7. Linear Algebra

Rank-nullity theorem

Q: FAQ 2: What is the most common reason students lose marks on Rank-Nullity questions in Linear Algebra exams?

Students frequently confuse whether to subtract the rank from the domain dimension or the codomain dimension. Always remember that the base equation is $\dim(\text{Domain}) = \text{rank} + \text{nullity}$. The codomain dimension is only used as an upper bound for the rank.

The Rank-Nullity Theorem establishes a fundamental structural bridge between the dimension of the domain, the rank (dimension of the image/range), and the nullity (dimension of the kernel) of a linear transformation between finite-dimensional vector spaces.

For: UP PGT Mathematics UP TGT Mathematics DSSSB PGT / TGT KVS / NVS PGT LT Grade

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Definition and statement

Rank-Nullity Theorem (Sylvester's Law of Nullity)

Let $V$ and $W$ be vector spaces over a field $F$, and let $T: V \to W$ be a linear transformation. If $V$ is finite-dimensional, then the kernel (null space) $\ker(T)$ and the range (image) $\text{Im}(T)$ are finite-dimensional, and: $$\dim(V) = \dim(\ker(T)) + \dim(\text{Im}(T))$$ or equivalently, $$\text{nullity}(T) + \text{rank}(T) = \dim(V)$$

📐 Geometric Interpretation

Geometrically, when a linear transformation maps a vector space $V$ to $W$, the domain space $V$ is partitioned into two complementary subspaces: the part that collapses to the zero vector (the null space, whose dimension is the nullity) and the part that is mapped injectively into the codomain (the range, whose dimension is the rank). The theorem shows that the sum of the degrees of freedom lost (nullity) and the degrees of freedom preserved (rank) precisely equals the total dimension of the domain.

⚙️ Algebraic & Structural Meaning

Algebraically, this theorem is the cornerstone for solving homogeneous and non-homogeneous linear equations, studying matrix transformations, determining invertibility, and analyzing the structural properties of linear operators without needing explicit bases.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition	Domain / Interval	Necessity	What Fails If Condition Dropped
Finite-Dimensional Domain ($V$)	`Vector Space $V$`	Strictly Mandatory	If $V$ is infinite-dimensional (e.g., polynomial space $\mathbb{R}[x]$), the sum relation breaks down or becomes mathematically undefined as infinite cardinalities cannot be equated algebraically.
Linearity of the Transformation ($T$)	`Mapping $T: V \to W$`	Strictly Mandatory	If $T$ is non-linear (e.g., $T(x, y) = (x^2, y)$), the concept of a kernel subspace fails, and the dimension relations do not hold.
Codomain ($W$) Dimensionality	`Vector Space $W$`	Not strictly finite	While $W$ can be infinite-dimensional, the rank (image dimension) is bounded by both $\dim(V)$ and the subspace dimension, but $W$ itself does not need to be finite-dimensional for the theorem to hold.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Matrix Rank-Nullity Dimension Check

$$ \text{Rank}(A) + \text{Nullity}(A) = n \quad \text{for an } m \times n \text{ matrix } A $$

Rule: If you know the rank of a matrix (number of non-zero rows in row echelon form) and the size of the matrix columns $n$, instantly find nullity as $n - \text{rank}$.

Application: undefined

Invertibility & Full Rank Condition

$$ \text{nullity}(T) = 0 \iff T \text{ is injective (one-to-one)} $$

Rule: For a linear operator $T: V \to V$ where $\dim(V) = n$, $\text{rank}(T) = n \iff \text{nullity}(T) = 0 \iff T$ is invertible (non-singular).

Application: Frequently used in KVS and NVS to quickly determine if a linear transformation is an isomorphism.

Dimension Inequality for Composite Maps

$$ \text{rank}(ST) \le \min(\text{rank}(S), \text{rank}(T)) $$

Rule: Use Sylvester's rank inequalities to bound the rank and nullity of composed transformations without matrix multiplication.

Application: Asked in DSSSB PGT to eliminate options involving composite linear transformations.

