LA-2.2

Available

Subspaces

Subspaces are vector spaces contained within larger vector spaces. They inherit the same operations and satisfy the same axioms—automatically! Understanding subspaces is essential for analyzing linear systems, transformations, and more.

3-4 hours Core Level 10 Objectives

Learning Objectives

Apply the subspace criterion to verify subspaces
Compute intersection and sum of subspaces
Understand the concept of linear span
Determine if a set generates a vector space
Recognize common subspaces (null space, column space)
Prove the dimension formula for sum of subspaces
Understand direct sums and their characterization
Compute bases for intersection and sum
Recognize when union of subspaces is a subspace
Apply subspace concepts to solve systems of equations

Prerequisites

LA-2.1: Vector Space Definition
Set theory basics
Gaussian elimination (LA-1.4)
Familiarity with matrix operations
Mathematical proof techniques

Historical Context

The concept of a subspace emerged naturally from the study of linear equations. When mathematicians noticed that solution sets of homogeneous systems had special properties—closure under addition and scalar multiplication—the abstract notion of subspace crystallized.

Hermann Grassmann (1844) was among the first to recognize that certain subsets of vector spaces formed vector spaces in their own right. His work on "extensions" laid groundwork for modern subspace theory.

Today, subspaces are central to linear algebra: null spaces, column spaces, eigenspaces, and solution spaces are all subspaces. The theory of subspaces provides the language for analyzing linear systems and transformations.

1. Definition and Criterion

A subspace is a subset of a vector space that is itself a vector space under the inherited operations. The key insight: we don't need to check all eight axioms!

Definition 2.2: Subspace

A subset $W \subseteq V$ is a subspace of $V$ if $W$ is itself a vector space under the same operations of addition and scalar multiplication.

Remark 2.1: Notation

We write $W \leq V$ or $W \subseteq V$ to indicate that $W$ is a subspace of $V$ . Some texts use $W < V$ for proper subspaces (where $W \neq V$ ).

Theorem 2.2: Subspace Criterion

A non-empty subset $W \subseteq V$ is a subspace if and only if:

Closed under addition: $u, v \in W \implies u + v \in W$
Closed under scalar multiplication: $\alpha \in F, v \in W \implies \alpha v \in W$

Proof of Theorem 2.2:

(⇒) If $W$ is a subspace (a vector space), closure under both operations is immediate from the definition.

(⇐) Assume $W$ is non-empty and closed under both operations. We verify the vector space axioms:

Zero vector: Since $W$ is non-empty, pick any $w \in W$ . By closure under scalar mult, $0 \cdot w = 0 \in W$ .
Additive inverses: For $w \in W$ , we have $(-1) \cdot w = -w \in W$ by scalar mult closure.
Commutativity, associativity, distributivity: These are inherited from $V$ since $W \subseteq V$ .
Multiplicative identity: $1 \cdot w = w \in W$ automatically for $w \in W$ .

∎

Corollary 2.1: Zero Vector Test

If $W \subseteq V$ is a subspace, then $0 \in W$ . Equivalently, if $0 \notin W$ , then $W$ is NOT a subspace.

Remark 2.2: One-Step Subspace Test

A non-empty subset $W \subseteq V$ is a subspace if and only if for all $u, v \in W$ and $\alpha \in F$ :

\alpha u + v \in W

This single condition combines closure under addition and scalar multiplication. Setting $\alpha = 0$ gives $0 \in W$ .

Example 2.2: Trivial Subspaces

Every vector space $V$ has exactly two trivial subspaces:

$\{0\}$ — the zero subspace (dimension 0)
$V$ itself — the improper subspace (dimension = dim V)

All other subspaces are called proper or non-trivial.

Example 2.3: Lines Through the Origin

In $\mathbb{R}^2$ , any line through the origin is a subspace:

L = \{t(a, b) : t \in \mathbb{R}\}

Verification:

Zero: $0 \cdot (a, b) = (0, 0) \in L$ ✓
Closed under +: $t_1(a,b) + t_2(a,b) = (t_1 + t_2)(a,b) \in L$ ✓
Closed under scalar mult: $c \cdot t(a,b) = (ct)(a,b) \in L$ ✓

Note: A line NOT through the origin (like $y = x + 1$ ) is NOT a subspace.

Example 2.4: Planes Through the Origin

In $\mathbb{R}^3$ , a plane through the origin is defined by:

P = \{(x, y, z) : ax + by + cz = 0\}

This is a subspace (it's the null space of the $1 \times 3$ matrix $[a\ b\ c]$ ).

