7. Linear Algebra

Rank of linear transformation

The rank of a linear transformation $T: V \to W$ is defined as the dimension of its range space (or image space), denoted by $\text{rank}(T) = \dim(\text{Im}(T))$.

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 (Dimension Theorem for Linear Transformations)

Let $V$ and $W$ be vector spaces over a field $\mathbb{F}$, with $V$ being finite-dimensional. Let $T: V \to W$ be a linear transformation. Then $\dim(V) = \dim(\ker(T)) + \dim(\text{Im}(T))$, which can also be written as $\text{nullity}(T) + \text{rank}(T) = \dim(V)$.

📐 Geometric Interpretation

Geometrically, the transformation $V$ collapses a subspace of dimension $\text{nullity}(T)$ to the zero vector in $W$, while the remaining dimensions of $V$ are faithfully mapped to an isomorphic subspace in $W$ of dimension $\text{rank}(T)$.

⚙️ Algebraic & Structural Meaning

It bridges abstract vector space dimensions with concrete matrix rank computations, guaranteeing that the column rank equals the row rank of the corresponding matrix representation.

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 $V$	`Domain space structure`	Strictly Mandatory	If $V$ is infinite-dimensional (e.g., space of polynomials $\mathbb{F}[x]$), the sum relation $\dim(\ker) + \dim(\text{Im}) = \dim(V)$ breaks down as dimensions are no longer finite integers.
Linearity of $T$	`Mapping property ($T(cx + y) = cTx + Ty$)`	Strictly Mandatory	Non-linear transformations do not preserve subspace structures, making the concepts of kernel and image subspaces invalid.
Codomain $W$ over the same Field $\mathbb{F}$	`Algebraic closure`	Essential for well-defined dimension	Mismatched fields prevent scalar multiplication consistency in the range and kernel spaces.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Sylvester's Inequality for Matrix/Transformation Rank

$$\text{rank}(AB) \ge \text{rank}(A) + \text{rank}(B) - n$$

Rule: For composite transformations $S \circ T$, the rank of the composition is bounded below by the sum of individual ranks minus the dimension of the intermediate space.

Application: Frequently asked in UP PGT and DSSSB PGT when finding minimum possible rank of a product.

Frobenius Inequality

$$\text{rank}(ABC) \ge \text{rank}(AB) + \text{rank}(BC) - \text{rank}(B)$$

Rule: Direct inequality for triple compositions or matrix chains tested in high-level screening exams.

Application: Standard PYQ appearance in NVS and KVS PGT Mathematics.

Rank of Transpose

$$\text{rank}(T^*) = \text{rank}(T)$$

Rule: The rank of the dual/transpose linear transformation equals the rank of the original transformation.

Dimension Bound by Codomain

$$\text{rank}(T) \le \min(\dim(V), \dim(W))$$

Rule: The rank can never exceed the dimension of the domain or the codomain.

Application: Used for instant elimination of impossible options in rank questions.

Matrix Rank via Echelon Form

$$\text{rank}(T) = \text{Number of non-zero rows in RREF}$$

Rule: Convert the standard matrix representation $[T]_B$ to row echelon form; the non-zero rows give the rank instantly.

Rapid Elimination via Nullity

$$\text{rank}(T) = \dim(V) - \text{nullity}(T)$$

Rule: If finding kernel dimension is easier than image dimension, calculate nullity first and subtract from $\dim(V)$.

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 nullity from $\dim(W)$ instead of the domain dimension $\dim(V)$.

Counterexample: For $T: \mathbb{R}^2 \to \mathbb{R}^4$, $\dim(V) = 2$, so $\text{rank}(T) \le 2$ regardless of codomain dimension 4.

Why it fails: The Rank-Nullity theorem explicitly states $\dim(V) = \text{nullity}(T) + \text{rank}(T)$, where $V$ is the domain.

🛡️ Exam Defense: Always check whether the domain or codomain dimension is given in the question stem.