Homogeneous System Free Variables

$$ \text{Nullity} = \text{Number of variables } (n) - \text{Rank of coefficient matrix } (r) $$

Rule: The dimension of the solution space (null space) of a homogeneous system $AX = 0$ is equal to the number of free variables, which is $n - r$.

Transformation via Basis Evaluation

$$ \dim(\text{Im}(T)) = \text{Number of linearly independent columns of standard matrix representation} $$

Rule: Do not compute the entire range space; simply construct the standard matrix and find its column rank using elementary row/column operations.

Application: Saves 2 minutes in UP TGT matrix transformation problems.

Rapid Option Elimination via Parity/Divisibility

$$ \text{rank}(A) + \text{nullity}(A) = n $$

Rule: If an option gives a nullity greater than the column dimension $n$ or non-integer rank, eliminate it instantly.

Common errors and counterexamples

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

⚠️ COMMON ERROR #01

Confusing Domain Dimension with Codomain Dimension

Common Pitfall: Candidates subtract rank from the dimension of the codomain $W$ instead of the domain $V$.

Counterexample: $T: \mathbb{R}^2 \to \mathbb{R}^5$

Why it fails: The formula is $\dim(V) = \text{rank} + \text{nullity}$, where $\dim(V) = 2$. Using $\dim(W) = 5$ will result in a completely incorrect nullity calculation.

🛡️ Exam Defense: Always check the domain space $V$ dimension. Nullity is bounded by $\dim(V)$, not $\dim(W)$.

⚠️ COMMON ERROR #02

Assuming Rank Equals Codomain Dimension

Common Pitfall: Assuming $\text{rank}(T) = \dim(W)$ for any arbitrary linear transformation.

Counterexample: Zero transformation $T: \mathbb{R}^3 \to \mathbb{R}^3$, where $T(v) = 0$ $\forall v$

Why it fails: Here, $\text{rank}(T) = 0$ and $\text{nullity}(T) = 3$, even though the codomain dimension is 3. Rank is only the dimension of the *image*, not the codomain.

🛡️ Exam Defense: Remember that $\text{rank}(T) \le \min(\dim(V), \dim(W))$. Equality holds if and only if the map is surjective.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Rank-Nullity Theorem	Finite-dimensional vector space $V$, Linear map $T: V \to W$	$\dim(V) = \text{rank}(T) + \text{nullity}(T)$	Domain splits into collapsed kernel and preserved image dimensions.	Keywords: 'linear transformation', 'dimension of range/kernel', 'matrix nullity'.
Sylvester's Law of Nullity	Matrix $A$ of order $m \times n$	$\text{rank}(A) + \text{nullity}(A) = n$	Number of pivot columns plus number of free variables equals total columns.	Keywords: 'homogeneous system', 'solution space dimension', 'matrix rank'.
Invertibility Theorem	Square matrix $A$ of order $n \times n$	$\text{rank}(A) = n \iff \text{nullity}(A) = 0$	Full dimensional mapping without collapsing any non-zero vectors.	Keywords: 'non-singular', 'isomorphism', 'one-to-one and onto'.
Dimension Theorem for Subspaces	Subspaces $U_1, U_2 \le V$	$\dim(U_1 + U_2) = \dim(U_1) + \dim(U_2) - \dim(U_1 \cap U_2)$	Accounting for overlapping dimensions when combining vector subspaces.	Keywords: 'sum of subspaces', 'intersection dimension'.

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

Let $T: \mathbb{R}^3 \to \mathbb{R}^3$ be a linear transformation defined by $T(x, y, z) = (x + y + z, 2x + y - z, 4x + 3y + z)$. What is the rank of $T$?

