Unit 2: Algebraic Foundations: Groups, Fields, Vector Spaces & Subspaces
Abstract algebraic structures, groups, fields (R, C, Z_p), vector space axioms over arbitrary fields, function and matrix spaces, subspace criteria, intersection, sum, and direct sum decompositions.
§2.1 Abstract Algebraic Foundations: Groups, Rings, and Fields
1. The Concept of a Group
A Group is an algebraic structure $(G, )$ consisting of a non-empty set $G$ together with a binary operation $: G \times G \to G$ satisfying the following four axioms:
1. Closure: For all $a, b \in G$, $a * b \in G$.
2. Associativity: For all $a, b, c \in G$, $(a * b) * c = a * (b * c)$.
3. Identity Element: There exists an element $e \in G$ such that for every $a \in G$:
4. Inverse Element: For each $a \in G$, there exists an element $a^{-1} \in G$ such that:
If, in addition, the operation satisfies commutativity:
then $(G, *)$ is called an Abelian (Commutative) Group.
2. Rings and Fields
Definition of a Field
A Field $(F, +, \cdot)$ is a set $F$ equipped with two binary operations called addition ($+$) and multiplication ($\cdot$) such that:
- $(F, +)$ is an abelian group with additive identity denoted by $0$ and additive inverse of $a$ denoted by $-a$.
- $(F \setminus \{0\}, \cdot)$ is an abelian group with multiplicative identity denoted by $1$ ($1 \ne 0$) and multiplicative inverse of $a \ne 0$ denoted by $a^{-1}$ or $1/a$.
- Distributivity: Multiplication distributes over addition:
Canonical Examples of Fields:
- The field of rational numbers $(\mathbb{Q}, +, \cdot)$.
- The field of real numbers $(\mathbb{R}, +, \cdot)$.
- The field of complex numbers $(\mathbb{C}, +, \cdot)$.
- The finite Galois field of prime order $(\mathbb{Z}_p, +, \cdot)$ where $p$ is prime and arithmetic is modulo $p$.
- Note: The integers $\mathbb{Z}$ do not form a field because non-zero integers other than $\pm 1$ do not possess multiplicative inverses in $\mathbb{Z}$.
Fundamental Field Properties:
From these axioms, several universal properties follow:
- $0 \cdot a = 0$ for all $a \in F$.
- $(-a) \cdot b = -(a \cdot b) = a \cdot (-b)$ for all $a, b \in F$.
- If $a \cdot b = 0$, then either $a = 0$ or $b = 0$ (no zero divisors).
§2.2 Axiomatic Vector Spaces over an Arbitrary Field
1. Axiomatic Definition of a Vector Space
Let $F$ be a field (whose elements are called scalars). A Vector Space over $F$ is a non-empty set $V$ (whose elements are called vectors), equipped with two operations:
1. Vector Addition: $+ : V \times V \to V$, assigning to each pair $(\vec{u}, \vec{v})$ a vector $\vec{u} + \vec{v} \in V$.
2. Scalar Multiplication: $\cdot : F \times V \to V$, assigning to each pair $(c, \vec{v})$ a vector $c\vec{v} \in V$.
such that for all $\vec{u}, \vec{v}, \vec{w} \in V$ and all $c, d \in F$, the following eight axioms are satisfied:
Axioms of Addition (Abelian Group $(V, +)$):
1. A1 (Commutativity): $\vec{u} + \vec{v} = \vec{v} + \vec{u}$
2. A2 (Associativity): $(\vec{u} + \vec{v}) + \vec{w} = \vec{u} + (\vec{v} + \vec{w})$
3. A3 (Additive Identity): There exists a vector $\vec{0} \in V$ such that $\vec{v} + \vec{0} = \vec{v}$ for all $\vec{v} \in V$.
4. A4 (Additive Inverse): For every $\vec{v} \in V$, there exists a vector $-\vec{v} \in V$ such that $\vec{v} + (-\vec{v}) = \vec{0}$.
