examples:

• Set e₁ = (1, 0, ··· 0), e₂ = (0, 1, ··· 0), ··· e_i = (0, 0, ··· 1 -ith position- ··· 0, 0), ··· e_n = (0, 0, ··· 1) ∈ ℝⁿ or ℂⁿ is the standard basis of ℝⁿ or ℂⁿ.
• {1, x, x², x³, ···} is the standard basis of F[x], F any field.
• {1, x, x²} is the standard basis of F₂[x] = $\wp_2$ (all polynomials of degree less or equal than two)
• V = {$(\begin{smallmatrix}a & a+b\\ a+b & b\end{smallmatrix})$ | a, b ∈ ℝ} is a vector space over ℝ, and B = { $(\begin{smallmatrix}1 & 1\\ 1 & 0\end{smallmatrix}), (\begin{smallmatrix}0 & 1\\ 1 & 1\end{smallmatrix})$} is a basis for V.

1. Linearly independent. Suppose ∃a, b ∈ ℝ s.t. $a(\begin{smallmatrix}1 & 1\\ 1 & 0\end{smallmatrix}) + b(\begin{smallmatrix}0 & 1\\ 1 & 1\end{smallmatrix}) = (\begin{smallmatrix}0 & 0\\ 0 & 0\end{smallmatrix}) ⇒ (\begin{smallmatrix}a & a+b\\ a+b & b\end{smallmatrix}) = (\begin{smallmatrix}0 & 0\\ 0 & 0\end{smallmatrix})$ ⇒ a = b = 0.
2. $(\begin{smallmatrix}a & a+b\\ a+b & b\end{smallmatrix}) = a(\begin{smallmatrix}1 & 1\\ 1 & 0\end{smallmatrix}) + b(\begin{smallmatrix}0 & 1\\ 1 & 1\end{smallmatrix})$ ⇒ B spans V.

Definition. Suppose V has a basis B = span{v₁, v₂, ···, v_n} = {α₁v₁ + α₂v₂ + ··· + α_nv_n| α_i ∈ F}. Let’s define the coordinate map, γ_B: V → Fⁿ, γ_B(v) = $(\begin{smallmatrix}α_1\\ α_2\\ ··· \\α_n\end{smallmatrix})$ if v = α₁v₁ + α₂v₂ + ··· + α_nv_n

Examples.

• {1, x, x²} is the standard basis of F₂[x] = $\wp_2$ (all polynomials of degree less or equal than two) γ_B: $\wp_2$ → F³, γ_B(α₀ + α₁x + α₂x₂) = $(\begin{smallmatrix}α_0\\ α_1\\α_2\end{smallmatrix})$.

• Is B = {$(\begin{smallmatrix}2\\ 3\end{smallmatrix}), (\begin{smallmatrix}1\\ -1\end{smallmatrix})$} a basis of ℝ²?

1. Linear independence. $x(\begin{smallmatrix}2\\ 3\end{smallmatrix}) + y(\begin{smallmatrix}1\\ -1\end{smallmatrix}) = (\begin{smallmatrix}0\\ 0\end{smallmatrix})$ ⇒ 2x + y = 0, 3x -y = 0 ⇒ x = y = 0.
2. v = $(\begin{smallmatrix}a\\ b\end{smallmatrix})$ ∈ ℝ², $x(\begin{smallmatrix}2\\ 3\end{smallmatrix}) + y(\begin{smallmatrix}1\\ -1\end{smallmatrix}) = (\begin{smallmatrix}a\\ b\end{smallmatrix})$ ⇒ 2x + y = a, 3x -y = b ⇒ x = (a+b)/5, y = (3/5)a -(2/5)b.

Theorem. Suppose V = span{v₁, v₂, ···, v_n} and {w₁, w₂, ···, w_m} is linearly independent. Then, m ≤ n.

Proof.

w₁ ∈ V = span{v₁, v₂, ···, v_n} ⇒ w₁ = α₁v₁ + α₂v₂ + ··· + α_nv_n, α_i ∈ F and not all α_i≠0 because if all α_i were zero, then w₁ = 0 ⇒ {w₁, w₂, ···, w_m} is lineal dependent.

Rename v₁, v₂, ···, v_n so that α₁ ≠ 0 ⇒ v₁ = -α₁^-1(α₂v₂ + α₃v₃ + ··· + α_nv_n - w₁) ∈ span{w₁, v₂, ···, v_n} ⇒ V = span{w₁, v₂, ···, v_n}

Let’s repeat the process, write w₂ = β₁w₁ + β₂v₂ + ··· + β_nv_n, where β₂,··· β_n not all zero because otherwise, w₂ = β₁w₁ and then {w₁, w₂, ···, w_m} is not linearly dependent.

