LA-2.5

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Direct Sums & Quotient Spaces

Direct sums combine spaces, quotients collapse subspaces. These constructions are fundamental for understanding the structure of linear maps.

Learning Objectives

Define internal and external direct sums
Prove the characterization of direct sums via unique decomposition
Find complementary subspaces and understand their non-uniqueness
Construct quotient spaces V/W and verify well-definedness of operations
Compute dimension of quotient spaces using the dimension formula
Understand cosets geometrically as parallel affine subspaces
Apply direct sums to decompose linear maps
Use the first isomorphism theorem for vector spaces
Recognize projections as maps arising from direct sum decompositions
Connect quotients to equivalence relations and partitions

Prerequisites

Subspaces and the subspace criterion (LA-2.2)
Linear independence and spanning (LA-2.3)
Basis and dimension (LA-2.4)
Linear combinations and linear span
Basic set theory (equivalence relations)

Historical Context

Direct sums and quotient spaces emerged in the late 19th and early 20th centuries as mathematicians sought to understand the structure of vector spaces. Emmy Noether(1882-1935) was instrumental in developing the abstract algebraic approach, where quotients became a central tool. The First Isomorphism Theorem, which connects kernels and images via quotients, appeared first in group theory and was later adapted to vector spaces and modules. These constructions are now fundamental in representation theory, homological algebra, and functional analysis.

1. Direct Sums

A direct sum decomposes a vector space into "independent" pieces. Just as integers factor into primes, vector spaces can sometimes be written as direct sums of simpler subspaces. This decomposition is fundamental for understanding linear operators.

Definition 1.1: Sum of Subspaces

Let $U, W$ be subspaces of $V$ . The sum is:

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

This is the smallest subspace containing both $U$ and $W$ .

Definition 1.2: Internal Direct Sum

We say $V$ is the internal direct sum of subspaces $U$ and $W$ , written:

V = U \oplus W

if both conditions hold:

$V = U + W$ (sum spans $V$ )
$U \cap W = \{0\}$ (trivial intersection)

Remark 1.1: Intuition

Direct sum means:

Completeness: Every vector can be written as a sum from $U$ and $W$
Uniqueness: There's only one way to do this decomposition

The condition $U \cap W = \{0\}$ prevents ambiguity.

Theorem 1.1: Direct Sum Characterization

$V = U \oplus W$ if and only if every $v \in V$ can be written uniquely as:

v = u + w \quad \text{with } u \in U, w \in W

Proof of Theorem 1.1:

(⇒) Suppose $V = U \oplus W$ .

Existence: Since $U + W = V$ , every $v$ equals some $u + w$ .

Uniqueness: If $v = u_1 + w_1 = u_2 + w_2$ , then:

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

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

(⇐) If decomposition is always unique:

Sum: Every $v$ is a sum, so $U + W = V$ .

Intersection: If $x \in U \cap W$ , then $x = x + 0 = 0 + x$ are two decompositions. By uniqueness, $x = 0$ .

∎

Example 1.1: Standard Decomposition of ℝ²

In $\mathbb{R}^2$ , let:

$U = \{(x, 0) : x \in \mathbb{R}\}$ (x-axis)
$W = \{(0, y) : y \in \mathbb{R}\}$ (y-axis)

Then $\mathbb{R}^2 = U \oplus W$ because:

Every $(a, b) = (a, 0) + (0, b)$
$U \cap W = \{(0, 0)\}$

Uniqueness: $(a, b)$ has only one decomposition as x-component + y-component.

Example 1.2: Non-Direct Sum

In $\mathbb{R}^2$ , let $U = \{(x, 0)\}$ and $W = \{(x, x)\}$ (line $y = x$ ).

Then $U + W = \mathbb{R}^2$ , but is this direct?

Check intersection: $(x, 0) = (y, y)$ requires $x = y$ and $0 = y$ , so $y = 0$ .

$U \cap W = \{(0, 0)\}$ . Yes, it's a direct sum!

Definition 1.3: Complement

A subspace $W$ is a complement of $U$ in $V$ if $V = U \oplus W$ .

We also say $U$ and $W$ are complementary subspaces.

Theorem 1.2: Existence of Complements

Every subspace $U$ of a finite-dimensional space $V$ has a complement.

Proof of Theorem 1.2:

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

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

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

Claim: $V = U \oplus W$ .

Sum: Any $v$ is a combination of all basis vectors, so $v = u + w$ with $u \in U$ , $w \in W$ .

Intersection: If $v \in U \cap W$ , then $v$ is a combination of $u_i$ 's and of $w_j$ 's. By independence of the full basis, all coefficients are 0.

∎

Remark 1.2: Non-Uniqueness of Complements

Warning: Complements are NOT unique! In $\mathbb{R}^2$ , the x-axis has infinitely many complements: any non-horizontal line through the origin.