Axioms of Scalar Multiplication:
5. M1 (Distributivity over Vector Addition): $c(\vec{u} + \vec{v}) = c\vec{u} + c\vec{v}$
6. M2 (Distributivity over Scalar Addition): $(c + d)\vec{v} = c\vec{v} + d\vec{v}$
7. M3 (Compatibility of Scalar Multiplication): $c(d\vec{v}) = (cd)\vec{v}$
8. M4 (Scalar Identity): $1\vec{v} = \vec{v}$, where $1 \in F$ is the multiplicative identity of $F$.
2. Canonical Examples of Vector Spaces
1. Coordinate Spaces $F^n$: Column vectors of length $n$ with entries in $F$.
2. Matrix Spaces $M_{m \times n}(F)$: The set of all $m \times n$ matrices with entries in $F$ under standard matrix addition and scalar multiplication.
3. Polynomial Spaces $\mathbb{P}_n(F)$ and $\mathbb{P}(F)$: The set $\mathbb{P}_n(F) = \{a_0 + a_1 t + \cdots + a_n t^n : a_i \in F\}$ of polynomials of degree $\le n$.
4. Function Spaces $C[a, b]$: The set of all continuous real-valued functions $f: [a, b] \to \mathbb{R}$ with $(f + g)(t) = f(t) + g(t)$ and $(cf)(t) = c f(t)$.
5. Sequence Spaces $\ell^2$ and $\ell^\infty$: The set of infinite sequences $(x_1, x_2, \dots)$ with $\sum |x_i|^2 < \infty$.
§2.3 Subspaces, Intersections, Sums & Direct Sum Decompositions
1. Vector Subspaces
Let $V$ be a vector space over field $F$. A subset $W \subseteq V$ is a subspace of $V$ (written $W \le V$) if $W$ is itself a vector space over $F$ under the operations of addition and scalar multiplication inherited from $V$.
Theorem 2.1 (The Subspace Criterion / Two-Step Test):
A non-empty subset $W \subseteq V$ is a subspace of $V$ if and only if:
1. Contains Zero: $\vec{0} \in W$.
2. Closure under Addition: For all $\vec{u}, \vec{v} \in W$, $\vec{u} + \vec{v} \in W$.
3. Closure under Scalar Multiplication: For all $c \in F$ and $\vec{u} \in W$, $c\vec{u} \in W$.
Equivalently (One-Step Test): $W \ne \emptyset$ and for all $c, d \in F$ and $\vec{u}, \vec{v} \in W$, $c\vec{u} + d\vec{v} \in W$.
2. Intersection and Sum of Subspaces
Let $W_1$ and $W_2$ be subspaces of $V$.
Theorem 2.2 (Intersection of Subspaces):
The intersection $W_1 \cap W_2 = \{\vec{v} \in V : \vec{v} \in W_1 \text{ and } \vec{v} \in W_2\}$ is always a subspace of $V$.
Proof:
- Since $\vec{0} \in W_1$ and $\vec{0} \in W_2$, $\vec{0} \in W_1 \cap W_2$.
- If $\vec{u}, \vec{v} \in W_1 \cap W_2$, then $\vec{u}, \vec{v} \in W_1 \implies \vec{u} + \vec{v} \in W_1$, and $\vec{u}, \vec{v} \in W_2 \implies \vec{u} + \vec{v} \in W_2$. Thus $\vec{u} + \vec{v} \in W_1 \cap W_2$.
- If $c \in F$ and $\vec{u} \in W_1 \cap W_2$, then $c\vec{u} \in W_1$ and $c\vec{u} \in W_2$, so $c\vec{u} \in W_1 \cap W_2$. $\blacksquare$
Warning on Unions: The union $W_1 \cup W_2$ is generally not a subspace unless one subspace is completely contained in the other ($W_1 \subseteq W_2$ or $W_2 \subseteq W_1$).
Definition: Sum of Subspaces
The sum of two subspaces $W_1, W_2$ is defined by:
$W_1 + W_2$ is the smallest subspace of $V$ containing both $W_1$ and $W_2$.
3. Direct Sums and Uniqueness of Decomposition
The sum $W_1 + W_2$ is called an Internal Direct Sum, denoted by:
if every vector $\vec{v} \in V$ can be written uniquely as $\vec{v} = \vec{w}_1 + \vec{w}_2$ with $\vec{w}_1 \in W_1$ and $\vec{w}_2 \in W_2$.