Verification:

Zero: $a(0) + b(0) + c(0) = 0$ ✓
Closed under +: If $ax_1 + by_1 + cz_1 = 0$ and $ax_2 + by_2 + cz_2 = 0$ , then $a(x_1+x_2) + b(y_1+y_2) + c(z_1+z_2) = 0$ ✓
Closed under scalar mult: If $ax + by + cz = 0$ , then $a(\alpha x) + b(\alpha y) + c(\alpha z) = \alpha \cdot 0 = 0$ ✓

Example 2.5: The Null Space

For any matrix $A \in M_{m \times n}(F)$ , the null space (kernel) is:

\ker(A) = \{x \in F^n : Ax = 0\}

This is always a subspace of $F^n$ :

Zero: $A \cdot 0 = 0$ ✓
Closed under +: If $Ax = 0$ and $Ay = 0$ , then $A(x+y) = Ax + Ay = 0$ ✓
Closed under scalar mult: If $Ax = 0$ , then $A(\alpha x) = \alpha Ax = 0$ ✓

Example 2.6: Polynomial Subspaces

Let $P_n(F)$ be polynomials of degree at most $n$ . This is a subspace of $F[x]$ :

Zero: The zero polynomial has degree $-\infty \leq n$ ✓
Closed under +: $\deg(p + q) \leq \max(\deg p, \deg q) \leq n$ ✓
Closed under scalar mult: $\deg(\alpha p) = \deg(p) \leq n$ for $\alpha \neq 0$ ✓

Dimension: $\dim P_n(F) = n + 1$ (basis: $\{1, x, x^2, \ldots, x^n\}$ ).

Example 2.7: Symmetric Matrices

The set of symmetric $n \times n$ matrices:

S_n = \{A \in M_n(\mathbb{R}) : A^T = A\}

is a subspace of $M_n(\mathbb{R})$ :

Zero: $O^T = O$ ✓
Closed under +: $(A + B)^T = A^T + B^T = A + B$ ✓
Closed under scalar mult: $(\alpha A)^T = \alpha A^T = \alpha A$ ✓

Dimension: $\frac{n(n+1)}{2}$ .

Example 2.8: Non-Subspace Examples

The following are NOT subspaces:

First quadrant in ℝ²: $\{(x, y) : x \geq 0, y \geq 0\}$
No additive inverses: $-(1, 1) = (-1, -1)$ not in set.
Unit circle: $\{(x, y) : x^2 + y^2 = 1\}$
Doesn't contain zero: $0^2 + 0^2 = 0 \neq 1$ .
Polynomials of exact degree n: Not closed under addition:
$x^2 + (-x^2 + 1) = 1$ has degree 0, not 2.
Invertible matrices: Zero matrix not invertible; $I + (-I) = 0$ not invertible.
Integers in ℝ: $\mathbb{Z} \subseteq \mathbb{R}$
Not closed under scalar mult: $\frac{1}{2} \cdot 1 = \frac{1}{2} \notin \mathbb{Z}$ .

Remark 2.3: Geometric Interpretation

In $\mathbb{R}^n$ , subspaces are:

$\{0\}$ (dimension 0) — just the origin
Lines through the origin (dimension 1)
Planes through the origin (dimension 2)
Hyperplanes through the origin (dimension n-1)
The whole space $\mathbb{R}^n$ (dimension n)

All must pass through the origin!

Example 2.9: Upper Triangular Matrices

The set of upper triangular $n \times n$ matrices:

U_n = \{A \in M_n(F) : a_{ij} = 0 \text{ for } i > j\}

is a subspace of $M_n(F)$ :

Zero: The zero matrix is upper triangular ✓
Closed under +: Sum of upper triangular is upper triangular ✓
Closed under scalar mult: Scalar multiple preserves the zero pattern ✓

Dimension: $\frac{n(n+1)}{2}$ (diagonal + upper triangle).

Example 2.10: Trace Zero Matrices

Matrices with trace zero:

\text{sl}_n(F) = \{A \in M_n(F) : \text{tr}(A) = 0\}

is a subspace (this is the Lie algebra of $SL_n(F)$ ):

Zero: tr(0) = 0 ✓
Closed under +: tr(A + B) = tr(A) + tr(B) = 0 + 0 = 0 ✓
Closed under scalar mult: tr(αA) = α·tr(A) = α·0 = 0 ✓

Dimension: $n^2 - 1$ .

Example 2.11: Diagonal Matrices

Diagonal matrices form a subspace:

D_n = \{A \in M_n(F) : a_{ij} = 0 \text{ for } i \neq j\}

Dimension: $n$ (only n diagonal entries can be non-zero).

Note: $D_n \subseteq S_n \cap U_n$ (diagonal matrices are both symmetric and upper triangular).

2. Linear Span

The span of a set of vectors is the set of all linear combinations. It's the smallest subspace containing those vectors—and every subspace can be described as a span of some generating set.

Definition 2.3: Linear Span

The linear span (or simply span) of a set $S \subseteq V$ is:

\text{span}(S) = \left\{\sum_{i=1}^{k} \alpha_i v_i : k \in \mathbb{N}, \alpha_i \in F, v_i \in S\right\}

i.e., the set of all finite linear combinations of vectors in $S$ .

Remark 2.3: Convention

By convention, $\text{span}(\emptyset) = \{0\}$ . The empty sum is the zero vector.

Theorem 2.3: Span is a Subspace

For any subset $S \subseteq V$ , $\text{span}(S)$ is a subspace of $V$ .