Example 1.3: Multiple Complements

In $\mathbb{R}^2$ , let $U = \{(x, 0)\}$ . Complements of $U$ include:

$W_1 = \{(0, y)\}$ (y-axis)
$W_2 = \{(t, t)\}$ (line $y = x$ )
$W_3 = \{(t, 2t)\}$ (line $y = 2x$ )
Any line through origin except the x-axis itself

Theorem 1.3: Dimension of Direct Sum

If $V = U \oplus W$ , then:

\dim(V) = \dim(U) + \dim(W)

Proof of Theorem 1.3:

By the dimension formula for sums:

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

Since $U \cap W = \{0\}$ , we have $\dim(U \cap W) = 0$ .

Thus $\dim(V) = \dim(U) + \dim(W)$ .

∎

Definition 1.4: External Direct Sum

Given vector spaces $V_1, V_2$ over the same field (not necessarily subspaces of a common space), the external direct sum is:

V_1 \oplus V_2 = \{(v_1, v_2) : v_1 \in V_1, v_2 \in V_2\}

with operations $(v_1, v_2) + (u_1, u_2) = (v_1 + u_1, v_2 + u_2)$ and $\alpha(v_1, v_2) = (\alpha v_1, \alpha v_2)$ .

Remark 1.3: Internal vs External

Internal: Subspaces of the same space, combined to give the whole space.
External: Separate spaces, combined into a larger product space.

When $V = U \oplus W$ (internal), the map $(u, w) \mapsto u + w$ is an isomorphism $U \oplus W \to V$ (external to internal).

Example 1.4: External Direct Sum

$\mathbb{R}^2 \oplus \mathbb{R}^3$ consists of pairs $((x, y), (a, b, c))$ .

Dimension: $\dim(\mathbb{R}^2 \oplus \mathbb{R}^3) = 2 + 3 = 5$ .

This is isomorphic to $\mathbb{R}^5$ via $((x, y), (a, b, c)) \mapsto (x, y, a, b, c)$ .

Definition 1.5: Projection

If $V = U \oplus W$ , the projection onto U along W is:

\pi_U: V \to U, \quad \pi_U(u + w) = u

where each $v \in V$ is uniquely written as $v = u + w$ .

Theorem 1.4: Properties of Projections

If $\pi$ is the projection onto $U$ along $W$ where $V = U \oplus W$ :

$\pi$ is linear
$\pi^2 = \pi$ (idempotent)
$\ker(\pi) = W$
$\text{im}(\pi) = U$
$I - \pi$ is the projection onto $W$ along $U$

Example 1.5: Projection in ℝ²

In $\mathbb{R}^2 = (\text{x-axis}) \oplus (\text{y-axis})$ :

Projection onto x-axis: $\pi_x(a, b) = (a, 0)$

Projection onto y-axis: $\pi_y(a, b) = (0, b)$

Note: $\pi_x + \pi_y = I$ and $\pi_x \circ \pi_y = 0$ .

Definition 1.6: Multiple Direct Sums

More generally, $V = U_1 \oplus U_2 \oplus \cdots \oplus U_k$ means:

$V = U_1 + U_2 + \cdots + U_k$
For each $i$ : $U_i \cap (U_1 + \cdots + U_{i-1} + U_{i+1} + \cdots + U_k) = \{0\}$

Equivalently: every $v$ has a unique representation $v = u_1 + \cdots + u_k$ .

Theorem 1.5: Multiple Direct Sum Dimension

If $V = U_1 \oplus \cdots \oplus U_k$ , then:

\dim(V) = \dim(U_1) + \cdots + \dim(U_k)

Example 1.6: Direct Sum of Eigenspaces

Let $T: \mathbb{R}^3 \to \mathbb{R}^3$ have eigenvalues $\lambda_1 = 1, \lambda_2 = 2, \lambda_3 = 3$ with 1-dimensional eigenspaces.

Then: $\mathbb{R}^3 = E_1 \oplus E_2 \oplus E_3$

This decomposition is the key to diagonalization!

Corollary 1.1: Projection Decomposition

If $V = U_1 \oplus \cdots \oplus U_k$ with projections $\pi_i$ onto $U_i$ :

$\pi_1 + \cdots + \pi_k = I$ (identity)
$\pi_i \circ \pi_j = 0$ for $i \neq j$
$\pi_i^2 = \pi_i$ (idempotent)

Remark 1.4: Direct Sum and Basis

If $V = U_1 \oplus \cdots \oplus U_k$ and $\mathcal{B}_i$ is a basis for $U_i$ , then:

\mathcal{B} = \mathcal{B}_1 \cup \mathcal{B}_2 \cup \cdots \cup \mathcal{B}_k

is a basis for $V$ . The bases "add up" nicely.

Example 1.7: Symmetric and Skew-Symmetric Decomposition

For $M_n(\mathbb{R})$ (n×n real matrices):

M_n = \text{Sym}_n \oplus \text{Skew}_n

Any matrix $A$ uniquely decomposes as:

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

symmetric part + skew-symmetric part.

