8.1 Vector Spaces, Subspaces, Span, and Linear Independence

8.1 Vector Spaces, Subspaces, Span, and Linear Independence

Linear algebra on the GRE Subject Test in Mathematics transitions from computational matrix manipulation to the abstract structural properties of vector spaces. Mastery of vector space axioms, subspace verification, linear spans, independence criteria, and subspace arithmetic (sums, intersections, and direct sums) is indispensable for rapid and accurate problem-solving.

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Axiomatic Foundations of Vector Spaces

A vector space over a field $\mathbb{F}$ (typically $\mathbb{R}$ or $\mathbb{C}$) is a set $V$ equipped with vector addition $+: V \times V \to V$ and scalar multiplication $\cdot: \mathbb{F} \times V \to V$ satisfying ten fundamental axioms:

1. Additive Closure: $u + v \in V$ for all $u, v \in V$.
2. Additive Commutativity: $u + v = v + u$ for all $u, v \in V$.
3. Additive Associativity: $(u + v) + w = u + (v + w)$ for all $u, v, w \in V$.
4. Additive Identity: There exists $0_V \in V$ such that $v + 0_V = v$ for all $v \in V$.
5. Additive Inverses: For each $v \in V$, there exists $-v \in V$ such that $v + (-v) = 0_V$.
6. Multiplicative Closure: $c v \in V$ for all $c \in \mathbb{F}$ and $v \in V$.
7. Scalar Distributivity over Vector Addition: $c(u + v) = cu + cv$ for all $c \in \mathbb{F}$ and $u, v \in V$.
8. Scalar Distributivity over Field Addition: $(a + b)v = av + bv$ for all $a, b \in \mathbb{F}$ and $v \in V$.
9. Scalar Compatibility: $a(bv) = (ab)v$ for all $a, b \in \mathbb{F}$ and $v \in V$.
10. Scalar Identity: $1_\mathbb{F} \cdot v = v$ for all $v \in V$.

Direct algebraic deductions include: the uniqueness of $0_V$ and $-v$; $0_\mathbb{F} \cdot v = 0_V$; $c \cdot 0_V = 0_V$; $(-1)v = -v$; and $c v = 0_V \implies c = 0$ or $v = 0_V$.

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The Subspace Test and Standard Function Spaces

A subset $W \subseteq V$ is a subspace of $V$ (written $W \le V$) if $W$ is itself a vector space under the operations inherited from $V$.

The Two-Step Subspace Criteria

To prove $W \subseteq V$ is a subspace, verify:

1. Nonemptiness / Zero Vector: $0_V \in W$ (or $W \neq \emptyset$).
2. Closure under Linear Combinations: For all $u, v \in W$ and $c \in \mathbb{F}$, $c u + v \in W$. (Equivalently, $u + v \in W$ and $c u \in W$).

Canonical Vector Spaces on the GRE

• Euclidean / Coordinate Spaces: $\mathbb{R}^n$ and $\mathbb{C}^n$.
• Polynomial Spaces: $P_n(\mathbb{F}) = {a_0 + a_1 x + \dots + a_n x^n \mid a_i \in \mathbb{F}}$, having dimension $n+1$; and the infinite-dimensional space of all polynomials $P(\mathbb{F}) = \mathbb{F}[x]$.
• Matrix Spaces: $M_{m \times n}(\mathbb{F})$ with dimension $mn$. Subspaces include symmetric matrices ($A^T = A$), skew-symmetric matrices ($A^T = -A$), and trace-zero matrices ($\operatorname{tr}(A) = 0$).
• Function Spaces: $C[a, b]$ (continuous functions on $[a, b]$), $C^k(I)$ ($k$-times continuously differentiable functions), and $C^\infty(I)$ (smooth functions).

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Span, Linear Independence, and the Wronskian

Linear Combinations and Span

For any subset $S \subseteq V$, the span of $S$, denoted $\operatorname{span}(S)$, is the set of all finite linear combinations: $\operatorname{span}(S) = \left\{ \sum_{i=1}^k c_i v_i \;\middle|\; k \in \mathbb{N}, \, c_i \in \mathbb{F}, \, v_i \in S \right\}.$ By definition, $\operatorname{span}(\emptyset) = {0_V}$. Crucially, $\operatorname{span}(S)$ is the unique smallest subspace of $V$ containing $S$.

Linear Independence

A finite set ${v_1, \dots, v_k} \subset V$ is linearly independent if: $c_1 v_1 + c_2 v_2 + \dots + c_k v_k = 0_V \implies c_1 = c_2 = \dots = c_k = 0.$ If non-trivial scalars exist producing $0_V$, the set is linearly dependent.

• Linear Dependence Lemma: A set ${v_1, \dots, v_k}$ ($k \ge 2$) is linearly dependent if and only if some $v_j$ is a linear combination of its predecessors ${v_1, \dots, v_{j-1}}$.

