7. Linear Algebra

Linear transformation

A linear transformation is a structure-preserving map between vector spaces that respects vector addition and scalar multiplication, forming the foundational core of linear algebra.

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)

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

📐 Geometric Interpretation

Geometrically, the theorem states that the dimension of the domain vector space $V$ is partitioned into two parts: the dimension of the subspace that gets collapsed to the zero vector (the nullity) and the dimension of the subspace that is successfully mapped into the codomain (the rank).

⚙️ Algebraic & Structural Meaning

It links the structural dimensions of domain, kernel, and image, allowing us to determine invertibility and injectivity/surjectivity instantly in matrix representations.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition	Domain / Interval	Necessity	What Fails If Condition Dropped
Linearity (Additivity and Homogeneity)	`Entire Domain $V$`	Strictly Mandatory / Necessary	If $T(x + y) \neq T(x) + T(y)$ or $T(cx) \neq cT(x)$, the mapping is non-linear, and matrix representation theorems or the Rank-Nullity theorem completely fail.
Finite-Dimensionality of Domain	`Domain Space $V$`	Mandatory for the standard Rank-Nullity dimension equality	In infinite-dimensional spaces (such as polynomial spaces without degree bounds), dimensions may not be well-defined or finite, requiring functional analytic generalizations.
Vector Space Field Consistency	`Base Field $\mathbb{F}$`	Essential for scalar multiplication consistency	If scalars do not belong to the same underlying field, scalar homogeneity fails.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Rank-Nullity Dimension Check

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

Rule: Always find the rank of the associated matrix $A$ via row reduction or determinant non-vanishing order, then subtract from domain dimension to get nullity instantly.

Application: Directly asked in UP PGT and KVS PGT to find the dimension of the null space of matrix transformations.

Matrix of Linear Transformation relative to Standard Bases

$$[T]_{B} = \begin{bmatrix} T(e_1) & T(e_2) & \cdots & T(e_n) \end{bmatrix}$$

Rule: Place the images of the standard basis vectors as the *columns* of the matrix representation.

Application: Crucial for quick matrix construction in DSSSB and UP TGT vector space problems.

Change of Basis Formula

$$[T]_{B'} = P^{-1} [T]_{B} P$$

Rule: Use the transition matrix $P$ formed by writing new basis elements in terms of the old basis to transform linear operators rapidly.

Invertibility & Determinant Condition

$$T \text{ is invertible} \iff \det(A) \neq 0 \iff \text{nullity}(T) = 0$$

Rule: An operator on a finite-dimensional space is one-to-one if and only if it is onto, which holds if and only if the matrix determinant is non-zero.

Application: Standard option-elimination tool in multiple-choice questions.

Eigenvalues of Polynomial Transformations

$$if \; T(v) = \lambda v, \; \text{then} \; P(T)(v) = P(\lambda)v$$

Rule: To find eigenvalues of a polynomial operator $P(T)$, simply substitute the eigenvalue $\lambda$ into the polynomial expression.

Rapid Dimension Exclusion

$$m < n \implies \text{rank} \le m, \; \text{nullity} \ge n - m$$

Rule: For a transformation from $\mathbb{R}^n$ to $\mathbb{R}^m$ where $n > m$, the nullity can never be zero (non-zero kernel always exists, i.e., non-injective).

Application: Instantaneous elimination of incorrect options in qualitative theory questions.

Common errors and counterexamples

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

⚠️ COMMON ERROR #01

Confusing Domain and Codomain Dimensions

Common Pitfall: Candidates subtract rank from the codomain dimension instead of the domain dimension in the Rank-Nullity theorem.

Counterexample: $T: \mathbb{R}^3 \to \mathbb{R}^5$ with $\text{rank}(T) = 2$. Nullity is $3 - 2 = 1$, NOT $5 - 2 = 3$.

Why it fails: The kernel is a subspace of the domain $V$, so its dimension is bounded by $\dim(V)$.

🛡️ Exam Defense: Remember: Dimension of domain = Nullity + Rank. Always start with the domain dimension.

⚠️ COMMON ERROR #02

Assuming Homogeneity Implies Linearity

Common Pitfall: Checking only $T(cv) = cT(v)$ or $T(u+v) = T(u) + T(v)$ alone, or testing with zero vector properties.

Counterexample: $T(x, y) = (x^2, y)$ on $\mathbb{R}^2$

Why it fails: $T(cx, cy) = (c^2x^2, cy) \neq c(x^2, y)$ for $c \neq 1$. It fails homogeneity despite passing some scalar checks.

🛡️ Exam Defense: Always verify both conditions simultaneously using the combined test: $T(au + bv) = aT(u) + bT(v)$.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Linear Transformation	$T(ax+by) = aT(x)+bT(y)$	Preserves vector space structure	Straight lines map to straight lines (origin fixed)	Check additivity & homogeneity
Rank-Nullity Theorem	Finite-dimensional domain $V$	$\dim(\ker) + \dim(\text{range}) = \dim(V)$	Splits domain into kernel collapse and range projection	Given matrix or transformation mapping dimensions
Isomorphism	Linear, bijective ($1-1$ and onto)	$T^{-1}$ exists, $\dim(V) = \dim(W)$	Bijective geometric copy (stretching/rotating without tearing)	Keywords: invertible, bijective, non-singular
Matrix Representation	Fixed bases in $V$ and $W$	$[T]_B = [ \; T(e_j) \; ]$	Coordinate coordinate-system translation	Expressed as coordinate vectors or matrices

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 nullity of $T$.