Dimension check: $n^2 = \frac{n(n+1)}{2} + \frac{n(n-1)}{2}$ ✓

2. Quotient Spaces

A quotient space "collapses" a subspace to a point, treating vectors that differ by an element of the subspace as equivalent. This construction is fundamental for the First Isomorphism Theorem and for understanding the structure of linear maps.

Definition 2.1: Coset

Let $W$ be a subspace of $V$ and $v \in V$ . The coset of $v$ modulo $W$ is:

v + W = \{v + w : w \in W\}

This is a "shifted copy" of $W$ passing through $v$ .

Remark 2.1: Geometric Intuition

If $W$ is a line through the origin, cosets are parallel lines:

The coset $0 + W = W$ is the original line
The coset $v + W$ is the line parallel to $W$ passing through $v$

All points on the same parallel line are "equivalent" in the quotient.

Definition 2.2: Equivalence Relation

Define $v \sim u$ if and only if $v - u \in W$ .

This is an equivalence relation:

Reflexive: $v - v = 0 \in W$
Symmetric: $v - u \in W \implies u - v = -(v - u) \in W$
Transitive: $v - u \in W$ and $u - w \in W \implies v - w = (v - u) + (u - w) \in W$

Definition 2.3: Quotient Space

The quotient space $V/W$ is the set of all cosets:

V/W = \{v + W : v \in V\}

with vector space operations:

(v + W) + (u + W) = (v + u) + W

\alpha(v + W) = (\alpha v) + W

Theorem 2.1: Well-Definedness

The operations on $V/W$ are well-defined: they don't depend on the choice of coset representatives.

Proof of Theorem 2.1:

Suppose $v + W = v' + W$ and $u + W = u' + W$ .

Then $v - v' \in W$ and $u - u' \in W$ .

Addition: $(v + u) - (v' + u') = (v - v') + (u - u') \in W$ .

So $(v + u) + W = (v' + u') + W$ . ✓

Scalar multiplication: $\alpha v - \alpha v' = \alpha(v - v') \in W$ .

So $\alpha v + W = \alpha v' + W$ . ✓

∎

Theorem 2.2: Quotient is a Vector Space

$V/W$ is a vector space over $F$ with:

Zero element: $0 + W = W$
Additive inverse: $-(v + W) = (-v) + W$

Theorem 2.3: Quotient Space Dimension

If $V$ is finite-dimensional:

\dim(V/W) = \dim(V) - \dim(W)

Proof of Theorem 2.3:

Let $\{w_1, \ldots, w_k\}$ be a basis of $W$ .

Extend to a basis $\{w_1, \ldots, w_k, v_1, \ldots, v_m\}$ of $V$ .

Claim: $\{v_1 + W, \ldots, v_m + W\}$ is a basis of $V/W$ .

Spanning: Any $v = \sum \alpha_i w_i + \sum \beta_j v_j$ . Then:

v + W = \sum \beta_j v_j + W = \sum \beta_j (v_j + W)

Independence: If $\sum \beta_j (v_j + W) = W$ , then $\sum \beta_j v_j \in W$ .

So $\sum \beta_j v_j = \sum \alpha_i w_i$ for some $\alpha_i$ .

By independence of the full basis, all $\beta_j = 0$ .

Thus $\dim(V/W) = m = n - k = \dim(V) - \dim(W)$ .

∎

Example 2.1: ℝ² / Line

Let $W = \{(t, 0) : t \in \mathbb{R}\}$ (x-axis) in $\mathbb{R}^2$ .

Cosets: $(a, b) + W = \{(a + t, b) : t \in \mathbb{R}\}$ = horizontal line at height $b$ .

Key insight: Two points are equivalent iff they have the same y-coordinate.

Dimension: $\dim(\mathbb{R}^2/W) = 2 - 1 = 1$ .

The quotient $\mathbb{R}^2/W \cong \mathbb{R}$ via $(a, b) + W \mapsto b$ .

Example 2.2: ℝ³ / Plane

Let $W = \{(x, y, 0) : x, y \in \mathbb{R}\}$ (xy-plane).

Cosets are horizontal planes at different heights:

(a, b, c) + W = \{(x, y, c) : x, y \in \mathbb{R}\}

Points are equivalent iff they have the same z-coordinate.

$\mathbb{R}^3/W \cong \mathbb{R}$ (1-dimensional).

Example 2.3: ℝ³ / Line

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

Cosets are lines parallel to the x-axis:

(a, b, c) + W = \{(t, b, c) : t \in \mathbb{R}\}

Points are equivalent iff they have the same y and z coordinates.

$\mathbb{R}^3/W \cong \mathbb{R}^2$ (2-dimensional).

Definition 2.4: Canonical Projection

The canonical projection (or quotient map) is:

\pi: V \to V/W, \quad \pi(v) = v + W

This map is linear, surjective, with $\ker(\pi) = W$ .

Theorem 2.4: First Isomorphism Theorem

If $T: V \to U$ is a linear map, then:

V / \ker(T) \cong \text{im}(T)

The isomorphism is $\bar{T}: v + \ker(T) \mapsto T(v)$ .