⚠️ COMMON ERROR #02

Assuming Rank of Product equals Product of Ranks

Common Pitfall: Assuming $\text{rank}(ST) = \text{rank}(S) \cdot \text{rank}(T)$ like scalar multiplication.

Counterexample: Take two nilpotent matrices of rank 1 whose product is the zero matrix of rank 0.

Why it fails: Rank satisfies submultiplicativity: $\text{rank}(ST) \le \min(\text{rank}(S), \text{rank}(T))$, never equality in general.

🛡️ Exam Defense: Never multiply ranks; use Sylvester's inequality instead.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Rank-Nullity Theorem	Finite-dimensional domain $V$	$\text{rank} + \text{nullity} = \dim(V)$	Dimension preservation vs kernel collapse	Given mapping between vector spaces
Sylvester's Inequality	Composable linear maps $S, T$	$\text{rank}(ST) \ge \text{rank}(S) + \text{rank}(T) - n$	Lower bound on composition image size	Questions on composite transformations
Full Rank Transformation	$\dim(V) = \dim(W) = n$	$\text{rank}(T) = n \iff \text{nullity}(T) = 0$	Bijective isomorphism	Invertibility or non-singularity keywords
Matrix Rank Theorem	Matrix representation of $T$	$\text{Column Rank} = \text{Row Rank}$	Dimension of span of row/column vectors	Given explicit matrix or basis matrix

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 $T(x, y, z) = (x+y+z, y+z, z)$. Find the rank of $T$.

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

📚 Method A: Formal Textbook Derivation

Step 1: Write matrix representation relative to standard basis: $A = \begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}$. Step 2: Since $A$ is upper triangular with non-zero diagonal entries, its determinant is non-zero ($\det(A) = 1$). Step 3: Thus $\text{rank}(A) = 3$.

⚡ Method B: 25-Second Exam Shortcut

An upper triangular matrix with all diagonal elements non-zero has full rank equal to its order ($n=3$).

💡 Tip: Look at the matrix representation diagonal entries if triangular.

CANONICAL PROBLEM 02 Dual-Method Breakdown

Let $T: P_2(\mathbb{R}) \to P_3(\mathbb{R})$ be a linear transformation defined by $T(p(x)) = x p(x)$. What is the rank of $T$?

(A) 2(B) 3(C) 4(D) 1

📚 Method A: Formal Textbook Derivation

Step 1: $\dim(P_2(\mathbb{R})) = 3$. Step 2: Find the kernel of $T$: $T(p(x)) = x p(x) = 0 \implies p(x) = 0$, so $\ker(T) = \{0\}$, meaning $\text{nullity}(T) = 0$. Step 3: By Rank-Nullity theorem, $\text{rank}(T) = \dim(P_2) - \text{nullity}(T) = 3 - 0 = 3$.

⚡ Method B: 25-Second Exam Shortcut

Multiplication by non-zero polynomial $x$ is injective, so nullity is 0 and rank equals domain dimension.

💡 Tip: For polynomial derivative or multiplication operators, check injectivity directly to determine nullity.

Practice questions

Choose an answer to check it and see the solution:

MCQ 1: If $T: V \to W$ is a linear transformation and $\dim(V) = 5$, $\dim(W) = 7$, what is the maximum possible rank of $T$?

A 5

B 7

C 2

D 12

Derivation & Analysis:

The rank cannot exceed the dimension of the domain $V$. Thus $\max(\text{rank}(T)) = \min(5, 7) = 5$.

MCQ 2: Let $T: \mathbb{R}^4 \to \mathbb{R}^3$ be a linear transformation. What is the minimum possible nullity of $T$?

A 0

B 1

C 3

D 4

Derivation & Analysis:

By Rank-Nullity theorem, $\text{rank}(T) + \text{nullity}(T) = 4$. Since $\text{rank}(T) \le \dim(\mathbb{R}^3) = 3$, we have $\text{nullity}(T) = 4 - \text{rank}(T) \ge 4 - 3 = 1$.

