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7. Linear Algebra

Quotient space

A quotient space $V/W$ is formed by partitioning a vector space $V$ using a subspace $W$, where vectors differing by an element in $W$ are treated as equivalent.

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Definition and statement

Fundamental Theorem of Quotient Spaces (First Isomorphism Theorem for Vector Spaces)
Let $V$ and $U$ be vector spaces over a field $\mathbb{F}$, and let $T: V \to U$ be a linear transformation. Then the induced map $\bar{T}: V/\ker(T) \to \text{Im}(T)$ defined by $\bar{T}(v + \ker(T)) = T(v)$ is a vector space isomorphism. Consequently, $V/\ker(T) \cong \text{Im}(T).$

📐 Geometric Interpretation

Geometrically, quotienting out by a subspace $W$ collapses $W$ to the origin, projecting the ambient space $V$ onto a complementary subspace of the same dimension as $V/W$. Parallel affine translates of $W$ become single points (equivalence classes) in the quotient space.

⚙️ Algebraic & Structural Meaning

It establishes the exact relation between the domain, kernel, and range of any linear map, reducing structural complexity by factoring out null components and providing the algebraic backbone for direct sum decompositions $V = W \oplus W'.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition Domain / Interval Necessity What Fails If Condition Dropped
Subspace Condition N/A Strictly Mandatory If $W$ is merely a subset or subgroup and not a vector subspace, the cosets $v + W$ do not form a well-defined vector space under induced operations.
Well-definedness of Coset Addition N/A Strictly Mandatory Without $W$ being a subspace, $(v_1 + W) + (v_2 + W)$ depends on the choice of representatives, destroying the linear structure.
Finite Dimensionality (Optional for general, mandatory for dimension formula) N/A Essential for $\dim(V/W) = \dim(V) - \dim(W)$ In infinite-dimensional spaces, dimensions cannot be subtracted finitely without invoking cardinal arithmetic or specific topological constraints.

Useful results and shortcuts

Use these identities and results when their conditions apply:

01

Dimension of Quotient Space

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

Rule: Whenever asked for the dimension of a quotient space of matrices, polynomials, or function spaces, subtract the dimension of the constraint subspace from the total space.

Application: Frequently asked in UP PGT and KVS PGT for spaces like $M_n(\mathbb{R})/\text{Sym}_n(\mathbb{R})$.
02

Symmetric vs Skew-Symmetric Matrix Quotient

$$\dim\left(\frac{M_n(\mathbb{R})}{\text{skew-sym}_n(\mathbb{R})}\right) = \frac{n(n+1)}{2}$$

Rule: Identify the subspace $W$. Here $W$ is the space of skew-symmetric matrices of dimension $n(n-1)/2$. Total dimension is $n^2$. Subtract to get $n(n+1)/2$ (dimension of symmetric matrices).

03

Polynomial Quotient Space Dimension

$$\dim\left(\frac{\mathbb{P}_n[x]}{\langle p(x) \rangle}\right) = \deg(p(x))$$

Rule: The dimension of the quotient of the polynomial space $\mathbb{P}_n$ by the principal ideal generated by a polynomial of degree $k$ (where $k \le n$) is exactly $k$, provided the base field allows division or degree matching.

04

Rank-Nullity Connection

$$V/\ker(T) \cong \text{Im}(T) \implies \dim(V) - \dim(\ker(T)) = \text{rank}(T)$$

Rule: Use the isomorphism directly to bypass constructing basis vectors for the quotient space when finding ranks or nullities.

05

Direct Sum Complement Dimension

$$V = W \oplus U \implies V/W \cong U$$

Rule: If a subspace $W$ has a complementary subspace $U$, the quotient space $V/W$ is naturally isomorphic to $U$, meaning $\dim(V/W) = \dim(U).$

Application: Used to quickly solve abstract vector space structure questions in NVS PGT.
06

Rapid Option Elimination via Dimension

$$W_1 \subset W_2 \implies \dim(V/W_1) > \dim(V/W_2)$$

Rule: If options contain dimensions, immediately eliminate any option violating monotonicity or basic dimension subtraction rules.

Common errors and counterexamples

These examples show how a conclusion can fail when a condition is missing:

⚠️ COMMON ERROR #01

Confusing Subspace with Arbitrary Subset

Common Pitfall: Assuming $V/S$ makes sense when $S$ is a subset that does not contain the zero vector or is not closed under addition/scalar multiplication.

Counterexample: Let $S = \{(x, y) \in \mathbb{R}^2 : xy = 0\}$, which is not a subspace.

Why it fails: Cosets $v + S$ fail to partition the space properly because addition of cosets is not well-defined.