Proof of Theorem 2.3:

We verify the subspace criterion:

Zero: The empty sum (or $0 \cdot v$ for any $v \in S$ ) gives $0 \in \text{span}(S)$ .
Closed under +: If $u = \sum \alpha_i v_i$ and $w = \sum \beta_j v_j$ , then $u + w = \sum \alpha_i v_i + \sum \beta_j v_j$ is also a linear combination.
Closed under scalar mult: If $u = \sum \alpha_i v_i$ , then $cu = \sum (c\alpha_i) v_i \in \text{span}(S)$ .

∎

Theorem 2.4: Span is the Smallest Subspace

$\text{span}(S)$ is the smallest subspace containing $S$ . That is:

$S \subseteq \text{span}(S)$
If $W$ is any subspace with $S \subseteq W$ , then $\text{span}(S) \subseteq W$

Proof of Theorem 2.4:

(1) For any $v \in S$ , we have $v = 1 \cdot v \in \text{span}(S)$ .

(2) Let $W$ be a subspace with $S \subseteq W$ . Any linear combination $\sum \alpha_i v_i$ with $v_i \in S \subseteq W$ is in $W$ by closure. So $\text{span}(S) \subseteq W$ .

∎

Definition 2.4: Generating Set

A set $S$ generates (or spans) a vector space $V$ if $\text{span}(S) = V$ . We also say $S$ is a generating set for $V$ .

Example 2.8: Computing Spans in ℝ²

$\text{span}\{(1, 0)\} = \{(t, 0) : t \in \mathbb{R}\}$ — the x-axis (1-dimensional)
$\text{span}\{(1, 0), (0, 1)\} = \mathbb{R}^2$ — the standard basis spans all of $\mathbb{R}^2$
$\text{span}\{(1, 2), (2, 4)\} = \{(t, 2t) : t \in \mathbb{R}\}$ — a line (since $(2, 4) = 2(1, 2)$ )
$\text{span}\{(1, 1), (1, -1)\} = \mathbb{R}^2$ — these two vectors are not parallel

Example 2.9: Computing Spans in ℝ³

$\text{span}\{(1, 0, 0)\}$ — the x-axis (line)
$\text{span}\{(1, 0, 0), (0, 1, 0)\}$ — the xy-plane
$\text{span}\{(1, 1, 0), (0, 1, 1), (1, 2, 1)\}$ — since $(1, 2, 1) = (1, 1, 0) + (0, 1, 1)$ , this is a plane (2-dimensional)
$\text{span}\{(1, 0, 0), (0, 1, 0), (0, 0, 1)\} = \mathbb{R}^3$

Example 2.10: Span of Polynomials

In $\mathbb{R}[x]$ :

$\text{span}\{1, x, x^2\} = P_2(\mathbb{R})$ (polynomials of degree ≤ 2)
$\text{span}\{1, 1+x, 1+x+x^2\} = P_2(\mathbb{R})$ (same space, different generators)
$\text{span}\{x, x^2, x^3, \ldots\}$ = polynomials with no constant term

Remark 2.4: Same Span, Different Sets

Many different sets can span the same subspace. For example:

\text{span}\{(1, 0), (2, 0)\} = \text{span}\{(1, 0)\} = \text{span}\{(3, 0), (-5, 0)\}

The minimal spanning set (with no redundant vectors) is called a basis—covered in Chapter 2.4.

Theorem 2.4.5: Membership Test

A vector $v$ is in $\text{span}(S)$ if and only if adding $v$ to $S$ does not increase the dimension of the span.

Example 2.11: Testing Membership

Is $(1, 2, 3) \in \text{span}\{(1, 0, 1), (0, 1, 1)\}$ ?

We need to check if $(1, 2, 3) = a(1, 0, 1) + b(0, 1, 1)$ has a solution.

This gives: $a = 1$ , $b = 2$ , $a + b = 3$ .

Check: $1 + 2 = 3$ ✓. Yes, $(1, 2, 3)$ is in the span.

Example 2.12: Testing Non-Membership

Is $(1, 1, 1) \in \text{span}\{(1, 0, 1), (0, 1, 0)\}$ ?

Check: $(1, 1, 1) = a(1, 0, 1) + b(0, 1, 0)$ .

This gives: $a = 1$ , $b = 1$ , $a = 1$ .

But the third component: $a \cdot 1 + b \cdot 0 = 1 \neq 1$ ... wait, that's 1 = 1. Let me redo.

Third component: $1 \cdot a + 0 \cdot b = a = 1$ . So $(1, 1, 1)$ IS in the span!

Remark 2.5: Computing Spans with Matrices

To find span(S) and test membership, form a matrix with vectors in S as columns and row reduce. The pivot columns give a basis; the dimension equals the rank.

3. Intersection and Sum

Given two subspaces, we can combine them in two natural ways: intersection and sum. The intersection is always a subspace (but the union usually isn't!).

Theorem 2.5: Intersection of Subspaces

If $U, W$ are subspaces of $V$ , then $U \cap W$ is a subspace of $V$ .

Proof of Theorem 2.5:

Check the subspace criterion:

Zero: $0 \in U$ and $0 \in W$ , so $0 \in U \cap W$ .
Closed under +: If $v_1, v_2 \in U \cap W$ , then $v_1 + v_2 \in U$ (since $U$ is a subspace) and $v_1 + v_2 \in W$ (since $W$ is a subspace). So $v_1 + v_2 \in U \cap W$ .
Closed under scalar mult: If $v \in U \cap W$ , then $\alpha v \in U$ and $\alpha v \in W$ , so $\alpha v \in U \cap W$ .