Proof of Theorem 2.4:

Well-defined: If $v + \ker(T) = v' + \ker(T)$ , then $v - v' \in \ker(T)$ .

So $T(v) - T(v') = T(v - v') = 0$ , hence $T(v) = T(v')$ . ✓

Linear: $\bar{T}((v + W) + (u + W)) = \bar{T}((v + u) + W) = T(v + u) = T(v) + T(u)$ . ✓

Injective: If $\bar{T}(v + W) = 0$ , then $T(v) = 0$ , so $v \in \ker(T)$ .

Thus $v + W = W$ , the zero in $V/\ker(T)$ . ✓

Surjective: For any $T(v) \in \text{im}(T)$ , we have $\bar{T}(v + W) = T(v)$ . ✓

∎

Remark 2.2: Significance of First Isomorphism Theorem

This theorem says:

Every linear map "factors" through a quotient by its kernel
The dimension of the image equals dim(V) - dim(ker T) (rank-nullity!)
The quotient "measures" what's left after collapsing the kernel

Example 2.4: First Isomorphism Theorem Application

Let $T: \mathbb{R}^3 \to \mathbb{R}^2$ be $T(x, y, z) = (x + y, y + z)$ .

Kernel: $x + y = 0$ and $y + z = 0$ , so $\ker(T) = \{t(-1, 1, -1) : t \in \mathbb{R}\}$ .

Image: $T$ is surjective onto $\mathbb{R}^2$ (check!).

By First Isomorphism Theorem: $\mathbb{R}^3/\ker(T) \cong \mathbb{R}^2$ .

Dimension check: $3 - 1 = 2$ . ✓

Theorem 2.5: Quotient and Complement

If $W$ is a subspace of $V$ and $U$ is any complement of $W$ , then:

V/W \cong U

Proof of Theorem 2.5:

Define $\phi: U \to V/W$ by $\phi(u) = u + W$ .

Linear: Clear from definition.

Injective: If $u + W = W$ , then $u \in W$ . But $u \in U$ and $U \cap W = \{0\}$ , so $u = 0$ .

Surjective: Any coset $v + W$ can be written $v = u + w$ where $u \in U$ , $w \in W$ .

Then $v + W = u + w + W = u + W = \phi(u)$ .

∎

Remark 2.3: Connection: Direct Sums and Quotients

Direct sums and quotients are "dual" constructions:

Direct sum: If $V = U \oplus W$ , then $V/W \cong U$
Quotient gives projection: The canonical projection $\pi: V \to V/W$ has kernel $W$
Complement gives section: The inclusion $U \hookrightarrow V$ composed with $\pi$ gives isomorphism $U \to V/W$

Example 2.5: Quotient in Polynomial Space

Consider $P_3(\mathbb{R})$ modulo $W = \{p : p(0) = 0\}$ (polynomials with zero constant term).

$W = \text{span}\{x, x^2, x^3\}$ , so $\dim(W) = 3$ .

$\dim(P_3/W) = 4 - 3 = 1$ .

Two polynomials $p, q$ are equivalent iff $p(0) = q(0)$ .

The quotient $P_3/W \cong \mathbb{R}$ via $p + W \mapsto p(0)$ .

Example 2.6: Quotient by Solution Space

Let $W = \{(x, y, z) : x + y + z = 0\}$ (a plane in $\mathbb{R}^3$ ).

$\dim(W) = 2$ , so $\dim(\mathbb{R}^3/W) = 1$ .

Points are equivalent iff they differ by a vector in $W$ .

Equivalently: $(a, b, c) \sim (a', b', c')$ iff $a + b + c = a' + b' + c'$ .

So $\mathbb{R}^3/W \cong \mathbb{R}$ via $(a, b, c) + W \mapsto a + b + c$ .

3. Worked Examples

Example 3.1: Verifying a Direct Sum

Problem: Is $\mathbb{R}^3 = U \oplus W$ where $U = \{(x, x, 0)\}$ and $W = \{(0, y, z)\}$ ?

Solution:

Sum: Any $(a, b, c) = (a, a, 0) + (0, b - a, c)$ ? Yes!

Intersection: $(x, x, 0) = (0, y, z)$ requires $x = 0, x = y, 0 = z$ .

So $x = y = z = 0$ , giving $U \cap W = \{0\}$ .

Answer: Yes, $\mathbb{R}^3 = U \oplus W$ .

Dimension check: $\dim(U) = 1$ , $\dim(W) = 2$ , $1 + 2 = 3$ . ✓

Example 3.2: Finding a Complement

Problem: Find a complement of $U = \text{span}\{(1, 1, 1)\}$ in $\mathbb{R}^3$ .

Solution: Need $W$ with $\dim(W) = 3 - 1 = 2$ and $U \cap W = \{0\}$ .

Extend $(1, 1, 1)$ to a basis. Add $(1, 0, 0)$ (independent).

Add $(0, 1, 0)$ (independent from both).