🛡️ Exam Defense: Always check that the denominator set $W$ is explicitly verified as a vector subspace before computing quotient dimensions.
⚠️ COMMON ERROR #02

Dimension Formula Misapplication in Infinite Dimensions

Common Pitfall: Applying $\dim(V/W) = \dim(V) - \dim(W)$ when both $V$ and $W$ are infinite-dimensional.

Counterexample: Let $V = \mathbb{R}^{\infty}$ and $W$ be a finite-codimensional subspace.

Why it fails: Infinity minus infinity is undefined; cardinal arithmetic must be used, or finite codimension must be specified.

🛡️ Exam Defense: Ensure $\dim(V) < \infty$ before using simple arithmetic subtraction for dimensions.

Comparison

Concept / TheoremHypotheses / AssumptionsKey Formula / OutcomeGeometric MeaningExam Identification Clue
Quotient Space $V/W$$W$ is a subspace of $V$$\dim(V/W) = \dim(V) - \dim(W)$Collapsing $W$ to origin, forming affine parallel planesKeywords: 'quotient space', 'modulo subspace', matrix spaces
First Isomorphism Theorem$T: V \to U$ is linear$V/\ker(T) \cong \text{Im}(T)$Bijective projection of reduced domain onto imageKeywords: 'isomorphic to image', 'kernel', 'linear transformation'
Direct Sum $V = W \oplus U$$W \cap U = \{0\}$ and $W+U = V$$\dim(V) = \dim(W) + \dim(U)$Space splits into independent orthogonal/complementary axesKeywords: 'direct sum', 'complementary subspace'
Codimension$W \le V$, $\dim(V) < \infty$$\text{codim}(W) = \dim(V/W)$Number of dimensions 'left over' after removing $W$Keywords: 'codimension of subspace'

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

Let $V = M_3(\mathbb{R})$ be the vector space of all $3 \times 3$ real matrices, and let $W$ be the subspace of all skew-symmetric $3 \times 3$ matrices. What is the dimension of the quotient space $V/W$?

(A) 3(B) 6(C) 9(D) 5
✓ Correct Answer: (B)
📚 Method A: Formal Textbook Derivation
Step 1: The total dimension of $M_3(\mathbb{R})$ is $3^2 = 9$. Step 2: The dimension of the space of skew-symmetric matrices $W$ of order 3 is $\frac{3(3-1)}{2} = 3$. Step 3: By the dimension formula for quotient spaces, $\dim(V/W) = \dim(V) - \dim(W) = 9 - 3 = 6$. Step 4: Note that this quotient space is isomorphic to the space of symmetric matrices of order 3, which has dimension 6.
⚡ Method B: 25-Second Exam Shortcut
Recognize that $M_n(\mathbb{R}) = \text{Sym}_n \oplus \text{Skew}_n$. Thus $M_n/\text{Skew}_n \cong \text{Sym}_n$, whose dimension is $\frac{n(n+1)}{2}$. For $n=3$, $\frac{3(4)}{2} = 6$.
💡 Tip: Quickly identify whether the quotient corresponds to symmetric or skew-symmetric matrices to write down the answer.
CANONICAL PROBLEM 02 Dual-Method Breakdown

Let $V$ be the vector space of all polynomials of degree at most 4 over $\mathbb{R}$, denoted $\mathbb{P}_4[x]$. Let $W$ be the subspace of polynomials divisible by $x^2$. What is the dimension of $V/W$?

(A) 5(B) 3(C) 2(D) 4
✓ Correct Answer: (C)
📚 Method A: Formal Textbook Derivation
Step 1: A general element of $\mathbb{P}_4[x]$ is $p(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + a_4 x^4$, so $\dim(V) = 5$. Step 2: The subspace $W$ consists of polynomials with $a_0 = 0$ and $a_1 = 0$, i.e., $p(x) = x^2(a_2 + a_3 x + a_4 x^2)$. Thus $\dim(W) = 3$. Step 3: $\dim(V/W) = \dim(V) - \dim(W) = 5 - 3 = 2$.
⚡ Method B: 25-Second Exam Shortcut
Moding out by multiples of $x^2$ essentially fixes the constant and linear terms, leaving 2 degrees of freedom (coefficients of $1$ and $x$). Thus the dimension of the quotient is 2.
💡 Tip: Count the number of independent conditions or remaining free coefficients after applying the subspace constraint.

Practice questions

Choose an answer to check it and see the solution:

Q1
MCQ 1: If $W$ is a subspace of a finite-dimensional vector space $V$, then $\dim(V/W) = 0$ if and only if:
A $\text{dim}(W) = \dim(V)$
B $W = \{0_V\}$
C $V = \{0_V\}$
D $W$ is a proper subspace of $V$
Derivation & Analysis:

Since $\dim(V/W) = \dim(V) - \dim(W)$, $\dim(V/W) = 0 \implies \dim(V) = \dim(W)$. Since $W \le V$, this implies $W = V$.