Rename or reorder, so that β₂ ≠ 0, and v₂ = -β₂^-1(β₁w₁ + β₃v₃ + ··· + β_nv_n -w₂) ∈ span{w₁, w₂, v₃ ···, v_n} ⇒ V = span{w₁, w₂, v₃ ···, v_n}

Let’s continue this process until we “use” all w_i ⇒ m ≤ n. V = spam{w₁,···, w_m, v_m+1, ···, v_n}

Definition. If B = {v₁, v₂, ···, v_n} is a basis for V, V is said to have dimension n.

Proposition. The notion of dimension is well defined. In other words, every basis has the same number of vectors., e.g., {1, i} is a basis for ℂ over ℝ, and the dimension is two.

Proof. Suppose B = {v₁, v₂, ···, v_n} and B’ = {w₁, w₂, ···, w_m} are basis for V.

B spans and B’ is linear independence ⇒ m ≤ n, but B’ spans and B is linear independence ⇒ n ≤ m. Therefore, m = n ∎.

Examples

• Find the basis and dimension V = {$(\begin{smallmatrix}x_1\\ x_2\\x_3\end{smallmatrix})|x_1+x_2+x_3=0$} ⊆ ℝ³

x₁ + x₂ + x₃ = 0 ⇒ x₁ = - x₂ - x₃ ⇒ $v=(\begin{smallmatrix}-x_2-x_3\\ x_2\\x_3\end{smallmatrix}) = x_2(\begin{smallmatrix}-1\\ 1\\0\end{smallmatrix})+x_3(\begin{smallmatrix}-1\\ 0\\1\end{smallmatrix})$, for some x₂, x₃ ∈ ℝ. Therefore, V = span{$(\begin{smallmatrix}-1\\ 1\\0\end{smallmatrix}), (\begin{smallmatrix}-1\\ 0\\1\end{smallmatrix})$}

We only need to prove that these two vectors are lineal independent. Suppose α, β ∈ ℝ,

$α(\begin{smallmatrix}-1\\ 1\\ 0\end{smallmatrix})+β(\begin{smallmatrix}-1\\ 0\\ 1\end{smallmatrix}) = (\begin{smallmatrix}0\\ 0\\ 0\end{smallmatrix})$ ⇒ -α-β=0, α=0, β=0, ∎. Therefore, we have a basis and dim(V) = 2.

• Find the basis and dimension {f(x) ∈ ℝ[x]| f(3)=0 and deg(f(x) ≤ 2)} ⊆ $\wp_2$. It is a subspace of the vector space of polynomials of degree equal or less than two.

Let an arbitrary f(x) ∈ {f(x) ∈ ℝ[x]| f(3)=0 and deg(f(x) ≤ 2)} ⊆ $\wp_2$ ⇒ f(3) = 0 ⇒ f(x) = (x-3)g(x) where g(x) is a polynomial of degree at most 1.

1. deg(f(x)) = 0 ⇒[f is arbitrary] f(x) = c ≠ 0 ⇒f(3) ≠ 0 ⊥
2. deg(f(x)) = 1 ⇒ deg(g(x)) = 0 ⇒ f(x) = c(x-3)
3. deg(f(x)) = 2 ⇒ deg(g(x)) = 1 ⇒ f(x) = (ax + b)(x - 3). This last form is the most general case of f(x), and a = 0 is a particular case (2).

f(x) = (ax + b)(x - 3) = ax(x -3) + b(x-3) = span{x(x-3), x-3}

{f(x) ∈ ℝ[x]| f(3)=0 and deg(f(x) ≤ 2)} = span{x(x-3), x-3}, so we only need to show that these two polynomials are linearly independent.

ax(x - 3) + b(x -3) = 0. If x = 0 ⇒ b(-3) = 0 ⇒ b = 0.

ax(x -3) = 0 ∀x ∈ ℝ. In particular, x = 1, a(-2) = 0 ⇒ a = 0. Therefore, it is a basis, and its dimension is equal to two.

Bibliography

1. NPTEL-NOC IITM, Introduction to Galois Theory.
2. Algebra, Second Edition, by Michael Artin.
3. LibreTexts, Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
4. Field and Galois Theory, by Patrick Morandi. Springer.
5. Michael Penn (Abstract Algebra), and MathMajor.
6. Contemporary Abstract Algebra, Joseph, A. Gallian.
7. Andrew Misseldine: College Algebra and Abstract Algebra.