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

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

So $a = b = c = 0$ . $U \cap W = \{0\}$ . ✓

Example 3.3: Computing in Quotient Space

Problem: In $\mathbb{R}^3/W$ where $W = \{(t, t, 0)\}$ , compute $[(1, 2, 3)] + [(0, 1, 1)]$ .

Solution:

[(1, 2, 3)] + [(0, 1, 1)] = [(1, 2, 3) + (0, 1, 1)] = [(1, 3, 4)]

What does $[(1, 3, 4)]$ represent? All vectors differing from $(1, 3, 4)$ by an element of $W$ :

[(1, 3, 4)] = \{(1 + t, 3 + t, 4) : t \in \mathbb{R}\}

Example 3.4: Quotient by Kernel

Problem: Let $T: P_2 \to P_1$ be $T(p) = p'(x)$ (derivative).

Describe $P_2/\ker(T)$ .

Solution:

Kernel: $p'(x) = 0$ means $p$ is constant. So $\ker(T) = P_0$ (constants).

Dimension: $\dim(P_2/\ker(T)) = 3 - 1 = 2$ .

Image: $\text{im}(T) = P_1$ (all derivatives of quadratics are linear).

By First Isomorphism Theorem: $P_2/P_0 \cong P_1$ .

Interpretation: Two polynomials are equivalent iff they differ by a constant.

Example 3.5: Multiple Direct Sums

Problem: Verify $\mathbb{R}^3 = X \oplus Y \oplus Z$ where $X$ , $Y$ , $Z$ are the coordinate axes.

Solution:

$X = \{(a, 0, 0)\}$ , $Y = \{(0, b, 0)\}$ , $Z = \{(0, 0, c)\}$
Sum: $(a, b, c) = (a, 0, 0) + (0, b, 0) + (0, 0, c)$ ✓
Intersections: $X \cap Y = X \cap Z = Y \cap Z = \{0\}$ ✓
Also: $X \cap (Y + Z) = \{0\}$ ✓

Dimension check: $1 + 1 + 1 = 3$ . ✓

Example 3.6: Projection Matrix

Problem: Find the projection matrix onto $U = \text{span}\{(1, 1)\}$ along $W = \text{span}\{(1, -1)\}$ .

Solution: Any $(a, b) = \alpha(1, 1) + \beta(1, -1)$ .

Solving: $\alpha = \frac{a + b}{2}$ , $\beta = \frac{a - b}{2}$ .

Projection onto $U$ : $\pi(a, b) = \alpha(1, 1) = \frac{a + b}{2}(1, 1)$ .

P = \frac{1}{2}\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}

Check: $P^2 = P$ ✓

Example 3.7: Basis of Quotient Space

Problem: Find a basis for $\mathbb{R}^4/W$ where $W = \text{span}\{(1, 0, 1, 0), (0, 1, 0, 1)\}$ .

Solution:

$\dim(\mathbb{R}^4/W) = 4 - 2 = 2$ .

Extend W's basis: add $e_1 = (1, 0, 0, 0)$ and $e_2 = (0, 1, 0, 0)$ .

Basis of quotient: $\{e_1 + W, e_2 + W\}$ .

Any $v = ae_1 + be_2 + ce_3 + de_4$ has $v + W = a(e_1 + W) + b(e_2 + W)$ .

Example 3.8: Non-Direct Sum

Problem: Show $\mathbb{R}^2 = U + W$ but $\mathbb{R}^2 \neq U \oplus W$ where $U = W = \{(x, x)\}$ .

Solution:

Sum: $U + W = \{(x, x) + (y, y)\} = \{(z, z)\} = U \neq \mathbb{R}^2$ .

Actually, $U + W = U$ , so it doesn't even span!

Also $U \cap W = U \neq \{0\}$ .

Conclusion: Not a direct sum (fails both conditions).

Example 3.9: Quotient of Matrices

Problem: Describe $M_2(\mathbb{R}) / \text{Sym}_2$ .

Solution:

$\dim(M_2) = 4$ , $\dim(\text{Sym}_2) = 3$ .

$\dim(M_2/\text{Sym}_2) = 4 - 3 = 1$ .

Two matrices are equivalent iff they have the same skew-symmetric part.

Isomorphism: $M_2/\text{Sym}_2 \cong \text{Skew}_2 \cong \mathbb{R}$ .

Example 3.10: Double Quotient

Problem: Let $W_1 \subseteq W_2 \subseteq V$ . What is $(V/W_1)/(W_2/W_1)$ ?

Solution: By the Third Isomorphism Theorem:

(V/W_1)/(W_2/W_1) \cong V/W_2

Example: $V = \mathbb{R}^4$ , $W_1 = \text{span}\{e_1\}$ , $W_2 = \text{span}\{e_1, e_2\}$ .

$(\mathbb{R}^4/\mathbb{R})/(\mathbb{R}^2/\mathbb{R}) \cong \mathbb{R}^3/\mathbb{R} \cong \mathbb{R}^2$ .