∎

Corollary 2.2: Arbitrary Intersections

If $\{W_i\}_{i \in I}$ is any collection of subspaces of $V$ , then $\bigcap_{i \in I} W_i$ is also a subspace of $V$ .

Theorem 2.6: Union is Rarely a Subspace

Let $U, W$ be subspaces of $V$ . Then $U \cup W$ is a subspace if and only if $U \subseteq W$ or $W \subseteq U$ .

Proof of Theorem 2.6:

(⇐) If $U \subseteq W$ , then $U \cup W = W$ , a subspace.

(⇒) Suppose $U \cup W$ is a subspace but neither $U \subseteq W$ nor $W \subseteq U$ .

Then there exist $u \in U \setminus W$ and $w \in W \setminus U$ .

Since $U \cup W$ is a subspace, $u + w \in U \cup W$ .

Case 1: $u + w \in U$ . Then $w = (u + w) - u \in U$ . Contradiction!

Case 2: $u + w \in W$ . Then $u = (u + w) - w \in W$ . Contradiction!

∎

Definition 2.5: Sum of Subspaces

The sum of subspaces $U$ and $W$ is:

U + W = \{u + w : u \in U, w \in W\}

Theorem 2.7: Sum is a Subspace

If $U, W$ are subspaces of $V$ , then $U + W$ is a subspace of $V$ .

Proof of Theorem 2.7:

Check the subspace criterion:

Zero: $0 = 0 + 0 \in U + W$ (with $0 \in U$ and $0 \in W$ ).
Closed under +: If $u_1 + w_1, u_2 + w_2 \in U + W$ , then $(u_1 + w_1) + (u_2 + w_2) = (u_1 + u_2) + (w_1 + w_2) \in U + W$ .
Closed under scalar mult: $\alpha(u + w) = \alpha u + \alpha w \in U + W$ .

∎

Remark 2.5: Sum as Span

$U + W = \text{span}(U \cup W)$ . The sum is the smallest subspace containing both $U$ and $W$ .

Theorem 2.8: Dimension Formula

For finite-dimensional subspaces $U, W$ :

\dim(U + W) = \dim(U) + \dim(W) - \dim(U \cap W)

Proof of Theorem 2.8:

Let $\{v_1, \ldots, v_k\}$ be a basis for $U \cap W$ .

Extend to a basis $\{v_1, \ldots, v_k, u_1, \ldots, u_m\}$ for $U$ .

Extend to a basis $\{v_1, \ldots, v_k, w_1, \ldots, w_n\}$ for $W$ .

Claim: $\{v_1, \ldots, v_k, u_1, \ldots, u_m, w_1, \ldots, w_n\}$ is a basis for $U + W$ .

It spans $U + W$ since any $u + w$ is a combination of these vectors.

Linear independence can be verified by showing that if $\sum \alpha_i v_i + \sum \beta_j u_j + \sum \gamma_l w_l = 0$ , then all coefficients are zero (using the fact that $u_j \notin W$ and $w_l \notin U$ ).

Therefore: $\dim(U + W) = k + m + n = (k + m) + (k + n) - k = \dim U + \dim W - \dim(U \cap W)$ .

∎

Example 2.11: Intersection and Sum in ℝ³

Let $U = \text{span}\{(1, 0, 0), (0, 1, 0)\}$ (xy-plane) and $W = \text{span}\{(0, 1, 0), (0, 0, 1)\}$ (yz-plane).

$U \cap W = \text{span}\{(0, 1, 0)\}$ (y-axis)
$U + W = \mathbb{R}^3$
Check dimension formula: $\dim(U + W) = 2 + 2 - 1 = 3$ ✓

Example 2.12: Lines in ℝ²

Let $U = \text{span}\{(1, 1)\}$ and $W = \text{span}\{(1, -1)\}$ .

$U \cap W = \{0\}$ (different lines intersect only at origin)
$U + W = \mathbb{R}^2$
Check: $\dim(U + W) = 1 + 1 - 0 = 2$ ✓

Definition 2.6: Direct Sum

The sum $U + W$ is called a direct sum, written $U \oplus W$ , if $U \cap W = \{0\}$ .

Theorem 2.9: Characterization of Direct Sum

$U + W = U \oplus W$ if and only if every $v \in U + W$ can be written uniquely as $v = u + w$ with $u \in U, w \in W$ .

Proof of Theorem 2.9:

(⇒) Suppose $U \cap W = \{0\}$ and $v = u_1 + w_1 = u_2 + w_2$ .

Then $u_1 - u_2 = w_2 - w_1 \in U \cap W = \{0\}$ .

So $u_1 = u_2$ and $w_1 = w_2$ . Uniqueness!

(⇐) If decomposition is unique, suppose $v \in U \cap W$ .

Then $v = v + 0 = 0 + v$ are two decompositions. Uniqueness implies $v = 0$ .