Theorem 2.3 (Direct Sum Equivalence Theorem):
Let $W_1, W_2$ be subspaces of $V$. Then $V = W_1 \oplus W_2$ if and only if:
- $V = W_1 + W_2$, and
- $W_1 \cap W_2 = \{\vec{0}\}$.
Proof: $(\implies)$ Assume $V = W_1 \oplus W_2$. Then $V = W_1 + W_2$ by definition. Suppose $\vec{x} \in W_1 \cap W_2$. Then we can express $\vec{0} \in V$ in two ways:
By the uniqueness of decomposition, we must have $\vec{x} = \vec{0}$. Hence $W_1 \cap W_2 = \{\vec{0}\}$.
$(\impliedby)$ Assume $V = W_1 + W_2$ and $W_1 \cap W_2 = \{\vec{0}\}$. Let $\vec{v} \in V$. Since $V = W_1 + W_2$, there exist $\vec{w}_1 \in W_1$ and $\vec{w}_2 \in W_2$ such that $\vec{v} = \vec{w}_1 + \vec{w}_2$. To prove uniqueness, suppose $\vec{v} = \vec{w}_1' + \vec{w}_2'$ with $\vec{w}_1' \in W_1$ and $\vec{w}_2' \in W_2$. Then:
The left side is in $W_1$ (since $W_1$ is a subspace), and the right side is in $W_2$. Thus:
This proves that the decomposition is strictly unique. $\blacksquare$
Step-by-Step Solved Examination Problems
Comprehensive analytical derivations, multi-tier solutions (Foundational, Intermediate Exam, and Honors/Proof Challenge) with complete line-by-line verification.
Let $V = M_{n \times n}(\mathbb{R})$ be the vector space of all $n \times n$ real matrices. Determine, with complete mathematical justification using the Subspace Criterion, whether each of the following subsets is a subspace of $V$: (a) $W_1 = \{A \in M_{n \times n}(\mathbb{R}) : A^T = A\}$ (the set of symmetric matrices). (b) $W_2 = \{A \in M_{n \times n}(\mathbb{R}) : \det(A) = 0\}$ (the set of singular matrices). (c) $W_3 = \{A \in M_{n \times n}(\mathbb{R}) : \text{tr}(A) = 0\}$ (the set of trace-zero matrices).
(a) Analysis of $W_1$ (Symmetric Matrices):
1. Contains Zero: The zero matrix $O_{n \times n}$ satisfies $O^T = O$, so $O \in W_1$.
2. Closure under Addition: Let $A, B \in W_1$. Then $A^T = A$ and $B^T = B$. Using the linearity of the transpose operation:
Thus $A + B \in W_1$.
3. Closure under Scalar Multiplication: For $c \in \mathbb{R}$ and $A \in W_1$:
Thus $c A \in W_1$. Therefore, $W_1$ is a subspace of $M_{n \times n}(\mathbb{R})$.
(b) Analysis of $W_2$ (Singular Matrices): $W_2$ is NOT a subspace for $n \ge 2$. While $O \in W_2$ and scalar multiples of singular matrices are singular, $W_2$ is not closed under addition. Counterexample for $n = 2$:
However:
Hence $W_2$ fails closure under addition and is not a subspace.
(c) Analysis of $W_3$ (Trace-Zero Matrices): The trace of a matrix is $\text{tr}(A) = \sum_{i=1}^n a_{ii}$.
1. Contains Zero: $\text{tr}(O) = 0$, so $O \in W_3$.
2. Closure under Addition & Scalar Multiplication: For $A, B \in W_3$ and $c, d \in \mathbb{R}$:
Thus $c A + d B \in W_3$. Therefore, $W_3$ is a subspace of $M_{n \times n}(\mathbb{R})$.
$W_1$ (symmetric matrices) and $W_3$ (trace-zero matrices) are subspaces of $M_{n \times n}(\mathbb{R})$. $W_2$ (singular matrices) is NOT a subspace because it is not closed under addition ($A = \text{diag}(1,0)$ and $B = \text{diag}(0,1)$ both have det 0, but $A+B = I$ has det 1).