(A) $0$(B) $1$(C) $2$(D) $3$

📚 Method A: Formal Textbook Derivation

Step 1: Write the standard matrix representation of $T$: $A = \begin{pmatrix} 1 & 1 & 1 \\ 2 & 1 & -1 \\ 4 & 3 & 1 \end{pmatrix}$. Step 2: Apply elementary row operations to reduce $A$ to row echelon form. $R_2 \to R_2 - 2R_1$, $R_3 \to R_3 - 4R_1$ gives $\begin{pmatrix} 1 & 1 & 1 \\ 0 & -1 & -3 \\ 0 & -1 & -3 \end{pmatrix}$. Step 3: $R_3 \to R_3 - R_2$ gives $\begin{pmatrix} 1 & 1 & 1 \\ 0 & -1 & -3 \\ 0 & 0 & 0 \end{pmatrix}$. Step 4: The number of non-zero rows is 2, hence $\text{rank}(T) = 2$.

⚡ Method B: 25-Second Exam Shortcut

Compute the determinant of matrix $A$. $\det(A) = 1(1 - (-3)) - 1(2 - (-4)) + 1(6 - 4) = 4 - 6 + 2 = 0$. Since $\det(A) = 0$, the rank is strictly less than 3. By inspection, the first two rows are linearly independent, so rank is 2.

💡 Tip: If the determinant of the matrix of a linear operator on $\mathbb{R}^n$ is zero, immediately rule out rank $n$.

CANONICAL PROBLEM 02 Dual-Method Breakdown

Let $V$ be a vector space of all polynomials of degree at most $3$. Let $D: V \to V$ be the differentiation operator $D(p(x)) = p'(x)$. What is the nullity of $D$?

(A) $0$(B) $1$(C) $2$(D) $3$

📚 Method A: Formal Textbook Derivation

Step 1: The dimension of $V$ (polynomials of degree $\le 3$) is $\dim(V) = 4$ (basis: ${1, x, x^2, x^3}$). Step 2: The kernel of $D$ consists of polynomials $p(x)$ such that $p'(x) = 0$. Step 3: The derivative is zero only for constant polynomials $p(x) = c$. Step 4: The space of constant polynomials is 1-dimensional (basis: ${1}$). Therefore, $\text{nullity}(D) = \dim(\ker(D)) = 1$.

⚡ Method B: 25-Second Exam Shortcut

Derivative of a polynomial is zero if and only if it is a constant. A constant space is spanned by ${1}$, dimension = 1. Done in 5 seconds.

💡 Tip: For differential operators, the nullity is always equal to the order of constant terms / polynomials mapped to zero (here, constants have dimension 1).

Practice questions

Choose an answer to check it and see the solution:

If $T: V \to W$ is a linear transformation and $\dim(V) = 7$, $\dim(W) = 5$, what is the minimum possible value of $\text{nullity}(T)$?

A 0

B 2

C 5

D 7

Derivation & Analysis:

By Rank-Nullity Theorem, $\text{rank}(T) + \text{nullity}(T) = 7$. Also, $\text{rank}(T) \le \dim(W) = 5$. Thus, $\text{nullity}(T) = 7 - \text{rank}(T) \ge 7 - 5 = 2$. Minimum nullity is 2.

Let $A$ be a $4 \times 6$ matrix of rank 3. What is the dimension of the null space of $A$?

A 1

B 2

C 3

D 4

Derivation & Analysis:

Using Sylvester's Rank-Nullity Law for matrices, $\text{rank}(A) + \text{nullity}(A) = n$, where $n$ is the number of columns ($n = 6$). Hence, $3 + \text{nullity}(A) = 6 \implies \text{nullity}(A) = 3$.

Which of the following statements is ALWAYS true for a linear transformation $T: V \to W$ where $\dim(V) = \dim(W) = n$?

A $T$ is one-to-one $\iff T$ is onto

B $T$ is always invertible

C $\text{rank}(T) = 0$

D $\text{nullity}(T) = n$

Derivation & Analysis:

For finite-dimensional vector spaces of equal dimension, a linear map is injective (one-to-one) if and only if it is surjective (onto), which is equivalent to being bijective (invertible).

If $T: \mathbb{R}^5 \to \mathbb{R}^3$ is a linear transformation, what is the maximum possible rank of $T$?