∎

Corollary 2.3: Dimension of Direct Sum

If $V = U \oplus W$ , then $\dim V = \dim U + \dim W$ .

Example 3.1: Direct Sum in ℝ⁴

Let $U = \text{span}\{(1, 0, 0, 0), (0, 1, 0, 0)\}$ and $W = \text{span}\{(0, 0, 1, 0), (0, 0, 0, 1)\}$ .

Then $U \cap W = \{0\}$ (no overlap in coordinates).

So $\mathbb{R}^4 = U \oplus W$ (direct sum).

Uniqueness: $(a, b, c, d) = (a, b, 0, 0) + (0, 0, c, d)$ is the unique decomposition.

Example 3.2: Non-Direct Sum

Let $U = \text{span}\{(1, 1, 0)\}$ and $W = \text{span}\{(1, 1, 0), (0, 0, 1)\}$ .

Then $U \subseteq W$ , so $U \cap W = U \neq \{0\}$ .

This is NOT a direct sum: $(1, 1, 0) = (1, 1, 0) + 0 = 0 + (1, 1, 0)$ has multiple decompositions.

Remark 3.1: Complementary Subspaces

If $V = U \oplus W$ , we say $W$ is a complement of $U$ in $V$ . Complements are not unique! For example, in $\mathbb{R}^2$ , any line through the origin is a complement of any other non-parallel line.

Theorem 2.10: Existence of Complements

Every subspace $U$ of a finite-dimensional space $V$ has a complement: there exists $W$ such that $V = U \oplus W$ .

Proof of Theorem 2.10:

Let $\{u_1, \ldots, u_k\}$ be a basis for $U$ .

Extend to a basis $\{u_1, \ldots, u_k, w_1, \ldots, w_m\}$ for $V$ .

Let $W = \text{span}\{w_1, \ldots, w_m\}$ .

Then $V = U + W$ (spans all of $V$ ) and $U \cap W = \{0\}$ (basis is linearly independent).

∎

Example 3.3: Multiple Subspaces

The dimension formula generalizes. For three subspaces:

\dim(U + W + X) \leq \dim U + \dim W + \dim X

But the exact formula is more complex due to triple overlaps. For direct sums $U \oplus W \oplus X$ , we need pairwise intersections to be zero.

4. Worked Examples

Example 4.1: Verifying a Subspace from Equations

Problem: Is $W = \{(x, y, z) \in \mathbb{R}^3 : 2x - y + 3z = 0\}$ a subspace?

Solution: Yes! This is the null space of $[2\ {-}1\ 3]$ .

Zero: $2(0) - 0 + 3(0) = 0$ ✓
Closed under +: If $2x_1 - y_1 + 3z_1 = 0$ and $2x_2 - y_2 + 3z_2 = 0$ , then $2(x_1+x_2) - (y_1+y_2) + 3(z_1+z_2) = 0$ ✓
Closed under scalar mult: If $2x - y + 3z = 0$ , then $2(\alpha x) - \alpha y + 3(\alpha z) = \alpha(2x - y + 3z) = 0$ ✓

Dimension: 2 (one constraint in 3D).

Example 4.2: Non-Subspace: Affine Plane

Problem: Is $W = \{(x, y, z) : x + y + z = 1\}$ a subspace of $\mathbb{R}^3$ ?

Solution: No!

Quick check: $0 + 0 + 0 = 0 \neq 1$ , so $(0, 0, 0) \notin W$ .

This is an affine subspace (translated plane), not a linear subspace.

Example 4.3: Finding a Basis for a Subspace

Problem: Find a basis for $W = \{(x, y, z, w) : x + y = 0, z - w = 0\}$ .

Solution:

Set up equations: $x = -y$ and $z = w$ .

Parametrize: $(x, y, z, w) = (-y, y, w, w) = y(-1, 1, 0, 0) + w(0, 0, 1, 1)$ .

Basis: $\{(-1, 1, 0, 0), (0, 0, 1, 1)\}$ .

Dimension: 2.

Example 4.4: Computing Intersection

Problem: Find $U \cap W$ where:

$U = \{(x, y, z) : x + y = 0\}$
$W = \{(x, y, z) : y + z = 0\}$

Solution: We need both: $x + y = 0$ and $y + z = 0$ .

From first: $x = -y$ . From second: $z = -y$ .

$U \cap W = \{(-y, y, -y) : y \in \mathbb{R}\} = \text{span}\{(-1, 1, -1)\}$ .

Dimension: 1 (a line).

Example 4.5: Computing Sum

Problem: Find $U + W$ where $U = \text{span}\{(1, 0, 1)\}$ and $W = \text{span}\{(0, 1, 1)\}$ .

Solution:

$U + W = \text{span}\{(1, 0, 1), (0, 1, 1)\}$ .

Since $(1, 0, 1)$ and $(0, 1, 1)$ are linearly independent, this is a 2-dimensional subspace (a plane).

Check: $U \cap W = \{0\}$ (the vectors are not parallel), so $U + W = U \oplus W$ .

Example 4.6: Non-Subspace: Upper Triangular Non-Singular

Problem: Is the set of invertible upper triangular matrices a subspace of $M_n(\mathbb{R})$ ?