Example 3.11: Direct Sum in Function Spaces

Problem: Show that continuous functions decompose as even + odd.

Solution: Define:

$E = \{f : f(-x) = f(x)\}$ (even functions)
$O = \{f : f(-x) = -f(x)\}$ (odd functions)

Any $f$ decomposes as:

f(x) = \frac{f(x) + f(-x)}{2} + \frac{f(x) - f(-x)}{2}

The first part is even, the second is odd.

Intersection: If $f$ is both even and odd, then $f(x) = f(-x) = -f(x)$ , so $f = 0$ .

Example 3.12: Using Dimension to Prove Direct Sum

Problem: In $\mathbb{R}^4$ , let $U = \text{span}\{(1,0,1,0), (0,1,0,1)\}$ and $W = \text{span}\{(1,1,0,0), (0,0,1,1)\}$ . Is $\mathbb{R}^4 = U \oplus W$ ?

Solution:

$\dim(U) = 2$ , $\dim(W) = 2$ .

If $U \cap W = \{0\}$ , then $\dim(U + W) = 2 + 2 = 4 = \dim(\mathbb{R}^4)$ .

Check intersection: Solve $a(1,0,1,0) + b(0,1,0,1) = c(1,1,0,0) + d(0,0,1,1)$ :

$a = c$ , $b = c$ , $a = d$ , $b = d$
So $a = b = c = d$ . Check: $a(1,0,1,0) + a(0,1,0,1) = (a,a,a,a)$ and $a(1,1,0,0) + a(0,0,1,1) = (a,a,a,a)$ ✓

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

Answer: No, $\mathbb{R}^4 \neq U \oplus W$ .

4. Common Mistakes

Mistake 1: Thinking sum equals direct sum

$U + W$ is NOT the same as $U \oplus W$ ! The sum always exists, but direct sum requires $U \cap W = \{0\}$ .

Mistake 2: Confusing cosets with subspaces

A coset $v + W$ (with $v \neq 0$ ) is NOT a subspace! It doesn't contain 0 (unless $v \in W$ ). Cosets are "shifted" subspaces.

Mistake 3: Assuming complement is unique

Complements are NOT unique! In $\mathbb{R}^2$ , any line (except the x-axis) complements the x-axis. There are infinitely many choices.

Mistake 4: Wrong quotient dimension

$\dim(V/W) = \dim(V) - \dim(W)$ , NOT $\dim(W)$ ! The quotient "removes" the dimension of $W$ .

Mistake 5: Forgetting well-definedness check

When defining operations on quotients, you must verify the result doesn't depend on the choice of representative. This is essential!

Mistake 6: Confusing V/W with a complement

$V/W$ is NOT a subspace of $V$ ! It's a different vector space whose elements are cosets. However, $V/W \cong U$ for any complement $U$ .

Mistake 7: Assuming U ⊕ W implies W ⊕ U

Direct sum IS commutative: $U \oplus W = W \oplus U$ (both equal $V$ ). But be careful with external direct sum ordering when identifying with $F^n$ .

Mistake 8: Not checking pairwise intersections for multiple sums

For $U \oplus V \oplus W$ , it's NOT enough that $U \cap V = U \cap W = V \cap W = \{0\}$ . You also need $U \cap (V + W) = \{0\}$ , etc.

✓ Verification Checklist

✓ For Direct Sum V = U ⊕ W:

Check U + W = V (every v = u + w)
Check U ∩ W = {0} (only overlap is zero)
Use dimension: dim(U) + dim(W) = dim(V)

✓ For Quotient V/W:

Identify W (what gets "collapsed")
Understand cosets as parallel copies of W
Check dim(V/W) = dim(V) - dim(W)

✓ For Well-Definedness:

If [v] = [v'], does the result depend only on [v]?
Show different representatives give same answer

5. Key Takeaways

Direct Sum = Unique Decomposition

$V = U \oplus W$ means every vector has a unique representation as $u + w$ .

Every Subspace Has Complements

But complements are not unique—there are typically infinitely many.

Quotient = Collapsing W

$V/W$ treats vectors differing by elements of $W$ as identical.

First Isomorphism Theorem

$V/\ker(T) \cong \text{im}(T)$ —the quotient "measures" the image.

Dimension Formulas

$\dim(U \oplus W) = \dim U + \dim W$ , $\dim(V/W) = \dim V - \dim W$

Projections from Direct Sums

$V = U \oplus W$ gives projections $\pi_U$ , $\pi_W$ with $\pi_U + \pi_W = I$ .

Quick Reference

Concept	Notation	Key Property
Sum of subspaces	U + W	Smallest subspace containing both
Direct sum	U ⊕ W	Sum with U ∩ W = {0}
Coset	v + W	Shifted copy of W
Quotient space	V/W	Set of all cosets
Canonical projection	π: V → V/W	π(v) = v + W, ker(π) = W

6. Applications

Eigenspace Decomposition

A diagonalizable operator decomposes $V$ as a direct sum of eigenspaces: $V = E_{\lambda_1} \oplus \cdots \oplus E_{\lambda_k}$ .