The Wronskian Test for Functions

For functions $f_1, \dots, f_k \in C^{k-1}(I)$, the Wronskian is: $W(f_1, \dots, f_k)(x) = \det \begin{pmatrix} f_1(x) & f_2(x) & \dots & f_k(x) \\ f_1'(x) & f_2'(x) & \dots & f_k'(x) \\ \vdots & \vdots & \ddots & \vdots \\ f_1^{(k-1)}(x) & f_2^{(k-1)}(x) & \dots & f_k^{(k-1)}(x) \end{pmatrix}.$

• Independence Criterion: If there exists at least one $x_0 \in I$ such that $W(f_1, \dots, f_k)(x_0) \neq 0$, then ${f_1, \dots, f_k}$ is linearly independent on $I$.
• GRE Caution: The converse fails in general! Two functions can be linearly independent even if their Wronskian is identically zero everywhere on $I$ (for example, $x^2$ and $x|x|$ on $(-1, 1)$). The converse holds only if the functions are solutions to an ordinary linear homogeneous differential equation with continuous coefficients.

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Intersections, Sums, and Direct Sums

Subspace Operations

Let $U, W \le V$:

• Intersection: $U \cap W$ is always a subspace of $V$.
• Union: $U \cup W$ is a subspace of $V$ if and only if $U \subseteq W$ or $W \subseteq U$.
• Sum: $U + W = {u + w \mid u \in U, w \in W} = \operatorname{span}(U \cup W)$, which is always a subspace.

The Grassmann Dimension Formula

For finite-dimensional subspaces $U, W \le V$: $\dim(U + W) = \dim U + \dim W - \dim(U \cap W).$ Since $\dim(U + W) \le \dim V$, this provides a sharp lower bound on the intersection dimension: $\dim(U \cap W) \ge \dim U + \dim W - \dim V.$

Internal Direct Sums

The sum $V = U + W$ is an internal direct sum, denoted $V = U \oplus W$, if every vector $v \in V$ can be written uniquely as $v = u + w$ with $u \in U, w \in W$.

• Equivalence: $V = U \oplus W \iff V = U + W \text{ and } U \cap W = {0_V}$.
• Dimension Criterion: If $\dim V < \infty$, then $V = U \oplus W \iff U \cap W = {0_V}$ and $\dim U + \dim W = \dim V$.

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Subspace Operations Comparison

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Step-by-Step Worked Problems

Problem 1: Subspace Verification and Dimension

Let $W = {p(x) \in P_3(\mathbb{R}) \mid p(1) = 0 \text{ and } p'(0) = 0}$. Prove that $W$ is a subspace of $P_3(\mathbb{R})$ and compute $\dim W$.

Solution:

1. Check zero polynomial: $0(1) = 0$ and $0'(0) = 0$, so $0 \in W$.
2. Check linear combinations: Let $p, q \in W$ and $c \in \mathbb{R}$. Then $(cp + q)(1) = c p(1) + q(1) = c(0) + 0 = 0$. Likewise, $(cp + q)'(0) = c p'(0) + q'(0) = c(0) + 0 = 0$. Hence $cp + q \in W$, confirming $W \le P_3(\mathbb{R})$.
3. Representation: Let $p(x) = ax^3 + bx^2 + cx + d$.
   • $p'(0) = c = 0 \implies c = 0$.
   • $p(1) = a + b + d = 0 \implies d = -a - b$.
   • Thus $p(x) = ax^3 + bx^2 - (a + b) = a(x^3 - 1) + b(x^2 - 1)$.
4. The set ${x^3 - 1, x^2 - 1}$ spans $W$ and is linearly independent. Thus $\dim W = 2$.

Problem 2: Bounding Intersection Dimension

Let $U$ and $W$ be subspaces of $\mathbb{R}^7$ with $\dim U = 4$ and $\dim W = 5$. Find all possible values of $\dim(U \cap W)$.

Solution:

1. Apply Grassmann's formula: $\dim(U \cap W) = \dim U + \dim W - \dim(U + W) = 4 + 5 - \dim(U + W) = 9 - \dim(U + W)$.
2. Since $U + W \le \mathbb{R}^7$, we have $\dim(U + W) \le 7$. Thus $\dim(U \cap W) \ge 9 - 7 = 2$.
3. Since $U \cap W \le U$, we have $\dim(U \cap W) \le \min(\dim U, \dim W) = \min(4, 5) = 4$.
4. Both bounds and all intermediate integers can be realized by selecting standard coordinate subspaces. Thus $\dim(U \cap W) \in {2, 3, 4}$.

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GRE Exam Traps & Pitfalls

Trap 1: Assuming the Union of Subspaces is a Subspace The union $U \cup W$ is almost never a subspace. For example, the $x$-axis and $y$-axis in $\mathbb{R}^2$ are subspaces, but their union does not contain $(1, 0) + (0, 1) = (1, 1)$. Remember: $U \cup W \le V \iff U \subseteq W$ or $W \subseteq U$.

Trap 2: Wronskian Converse Fallacy If $W(f_1, \dots, f_k)(x_0) \neq 0$ for some $x_0$, independence is guaranteed. But if $W(x) \equiv 0$ everywhere, functions need not be dependent unless they solve an $n$-th order linear homogeneous ODE.

Trap 3: Direct Sum Confusions Having $\dim U + \dim W = \dim V$ is insufficient for $V = U \oplus W$; the intersection $U \cap W$ must be strictly ${0}$.