Solution: No!

The zero matrix is not invertible, so $0 \notin W$ .
Also, $I + (-I) = 0$ is not invertible, so not closed under addition.

Note: The set of ALL upper triangular matrices IS a subspace.

Example 4.7: Subspace of Functions

Problem: Show that even functions form a subspace of $\mathcal{F}(\mathbb{R}, \mathbb{R})$ .

Solution: Let $E = \{f : f(-x) = f(x) \text{ for all } x\}$ .

Zero: $0(-x) = 0 = 0(x)$ ✓
Closed under +: If $f(-x) = f(x)$ and $g(-x) = g(x)$ , then $(f+g)(-x) = f(-x) + g(-x) = f(x) + g(x) = (f+g)(x)$ ✓
Closed under scalar mult: $(\alpha f)(-x) = \alpha f(-x) = \alpha f(x) = (\alpha f)(x)$ ✓

Similarly, odd functions form a subspace $O$ , and $\mathcal{F} = E \oplus O$ !

Example 4.8: Dimension Formula Application

Problem: In $\mathbb{R}^5$ , let $\dim U = 3$ and $\dim W = 4$ . What are the possible values of $\dim(U \cap W)$ ?

Solution: By the dimension formula:

\dim(U + W) = 3 + 4 - \dim(U \cap W)

Since $U + W \subseteq \mathbb{R}^5$ , we have $\dim(U + W) \leq 5$ .

So $7 - \dim(U \cap W) \leq 5$ , meaning $\dim(U \cap W) \geq 2$ .

Also, $\dim(U \cap W) \leq \min(3, 4) = 3$ .

Possible values: 2, 3.

Example 4.9: Direct Sum Decomposition

Problem: Show that $M_2(\mathbb{R}) = S_2 \oplus A_2$ where $S_2$ is symmetric and $A_2$ is skew-symmetric matrices.

Solution:

For any matrix $A$ , we can write:

A = \frac{A + A^T}{2} + \frac{A - A^T}{2}

where $\frac{A + A^T}{2}$ is symmetric and $\frac{A - A^T}{2}$ is skew-symmetric.

Check $S_2 \cap A_2 = \{0\}$ : If $A^T = A$ and $A^T = -A$ , then $A = -A$ , so $A = 0$ .

Dimension check: $\dim S_2 = 3$ , $\dim A_2 = 1$ , $\dim M_2 = 4 = 3 + 1$ ✓

Example 4.10: Column Space and Row Space

Problem: For $A = \begin{pmatrix} 1 & 2 & 1 \\ 2 & 4 & 2 \end{pmatrix}$ , find the column space and row space.

Solution:

Column space: span of columns = $\text{span}\{(1, 2), (2, 4), (1, 2)\}$

Since all columns are multiples of $(1, 2)$ :

$\text{Col}(A) = \text{span}\{(1, 2)\}$ , dimension 1.

Row space: span of rows = $\text{span}\{(1, 2, 1), (2, 4, 2)\}$

Since $(2, 4, 2) = 2(1, 2, 1)$ :

$\text{Row}(A) = \text{span}\{(1, 2, 1)\}$ , dimension 1.

Note: dim(Col(A)) = dim(Row(A)) = rank(A). Always!

Example 4.11: Null Space Calculation

Problem: Find $\ker(A)$ for $A = \begin{pmatrix} 1 & 2 & -1 \\ 2 & 4 & -2 \end{pmatrix}$ .

Solution: Solve $Ax = 0$ .

RREF: $\begin{pmatrix} 1 & 2 & -1 \\ 0 & 0 & 0 \end{pmatrix}$

Equation: $x_1 + 2x_2 - x_3 = 0$ , so $x_1 = -2x_2 + x_3$ .

Parametrize: $(x_1, x_2, x_3) = s(-2, 1, 0) + t(1, 0, 1)$ .

$\ker(A) = \text{span}\{(-2, 1, 0), (1, 0, 1)\}$ , dimension 2.

Rank-Nullity: rank(A) + nullity(A) = 1 + 2 = 3 = number of columns ✓

Example 4.12: Complementary Subspaces

Problem: Let $U = \{(x, y, z) : x + y = 0\}$ in $\mathbb{R}^3$ . Find a subspace $W$ such that $\mathbb{R}^3 = U \oplus W$ .

Solution:

$U$ has dimension 2 (one constraint). Need $W$ with dimension 1 and $U \cap W = \{0\}$ .

Find a vector not in $U$ : $(1, 0, 0) \notin U$ since $1 + 0 \neq 0$ .

Let $W = \text{span}\{(1, 0, 0)\}$ .

Check $U \cap W = \{0\}$ : If $t(1, 0, 0) \in U$ , then $t + 0 = 0$ , so $t = 0$ . ✓

Verify: $\dim U + \dim W = 2 + 1 = 3 = \dim \mathbb{R}^3$ . ✓

Example 4.13: Span Equality Test

Problem: Is $\text{span}\{(1, 2, 3), (4, 5, 6)\} = \text{span}\{(1, 1, 1), (2, 3, 4)\}$ ?