Rank-Nullity Theorem

The First Isomorphism Theorem gives: $\dim V = \dim(\ker T) + \dim(\text{im } T)$ .

Signal Processing

Signals decompose into frequency components: direct sum of eigenspaces of shift operators (Fourier decomposition).

Homological Algebra

Quotients define cohomology groups: $H^n = \ker(d^n)/\text{im}(d^{n-1})$ .

Example 6.1: Jordan Decomposition Preview

Every linear operator on a complex vector space decomposes as:

T = D + N

where $D$ is diagonalizable, $N$ is nilpotent, and $DN = ND$ .

This requires understanding generalized eigenspaces and their direct sum structure.

Example 6.2: Quotient in Function Spaces

Consider continuous functions $C[0, 1]$ modulo constants:

C[0, 1] / \mathbb{R}

Two functions are equivalent iff they differ by a constant.

This is useful when we only care about "shape" of functions, not vertical shift.

Remark 6.1: Why These Constructions Matter

Direct sums and quotients are the two fundamental ways to build/decompose vector spaces:

Direct sum: Combines independent pieces (like Cartesian product)
Quotient: Collapses redundancy (like modular arithmetic)

Together, they form the foundation for structural theorems in linear algebra.

Example 6.3: Primary Decomposition

If $T$ has characteristic polynomial $p(x) = (x - \lambda_1)^{n_1} \cdots (x - \lambda_k)^{n_k}$ , then:

V = \ker(T - \lambda_1 I)^{n_1} \oplus \cdots \oplus \ker(T - \lambda_k I)^{n_k}

These are the generalized eigenspaces. This decomposition is key to Jordan normal form.

Example 6.4: Quotient in Affine Geometry

Affine space can be viewed as a quotient:

\mathbb{A}^n \cong \mathbb{R}^{n+1} / \text{(scaling)}

Points in projective space are lines through origin in $\mathbb{R}^{n+1}$ .

Homogeneous coordinates: $[x_0 : x_1 : \cdots : x_n]$ where $[\lambda x_0 : \cdots : \lambda x_n] = [x_0 : \cdots : x_n]$ .

Applications Summary Table

Area	Construction	Application
Spectral Theory	Direct sum of eigenspaces	Diagonalization
Linear Maps	V/ker(T)	Rank-nullity theorem
Signal Processing	Frequency decomposition	Fourier analysis
Matrix Theory	Symmetric ⊕ Skew	Matrix classification
Topology	Cohomology = ker/im	Invariants of spaces

7. Chapter Summary

Direct Sums

$V = U \oplus W$ ⟺ $V = U + W$ and $U \cap W = \{0\}$
Equivalent to unique decomposition: $v = u + w$
Dimension adds: $\dim(U \oplus W) = \dim U + \dim W$
Every subspace has complements (not unique)
Direct sums give projections: $\pi_U + \pi_W = I$

Quotient Spaces

$V/W$ = cosets $v + W$ , where $v \sim u$ iff $v - u \in W$
Operations well-defined: $[v] + [u] = [v + u]$ , $\alpha[v] = [\alpha v]$
Dimension subtracts: $\dim(V/W) = \dim V - \dim W$
First Isomorphism: $V/\ker T \cong \text{im } T$

Key Connections

$V/W \cong U$ for any complement $U$ of $W$
Rank-nullity follows from First Isomorphism Theorem
Projections ↔ direct sum decompositions

Remark 7.1: Completing Part II: Vector Spaces

This concludes Part II. We've built the foundation: vector spaces, subspaces, linear independence, basis, dimension, direct sums, and quotients. In Part III: Linear Mappings, we'll study functions between vector spaces that preserve the structure—these connect everything together.

Part II: Vector Spaces Complete!

You've learned:

Vector Space Definition: Axioms and fundamental examples
Subspaces: Subspace criterion, span, sum and intersection
Linear Independence: Definition, tests, Steinitz exchange
Basis & Dimension: Coordinates, isomorphism to Fⁿ
Direct Sums & Quotients: Decomposition and collapsing

Next: Part III - Linear Mappings will show how these structures interact!

8. Quick Reference

Checking Direct Sum

Verify $U + W = V$ (spans)
Verify $U \cap W = \{0\}$
OR: dim(U) + dim(W) = dim(V) and spans

Finding a Complement

Take basis of W
Extend to basis of V
Span of new vectors = complement

Quotient Operations

• $[v] + [u] = [v + u]$
• $\alpha[v] = [\alpha v]$
• Zero: $[0] = W$
• $[v] = [u]$ iff $v - u \in W$

Key Dimension Formulas

• $\dim(U \oplus W) = \dim U + \dim W$
• $\dim(V/W) = \dim V - \dim W$
• $\dim(U + W) = \dim U + \dim W - \dim(U \cap W)$