MCQ 3: The rank of the zero transformation $T: V \to W$ where $\dim(V) = n$ is:

A $n$

B $1$

C $0$

D Undefined

Derivation & Analysis:

The zero transformation maps every vector to the zero vector, so its image space contains only the zero vector, giving rank 0.

MCQ 4: Let $T: \mathbb{R}^3 \to \mathbb{R}^3$ be given by $T(x,y,z) = (x+y, y+z, x+z)$. The nullity of $T$ is:

A 0

B 1

C 2

D 3

Derivation & Analysis:

The matrix of $T$ is $\begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{pmatrix}$ with determinant $2 \neq 0$. Thus $\text{rank} = 3$ and nullity $= 3 - 3 = 0$.

MCQ 5: If $S$ and $T$ are two linear transformations on a finite-dimensional space $V$ such that $ST = 0$, then:

A $\text{rank}(S) + \text{rank}(T) \le \dim(V)$

B $\text{rank}(S) = \text{rank}(T)$

C $S = 0$ or $T = 0$

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

Derivation & Analysis:

By Sylvester's rank inequality or properties of null spaces, $\text{Im}(T) \subseteq \ker(S)$, which implies $\text{rank}(T) \le \text{nullity}(S) = \dim(V) - \text{rank}(S)$.

MCQ 6: Let $V$ be the vector space of all $3 \times 3$ skew-symmetric matrices. The dimension of the range of a non-trivial linear transformation on $V$ cannot exceed:

A 9

B 3

C 6

D 1

Derivation & Analysis:

The dimension of the space of $3 \times 3$ skew-symmetric matrices is $\frac{3(2)}{2} = 3$. Hence any rank is bounded by 3.

MCQ 7: Which of the following cannot be the rank of a $4 \times 4$ real matrix?

A 0

B 2

C 4

D 5

Derivation & Analysis:

The rank of a $4 \times 4$ matrix cannot exceed its dimension 4. Therefore, 5 is impossible.

MCQ 8: Assertion (A): If $\text{rank}(T) = \dim(V)$, then $T$ is one-to-one. Reason (R): $\text{nullity}(T) = 0$ when rank equals domain dimension.

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, (R) is false

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

Derivation & Analysis:

By Rank-Nullity theorem, $\text{rank}(T) + \text{nullity}(T) = \dim(V)$. If $\text{rank}(T) = \dim(V)$, then $\text{nullity}(T) = 0$, meaning $\ker(T) = \{0\}$, so $T$ is injective.

MCQ 9: Let $T: \mathbb{R}^5 \to \mathbb{R}^5$ be a linear transformation with $\text{rank}(T) = 3$. Then the dimension of $\ker(T - I)$ is at most:

A 3

B 5

C 2

D 0

Derivation & Analysis:

The dimension of the eigenspace corresponding to eigenvalue 1 is bounded by the algebraic/geometric constraints, but directly nullity is 2, and rank-nullity principles apply.

Q10

MCQ 10: If $A$ is an $m \times n$ matrix and $\text{rank}(A) = r$, then the dimension of the null space of $A^T$ is:

A $n - r$

B $m - r$

C $r$

D $m + n - r$

Derivation & Analysis:

The null space of $A^T$ is the left null space of $A$, whose dimension is $m - \text{rank}(A) = m - r$.

Common questions

Yes! The rank of a linear transformation is defined independently of any basis as the dimension of its image space, and it is precisely equal to the rank of any matrix representation of the transformation relative to chosen bases.

Confusing the domain dimension with the codomain dimension in the Rank-Nullity formula, or mistakenly assuming rank is multiplicative like matrix determinants.

Use row operations to quickly convert matrices to echelon form, check for determinant non-zero conditions for square matrices, or apply dimension inequalities to eliminate incorrect options.

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