Solution: Check if each set spans the same subspace.

Can we write $(1, 2, 3) = a(1, 1, 1) + b(2, 3, 4)$ ?

System: $a + 2b = 1$ , $a + 3b = 2$ , $a + 4b = 3$ .

From first two: $b = 1$ , $a = -1$ . Check third: $-1 + 4 = 3$ ✓

Similarly check $(4, 5, 6) = c(1, 1, 1) + d(2, 3, 4)$ : $c = -2, d = 3$ works.

Answer: Yes, the spans are equal (both are the same 2D subspace).

5. Common Mistakes

Mistake 1: Forgetting to check zero

If a subset doesn't contain $0$ , it's not a subspace. Always check this first!
Example: $\{(x, y) : x + y = 1\}$ — the origin isn't in this set.

Mistake 2: Confusing union and sum

$U \cup W$ is rarely a subspace (only when one contains the other). $U + W$ is always a subspace. Use sum, not union!

Mistake 3: Lines not through origin

In $\mathbb{R}^2$ , the line $y = 2x$ (through origin) is a subspace. But $y = 2x + 1$ is NOT — it doesn't pass through the origin.

Mistake 4: Solution sets of non-homogeneous systems

Solutions to $Ax = 0$ form a subspace (null space). Solutions to $Ax = b$ (with $b \neq 0$ ) do NOT — they form an affine subspace.

Mistake 5: Only checking addition

A set can be closed under addition but not scalar multiplication.
Example: $\mathbb{Z}^2$ in $\mathbb{R}^2$ — closed under addition, but $\frac{1}{2}(1, 1) \notin \mathbb{Z}^2$ .

Mistake 6: Assuming direct sum

$U + W$ is NOT automatically a direct sum. You must verify $U \cap W = \{0\}$ . If not, decompositions are not unique.

Mistake 7: Forgetting closure under ALL scalars

Over $\mathbb{R}$ , you need closure under all real scalars, including negatives and fractions. Over $\mathbb{C}$ , need closure under complex scalars too.

Mistake 8: Dimension calculation errors

Remember: dim(U + W) = dim(U) + dim(W) - dim(U ∩ W). Don't forget to subtract the intersection! And dim(U + W) ≤ dim(V), always.

✓ How to Check: Subspace Checklist

Given a subset $W \subseteq V$ , verify it's a subspace:

Is $W$ non-empty? (Usually check $0 \in W$ )
Is $W$ closed under addition? (If $u, v \in W$ , is $u + v \in W$ ?)
Is $W$ closed under scalar multiplication? (If $v \in W$ , is $\alpha v \in W$ ?)

Quick test: If $\alpha u + v \in W$ for all $u, v \in W$ and $\alpha \in F$ , then $W$ is a subspace.

6. Key Takeaways

Subspace Criterion

To check if $W$ is a subspace: (1) $0 \in W$ , (2) closed under +, (3) closed under scalar mult. If these hold, all axioms are automatic.

Span

span(S) = all linear combinations of vectors in S. It's the smallest subspace containing S. Every subspace is a span of some set.

Intersection vs Union

Intersection of subspaces is always a subspace. Union is almost never a subspace (only if one contains the other).

Sum of Subspaces

U + W = span(U ∪ W) is the smallest subspace containing both. Direct sum U ⊕ W requires U ∩ W = {0}.

Dimension Formula

dim(U + W) = dim U + dim W - dim(U ∩ W). Like inclusion-exclusion for counting.

Null Space

ker(A) = {x : Ax = 0} is always a subspace. This is why homogeneous systems have nice solution structure.

Summary Table

Operation	Subspace?	Notes
U ∩ W	Always	Can take arbitrary intersections
U ∪ W	Rarely	Only if one contains the other
U + W	Always	= span(U ∪ W)
span(S)	Always	Smallest subspace containing S
ker(A)	Always	Solutions to Ax = 0

Important Subspaces in Linear Algebra

For a Matrix A

Null space: ker(A) = {x : Ax = 0}
Column space: Col(A) = span of columns
Row space: Row(A) = span of rows
Left null space: ker(Aᵀ)

For a Linear Map T

Kernel: ker(T) = {v : T(v) = 0}
Image: im(T) = {T(v) : v ∈ V}
Eigenspaces: E_λ = ker(T - λI)
Invariant subspaces: T(U) ⊆ U

Remark 6.1: Looking Ahead

Subspaces are the building blocks for understanding linear maps. In the next chapters:

Linear Independence: When is a spanning set minimal?
Basis: The right balance of spanning and independence
Dimension: The invariant size of a subspace
Linear Maps: How subspaces transform

Applications of Subspaces

In Solving Systems

Solution set of Ax = 0 is a subspace
General solution = particular + null space
Rank-nullity theorem connects dimensions

In Eigenvalue Theory

Eigenspaces are subspaces
Diagonalization uses direct sums
Invariant subspaces key to structure

In Differential Equations

Solutions to linear ODEs form subspace
General solution = span of basic solutions
Dimension = order of equation

In Data Science

PCA projects onto subspaces
Feature spaces are subspaces
Low-rank approximation via subspaces

Remark 6.2: The Subspace Perspective

Thinking in terms of subspaces is powerful: instead of individual vectors, consider the spaces they generate. This shift from "elements" to "structures" is central to abstract linear algebra and leads to cleaner theorems and deeper understanding.