A 2

B 3

C 5

D 8

Derivation & Analysis:

The rank of $T$ is bounded by $\min(\dim(V), \dim(W)) = \min(5, 3) = 3$. Therefore, the maximum possible rank is 3.

Let $W_1$ and $W_2$ be subspaces of a finite-dimensional vector space $V$. If $\dim(W_1) = 3$, $\dim(W_2) = 4$, and $\dim(W_1 \cap W_2) = 1$, what is $\dim(W_1 + W_2)$?

A 5

B 6

C 7

D 8

Derivation & Analysis:

Using the dimension formula for sums of subspaces: $\dim(W_1 + W_2) = \dim(W_1) + \dim(W_2) - \dim(W_1 \cap W_2) = 3 + 4 - 1 = 6$.

What is the nullity of the zero transformation $T: \mathbb{R}^n \to \mathbb{R}^m$ defined by $T(x) = 0$?

A 0

B 1

C $m$

D $n$

Derivation & Analysis:

The zero transformation maps every vector in $\mathbb{R}^n$ to 0. Thus, $\ker(T) = \mathbb{R}^n$, so its dimension (nullity) is $n$.

If a square matrix $A$ of order $n$ is invertible, what is its nullity?

A 0

B 1

C $n$

D Depends on $n$

Derivation & Analysis:

An invertible matrix has $\text{rank}(A) = n$. By Rank-Nullity, $\text{nullity}(A) = n - \text{rank}(A) = n - n = 0$.

Assertion (A): If $\text{rank}(T) = \dim(V)$ for $T: V \to W$, then $T$ is one-to-one. Reason (R): $\text{nullity}(T) = 0$ implies injectivity.

A Both (A) and (R) are true and (R) is the correct explanation of (A)

B Both (A) and (R) are true but (R) is not the correct explanation

C (A) is true but (R) is false

D (A) is false but (R) is true

Derivation & Analysis:

By Rank-Nullity, $\text{rank}(T) = \dim(V) \implies \text{nullity}(T) = 0$. A zero nullity means $\ker(T) = \{0\}$, which is the exact condition for $T$ to be injective (one-to-one).

Let $T$ be a linear operator on $\mathbb{R}^3$ such that $\text{rank}(T) = 2$. What is the dimension of the range space of $T$?

A 0

B 1

C 2

D 3

Derivation & Analysis:

By definition, $\text{rank}(T)$ is precisely the dimension of the range space (image) of $T$. Hence, the dimension is 2.

Q10

If $S$ and $T$ are two linear transformations on a vector space $V$ of dimension $n$, what is the relation between $\text{nullity}(ST)$ and individual nullities?

A $\text{nullity}(ST) \le \text{nullity}(S) + \text{nullity}(T)$

B $\text{nullity}(ST) = \text{nullity}(S)\text{nullity}(T)$

C $\text{nullity}(ST) \ge \text{nullity}(S)$

D $\text{nullity}(ST) = \text{nullity}(T)$

Derivation & Analysis:

By Sylvester's inequality for nullities of products, the nullity of a composite linear transformation is bounded by the sum of the nullities: $\text{nullity}(ST) \le \text{nullity}(S) + \text{nullity}(T)$.

Common questions

No. Since $\text{Im}(T)$ is a subspace of the codomain $W$ and is also isomorphic to $V / \ker(T)$, the rank cannot exceed $\dim(V)$ nor can it exceed $\dim(W)$. Therefore, $\text{rank}(T) \le \min(\dim(V), \dim(W))$.

Candidates frequently confuse whether to subtract the rank from the domain dimension or the codomain dimension. Always remember that the base equation is $\dim(\text{Domain}) = \text{rank} + \text{nullity}$. The codomain dimension is only used as an upper bound for the rank.

Instead of finding the full row echelon form, calculate the determinant (for square matrices) or inspect pivot columns after quick row reductions. Use the formula $\text{Nullity} = n - \text{Rank}$ immediately once the rank is established.

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