Q2
MCQ 2: Let $V = \mathbb{R}^3$ and $W = \{(x, y, z) \in \mathbb{R}^3 : x + y + z = 0\}$. The dimension of the quotient space $V/W$ is:
A 0
B 1
C 2
D 3
Derivation & Analysis:

$V = \mathbb{R}^3$ has dimension 3. $W$ is defined by 1 independent linear homogeneous equation, so $\dim(W) = 3 - 1 = 2$. Hence $\dim(V/W) = 3 - 2 = 1$.

Q3
MCQ 3: Let $T: V \to U$ be a linear transformation between vector spaces. By the First Isomorphism Theorem, $V / \ker(T)$ is isomorphic to:
A $\ker(T)$
B $U$
C $\text{Im}(T)$
D $V$
Derivation & Analysis:

The First Isomorphism Theorem states that $V/\ker(T) \cong \text{Im}(T)$.

Q4
MCQ 4: What is the dimension of the quotient space $\frac{M_n(\mathbb{R})}{\text{scalar matrices}}$?
A $n^2 - 1$
B $n^2 - n$
C $n$
D $n^2$
Derivation & Analysis:

The space $M_n(\mathbb{R})$ has dimension $n^2$. Scalar matrices form a 1-dimensional subspace (spanned by the identity matrix $I_n$). Thus $\dim = n^2 - 1$.

Q5
MCQ 5: Let $W_1 \subseteq W_2$ be subspaces of a finite-dimensional vector space $V$. Then:
A $\dim(V/W_1) \le \dim(V/W_2)$
B $\dim(V/W_1) \ge \dim(V/W_2)$
C $\dim(V/W_1) = \dim(V/W_2)$
D No relation can be established
Derivation & Analysis:

Since $W_1 \subseteq W_2$, $\dim(W_1) \le \dim(W_2)$. Therefore, $\dim(V/W_1) = \dim(V) - \dim(W_1) \ge \dim(V) - \dim(W_2) = \dim(V/W_2)$.

Q6
MCQ 6: If $\dim(V) = 7$ and $\dim(W) = 4$, what is the codimension of $W$ in $V$?
A 3
B 4
C 7
D 11
Derivation & Analysis:

Codimension of $W$ in $V$ is defined as $\dim(V/W) = \dim(V) - \dim(W) = 7 - 4 = 3$.

Q7
MCQ 7: Which of the following is true regarding the elements of a quotient space $V/W$?
A Elements are individual vectors of $V$
B Elements are scalars in the base field $\mathbb{F}$
C Elements are affine subsets of the form $v + W$
D Elements are linear transformations from $V$ to $W$
Derivation & Analysis:

The elements of a quotient space $V/W$ are cosets of the form $v + W$, which are affine subsets of $V$.

Q8
MCQ 8: Let $V$ be the vector space of all differentiable functions on $\mathbb{R}$, and let $W$ be the subspace of constant functions. Two functions $f, g \in V$ belong to the same coset in $V/W$ if and only if:
A $f(x) = g(x)$ for all $x$
B $f(x) - g(x) = C$ for some constant $C$
C $f'(x) = g'(x)$ for all $x$
D $\int f = \int g$
Derivation & Analysis:

By definition, $f + W = g + W \iff f - g \in W \iff f(x) - g(x) = C$ (a constant function).

Q9
MCQ 9: If $V = U \oplus W$, then the quotient space $V/W$ is canonically isomorphic to:
A $W$
B $U$
C $V$
D \{0\}
Derivation & Analysis:

If $V = U \oplus W$, every vector $v \in V$ can be uniquely written as $u + w$. Mapping $u + w + W$ to $u$ gives a natural vector space isomorphism between $V/W$ and $U$.

Q10
MCQ 10: Let $V$ be a vector space over $\mathbb{R}$ of dimension 5. If $W_1$ has dimension 2 and $W_2$ has dimension 3, and $W_1 \cap W_2 = \{0\}$, then $\dim(V/(W_1 \oplus W_2))$ is:
A 0
B 1
C 2
D 5
Derivation & Analysis:

Since $W_1 \cap W_2 = \{0\}$, $\dim(W_1 \oplus W_2) = \dim(W_1) + \dim(W_2) = 2 + 3 = 5$. Thus $\dim(V/(W_1 \oplus W_2)) = 5 - 5 = 0$.

Common questions

Because a quotient space identifies all vectors that differ by an element of $W$ as equivalent. A coset $v + W$ represents the entire equivalence class of vectors sharing the same 'remainder' relative to the subspace $W$.

Candidates often confuse the dimension of the subspace $W$ with the dimension of the quotient space $V/W$, forgetting to subtract it from the total dimension $\dim(V)$.

Instantly apply standard direct sum decompositions (e.g., $M_n = \text{Sym} \oplus \text{Skew}$) to read off dimensions without setting up explicit bases.

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