Algorithm: Computing Projection

Input: Direct sum $V = U \oplus W$ , vector $v \in V$

Choose bases $\mathcal{B}_U$ and $\mathcal{B}_W$
Express $v$ in combined basis: $v = u + w$
Output: $\pi_U(v) = u$ , $\pi_W(v) = w$

Algorithm: Basis of Quotient Space

Input: Subspace $W$ of $V$

Find basis $\{w_1, \ldots, w_k\}$ of $W$
Extend to basis $\{w_1, \ldots, w_k, v_1, \ldots, v_m\}$ of $V$
Output: $\{v_1 + W, \ldots, v_m + W\}$ is a basis of $V/W$

Remark 8.1: What Comes Next

In Part III, you'll learn:

Linear maps: Structure-preserving functions between vector spaces
Kernel and image: Measuring injectivity and surjectivity
Rank-nullity theorem: The fundamental dimension equation
Isomorphisms: When two spaces are "the same"
Matrix representation: Coordinates for linear maps

Conceptual Summary: Building vs Collapsing

Think of vector spaces like LEGO structures:

Direct Sum (Building)

Combines independent pieces
Like stacking LEGO blocks
Total = sum of parts
Each piece is recoverable (projection)

Quotient (Collapsing)

Identifies points that differ by W
Like crushing a dimension flat
Total = original minus collapsed
Loses information (only equivalence classes remain)

Summary of Key Results

Theorem	Statement
Direct Sum Characterization	$V = U \oplus W$ ⟺ unique decomposition
Existence of Complements	Every subspace has a complement
Direct Sum Dimension	$\dim(U \oplus W) = \dim U + \dim W$
Quotient Dimension	$\dim(V/W) = \dim V - \dim W$
First Isomorphism	$V/\ker T \cong \text{im } T$
Quotient ≅ Complement	$V/W \cong U$ for any complement $U$

Direct Sums & Quotients Practice

Questions

Correct

Accuracy

V = U \oplus W

, what is

U \cap W

Easy

Not attempted

\dim(V) = 5

and

\dim(W) = 2

where

W \subseteq V

, what is

\dim(V/W)

Easy

Not attempted

\mathbb{R}^3

, let

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

. What does

V/W

look like?

Medium

Not attempted

U

and

W

are subspaces with

V = U + W

and

\dim(U \cap W) = 0

, then:

Medium

Not attempted

Does every subspace have a complement?

Medium

Not attempted

What is

[v] + [u]

V/W

Easy

Not attempted

V = U \oplus W

and

\dim U = 3

\dim W = 4

, what is

\dim V

Easy

Not attempted

\mathbb{R}^2

, which pairs are complementary subspaces?

Hard

Not attempted

What is the zero element of

V/W

Medium

Not attempted

v - u \in W

, what does this mean in

V/W

Medium

Not attempted

V = U_1 \oplus U_2 \oplus U_3

with

\dim U_i = 2

, what is

\dim V

Easy

Not attempted

What is the projection onto

U

along

W

V = U \oplus W

Hard

Not attempted

Frequently Asked Questions

What's the intuition for quotient spaces?

V/W 'collapses' W to a point. Elements of V/W are parallel copies of W. If W is a line through origin, V/W contains all lines parallel to W, each as a single 'point' in the quotient.

How is V/W different from a complement of W?

A complement U is a subspace of V with V = W ⊕ U. The quotient V/W is a different space altogether. But V/W ≅ U for any complement U.

Why do we need quotient spaces?

They're essential for the First Isomorphism Theorem: if T: V → W is linear, then V/ker(T) ≅ im(T). Quotients also appear in defining cosets in group theory and factor rings in algebra.

Is the complement of a subspace unique?

No! A subspace typically has infinitely many complements. For example, in ℝ², any line not equal to the x-axis is a complement of the x-axis.

What's the external direct sum?

Given spaces V and W (not necessarily subspaces of a common space), V ⊕ W is the set of pairs (v, w) with componentwise operations. Its dimension is dim(V) + dim(W).

How do I verify that V = U ⊕ W?

Check two things: (1) U + W = V (every vector is a sum), and (2) U ∩ W = {0} (only overlap is zero). Equivalently, show every v has a UNIQUE decomposition as u + w.

What is a projection?

If V = U ⊕ W, the projection onto U along W is the map π: V → U defined by π(u + w) = u. It satisfies π² = π (idempotent) and ker(π) = W, im(π) = U.

Why check well-definedness for quotient operations?

Operations on V/W are defined using representatives: [v] + [u] = [v + u]. We must verify this doesn't depend on which representatives we choose—different v', u' with [v'] = [v], [u'] = [u] must give [v' + u'] = [v + u].

How are direct sums and products related?

For finite families of spaces, the direct sum and direct product coincide. For infinite families, they differ: direct sum requires only finitely many non-zero components, while products allow all components to be non-zero.

What's the connection to eigenspace decomposition?

If a linear operator has enough eigenvectors, V decomposes as a direct sum of eigenspaces: V = E_{λ₁} ⊕ E_{λ₂} ⊕ ... This is the foundation of diagonalization.