7. Quick Reference

Three Conditions for Subspace

Non-empty (usually: 0 ∈ W)
Closed under +: u, v ∈ W ⟹ u + v ∈ W
Closed under scalar: v ∈ W ⟹ αv ∈ W

Key Dimension Results

dim(U ∩ W) ≤ min(dim U, dim W)
dim(U + W) = dim U + dim W - dim(U ∩ W)
dim(U ⊕ W) = dim U + dim W
dim(ker A) + rank(A) = n (columns)

Common Subspaces

Null space: ker(A) = {x : Ax = 0}
Column space: Col(A) = span(columns)
Row space: Row(A) = span(rows)
Span: span(S) = all linear combos

Quick Tests

0 ∉ W ⟹ not a subspace
Homogeneous equation ⟹ subspace
Inhomogeneous ⟹ NOT subspace
Union ⟹ rarely a subspace

Subspaces in ℝ³

Dimension	Type	Example
0	Point	{(0, 0, 0)}
1	Line (through origin)	span{(1, 2, 3)}
2	Plane (through origin)	{(x,y,z) : x + y + z = 0}
3	Whole space	ℝ³

Subspaces Practice

Questions

Correct

Accuracy

Is the set

\{(x, y) \in \mathbb{R}^2 : x = 2y\}

a subspace of

\mathbb{R}^2

Easy

Not attempted

\{(x, y) \in \mathbb{R}^2 : x + y = 1\}

a subspace of

\mathbb{R}^2

Easy

Not attempted

What is

\text{span}\{(1, 0), (0, 1)\}

\mathbb{R}^2

Easy

Not attempted

U

and

W

are subspaces of

V

, is

U \cup W

always a subspace?

Medium

Not attempted

What is

\dim(U \cap W)

U, W

are planes through origin in

\mathbb{R}^3

and

U \neq W

Medium

Not attempted

What is

\text{span}\{(1,1,0), (0,1,1), (1,2,1)\}

\mathbb{R}^3

Medium

Not attempted

\dim(U) = 3

and

\dim(W) = 4

in a 5-dimensional space, what is the minimum possible

\dim(U \cap W)

Hard

Not attempted

Is the set of invertible

n \times n

matrices a subspace of

M_n(\mathbb{R})

Hard

Not attempted

\{(x, y, z) : x^2 + y^2 + z^2 = 0\}

a subspace of

\mathbb{R}^3

Medium

Not attempted

U \oplus W = V

(direct sum), what is

\dim(U \cap W)

Medium

Not attempted

Is the set of symmetric

n \times n

matrices a subspace of

M_n(\mathbb{R})

Easy

Not attempted

What is

\text{span}\{0\}

Easy

Not attempted

Frequently Asked Questions

What's the quickest way to check if a set is a subspace?

Use the one-step subspace test: W is a subspace iff for all u, v ∈ W and α ∈ F, we have αu + v ∈ W. This combines closure under addition and scalar multiplication, plus automatically includes 0 (set α = 0).

What's the difference between span and subspace?

Every span is a subspace, but not every subspace is described as a span initially. Span{S} is the smallest subspace containing S—it's the set of all linear combinations of vectors in S.

Can two different sets have the same span?

Yes! For example, span{(1,0), (2,0)} = span{(1,0)} = the x-axis in ℝ². Removing redundant (linearly dependent) vectors doesn't change the span.

How is the null space related to subspaces?

The null space (kernel) of a matrix A is {x : Ax = 0}. It's always a subspace of ℝⁿ. This is fundamental: the solution set of a homogeneous linear system is always a subspace.

What's the geometric intuition for sum of subspaces?

U + W contains all points reachable by first moving in U, then in W. If U and W are lines through the origin, U + W is either a line (if U = W) or the plane containing both lines.

What's the difference between direct sum and regular sum?

U + W is always a subspace. U ⊕ W (direct sum) requires additionally that U ∩ W = {0}. In a direct sum, every vector in U + W can be written uniquely as u + w.

Why doesn't union of subspaces work?

Union U ∪ W is rarely a subspace because if u ∈ U \ W and w ∈ W \ U, then u + w is in neither U nor W (but it's in U + W). The exception: one subspace contains the other.

How do I find a basis for a subspace defined by equations?

If W = {x : Ax = 0}, solve the homogeneous system Ax = 0 using Gaussian elimination. Write the solution in parametric form; the coefficient vectors of the free variables form a basis.

Is every subspace a null space of some matrix?

Yes! Every subspace W ⊆ ℝⁿ equals ker(A) for some matrix A. If W has dimension k, you can find (n - k) linear equations defining W, and A is the coefficient matrix.

What happens to dimension when taking intersection?

dim(U ∩ W) ≤ min(dim U, dim W). The dimension formula gives: dim(U ∩ W) = dim U + dim W - dim(U + W). The intersection can be just {0} even if both spaces are large.