Let $V = M_{n \times n}(\mathbb{R})$. Let $W_{\text{sym}} = \{A \in V : A^T = A\}$ be the subspace of symmetric matrices, and let $W_{\text{skew}} = \{A \in V : A^T = -A\}$ be the subspace of skew-symmetric matrices.
(a) Prove that $W_{\text{sym}} \cap W_{\text{skew}} = \{O\}$. (b) Prove that every matrix $A \in V$ can be decomposed into a sum $A = A_{\text{sym}} + A_{\text{skew}}$ with $A_{\text{sym}} \in W_{\text{sym}}$ and $A_{\text{skew}} \in W_{\text{skew}}$. (c) Conclude that $M_{n \times n}(\mathbb{R}) = W_{\text{sym}} \oplus W_{\text{skew}}$, and compute the unique decomposition for the matrix:
Step 1: Prove $W_{\text{sym}} \cap W_{\text{skew}} = \{O\}$: Let $A \in W_{\text{sym}} \cap W_{\text{skew}}$.
- Since $A \in W_{\text{sym}}$, $A^T = A$.
- Since $A \in W_{\text{skew}}$, $A^T = -A$.
Therefore:
Thus $W_{\text{sym}} \cap W_{\text{skew}} = \{O\}$.
Step 2: Prove $V = W_{\text{sym}} + W_{\text{skew}}$: For any arbitrary matrix $A \in M_{n \times n}(\mathbb{R})$, write:
Define:
Verify membership:
Since $A = A_{\text{sym}} + A_{\text{skew}}$, every matrix in $V$ is in $W_{\text{sym}} + W_{\text{skew}}$.
Step 3: Direct Sum Conclusion: By Theorem 2.3, since $V = W_{\text{sym}} + W_{\text{skew}}$ and $W_{\text{sym}} \cap W_{\text{skew}} = \{O\}$, we have:
Step 4: Explicit computation for given matrix $A$:
Check: $A_{\text{sym}} + A_{\text{skew}} = \begin{pmatrix} 2+0 & 3+2 & 1-2 \\ 3-2 & 4+0 & 2+4 \\ 1+2 & 2-4 & 7+0 \end{pmatrix} = \begin{pmatrix} 2 & 5 & -1 \\ 1 & 4 & 6 \\ 3 & -2 & 7 \end{pmatrix} = A$. Verified!
$M_{n \times n}(\mathbb{R}) = W_{\text{sym}} \oplus W_{\text{skew}}$. For the given matrix: $A_{\text{sym}} = \begin{pmatrix} 2 & 3 & 1 \\ 3 & 4 & 2 \\ 1 & 2 & 7 \end{pmatrix}$ and $A_{\text{skew}} = \begin{pmatrix} 0 & 2 & -2 \\ -2 & 0 & 4 \\ 2 & -4 & 0 \end{pmatrix}$.
Let $W_1$ and $W_2$ be two vector subspaces of a vector space $V$ over a field $F$.
(a) Prove rigorously that the union $W_1 \cup W_2$ is a subspace of $V$ if and only if either $W_1 \subseteq W_2$ or $W_2 \subseteq W_1$. (b) Provide a concrete geometric counterexample in $\mathbb{R}^2$ demonstrating the failure of closure under vector addition when neither subspace contains the other. (c) Generalization: If $F$ is an infinite field, prove that a vector space $V$ cannot be written as the union of a finite number of proper subspaces $V = \bigcup_{k=1}^m W_k$ with $W_k \subsetneq V$.
Part (a): Proof of Subspace Union Equivalence: We wish to prove: $W_1 \cup W_2 \le V \iff W_1 \subseteq W_2 \text{ or } W_2 \subseteq W_1$.
$(\impliedby)$ If $W_1 \subseteq W_2$, then $W_1 \cup W_2 = W_2$. Since $W_2$ is a subspace, $W_1 \cup W_2$ is a subspace. The same holds if $W_2 \subseteq W_1$, where $W_1 \cup W_2 = W_1$.
$(\implies)$ Suppose $W_1 \cup W_2 \le V$. We proceed by contradiction. Assume that $W_1 \not\subseteq W_2$ and $W_2 \not\subseteq W_1$. Then:
- There exists a vector $\vec{w}_1 \in W_1$ such that $\vec{w}_1 \notin W_2$.
- There exists a vector $\vec{w}_2 \in W_2$ such that $\vec{w}_2 \notin W_1$.
Both $\vec{w}_1, \vec{w}_2 \in W_1 \cup W_2$. Since $W_1 \cup W_2$ is assumed to be a subspace, it must be closed under addition:
This implies that either $\vec{w}_1 + \vec{w}_2 \in W_1$ or $\vec{w}_1 + \vec{w}_2 \in W_2$.
- Case 1: Suppose $\vec{w}_1 + \vec{w}_2 \in W_1$. Since $W_1$ is a subspace and $\vec{w}_1 \in W_1$, $(-\vec{w}_1) \in W_1$. Then:
This contradicts our premise that $\vec{w}_2 \notin W_1$.
- Case 2: Suppose $\vec{w}_1 + \vec{w}_2 \in W_2$. Since $W_2$ is a subspace and $\vec{w}_2 \in W_2$, $(-\vec{w}_2) \in W_2$. Then:
This contradicts our premise that $\vec{w}_1 \notin W_2$.
Both cases yield a contradiction. Hence, we must have $W_1 \subseteq W_2$ or $W_2 \subseteq W_1$. $\blacksquare$
Part (b): Geometric Counterexample in $\mathbb{R}^2$: In $\mathbb{R}^2$, let $W_1 = \text{span}\{(1, 0)\} = \{(x, 0) : x \in \mathbb{R}\}$ ($x$-axis) and $W_2 = \text{span}\{(0, 1)\} = \{(0, y) : y \in \mathbb{R}\}$ ($y$-axis). Both are 1D subspaces of $\mathbb{R}^2$. $\vec{u} = (1, 0) \in W_1 \subseteq W_1 \cup W_2$ and $\vec{v} = (0, 1) \in W_2 \subseteq W_1 \cup W_2$. Their sum is $\vec{u} + \vec{v} = (1, 1)$. Since $(1, 1) \notin W_1$ and $(1, 1) \notin W_2$, $(1, 1) \notin W_1 \cup W_2$. Closure under addition fails!
Part (c): Proof for Union of Finitely Many Subspaces over Infinite Field: Let $V = \bigcup_{k=1}^m W_k$ with proper subspaces $W_k \subsetneq V$. Assume $m$ is minimal, so $m \ge 2$ and $V \ne \bigcup_{k \ne j} W_k$. Pick $\vec{u} \in W_1 \setminus \bigcup_{k=2}^m W_k$, and pick $\vec{v} \in V \setminus W_1$. Consider the infinite family of vectors $\{\vec{v} + c\vec{u} : c \in F\}$. Since $F$ is infinite, this family is infinite. If two distinct scalars $c_1 \ne c_2$ give vectors in the same subspace $W_j$:
Then their difference $(c_1 - c_2)\vec{u} \in W_j \implies \vec{u} \in W_j$ (since $c_1 - c_2 \ne 0$).
- If $j = 1$, then $\vec{v} = (\vec{v} + c_1\vec{u}) - c_1\vec{u} \in W_1$, contradicting $\vec{v} \notin W_1$.
- If $j \ge 2$, then $\vec{u} \in W_j$, contradicting $\vec{u} \notin \bigcup_{k=2}^m W_k$.
Thus, each subspace $W_j$ can contain at most ONE vector from the infinite set $\{\vec{v} + c\vec{u} : c \in F\}$. Since $m$ is finite and $F$ is infinite, the union $\bigcup_{j=1}^m W_j$ cannot cover the set, contradicting $V = \bigcup W_k$. $\blacksquare$
$W_1 \cup W_2$ is a subspace if and only if $W_1 \subseteq W_2$ or $W_2 \subseteq W_1$. Geometrically in $\mathbb{R}^2$, the union of the $x$-axis and $y$-axis contains $(1,0)$ and $(0,1)$ but not their sum $(1,1)$. Furthermore, an infinite-dimensional or finite-dimensional vector space over an infinite field cannot be covered by a finite union of proper subspaces.