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

📚 Method A: Formal Textbook Derivation

Step 1: Set $T(x, y, z) = (0, 0, 0)$. Step 2: Solve $x+y+z=0, y+z=0, z=0$. Step 3: Backward substitution gives $z=0, y=0, x=0$. Step 4: $\ker(T) = \{(0,0,0)\}$, so $\dim(\ker(T)) = 0$.

⚡ Method B: 25-Second Exam Shortcut

Write the matrix of $T$: $A = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix}$. Since $\det(A) = 1 \neq 0$, the matrix is non-singular, rank = 3, and nullity = $3 - 3 = 0$.

💡 Tip: Upper triangular matrices with non-zero diagonal entries have full rank and zero nullity instantly.

CANONICAL PROBLEM 02 Dual-Method Breakdown

If $T: \mathbb{R}^2 \to \mathbb{R}^2$ is a linear transformation such as $T(1, 0) = (2, 3)$ and $T(0, 1) = (1, 4)$, then $T(5, 2)$ is equal to:

(A) (12, 23)(B) (11, 23)(C) (10, 20)(D) (12, 19)

📚 Method A: Formal Textbook Derivation

Express $(5, 2)$ as a linear combination of standard basis: $(5, 2) = 5(1, 0) + 2(0, 1)$. By linearity, $T(5, 2) = 5T(1, 0) + 2T(0, 1) = 5(2, 3) + 2(1, 4) = (10+2, 15+8) = (12, 23)$.

⚡ Method B: 25-Second Exam Shortcut

Form the matrix $A = \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix}$ and multiply by column vector $\begin{bmatrix} 5 \\ 2 \end{bmatrix}$: $\begin{bmatrix} 2(5) + 1(2) \\ 3(5) + 4(2) \end{bmatrix} = \begin{bmatrix} 12 \\ 23 \end{bmatrix}$.

💡 Tip: When standard basis images are given, matrix multiplication is faster than linear combination expansion.

Practice questions

Choose an answer to check it and see the solution:

MCQ 1: Which of the following mappings $T: \mathbb{R}^2 \to \mathbb{R}^2$ is a linear transformation?

A $T(x, y) = (x^2, y)$

B $T(x, y) = (x+1, y)$

C $T(x, y) = (x-y, 2x+y)$

D $T(x, y) = (xy, 0)$

Derivation & Analysis:

Option (C) satisfies $T(ax_1+bx_2, ay_1+by_2) = aT(x_1, y_1) + bT(x_2, y_2)$ component-wise as each component is a homogeneous linear polynomial. Options A, B, and D fail due to degrees greater than 1 or constant offsets.

MCQ 2: Let $T: V \to W$ be a linear transformation where $\dim(V) = 5$ and $\dim(W) = 4$. What is the maximum possible rank of $T$?

A 5

B 4

C 1

D 9

Derivation & Analysis:

The range of $T$ is a subspace of $W$, so $\dim(\text{range}(T)) \le \dim(W) = 4$. Also, $\dim(\text{range}(T)) \le \dim(V) = 5$. Thus, the maximum possible rank is $\min(5, 4) = 4$.

MCQ 3: If $T: \mathbb{R}^3 \to \mathbb{R}^3$ has a nullity of 2, then the dimension of the range space of $T$ is:

A 3

B 2

C 1

D 0

Derivation & Analysis:

By the Rank-Nullity Theorem, $\dim(\ker(T)) + \dim(\text{range}(T)) = \dim(\mathbb{R}^3) \implies 2 + \text{rank} = 3 \implies \text{rank} = 1$.

MCQ 4: Let $T: P_n(\mathbb{R}) \to P_{n-1}(\mathbb{R})$ be the differentiation operator $T(p(x)) = p'(x)$. The nullity of $T$ is:

A 0

B 1

C $n$

D $n-1$

Derivation & Analysis:

The derivative of a polynomial is zero if and only if the polynomial is a constant. The space of constant polynomials has dimension 1, so $\dim(\ker(T)) = 1$.

MCQ 5: A linear transformation $T: V \to W$ is one-to-one if and only if:

A $\ker(T) = V$

B $\ker(T) = \{0\}$

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

D $\dim(\text{range}(T)) = 0$

Derivation & Analysis:

A linear transformation is injective (one-to-one) if and only if its kernel contains only the zero vector, i.e., $\ker(T) = \{0\}$. This is a fundamental theorem in linear algebra.

MCQ 6: If $A$ is an $m \times n$ matrix, the rank of $A$ plus the nullity of $A$ equals:

A $m$

B $n$

C $m+n$

D $|m-n|$

Derivation & Analysis:

By the Rank-Nullity theorem applied to the matrix transformation $T(x) = Ax$ where $A: \mathbb{R}^n \to \mathbb{R}^m$, the sum of rank and nullity equals the dimension of the domain, which is $n$.

MCQ 7: Which of the following is NOT a linear transformation?

A Zero transformation

B Identity transformation

C Translation by a non-zero vector $v_0$

D Scaling transformation $T(x) = c x$

Derivation & Analysis:

Translation $T(x) = x + v_0$ where $v_0 \neq 0$ maps the zero vector to $v_0 \neq 0$, failing $T(0) = 0$ which is required for all linear transformations.

MCQ 8: Let $T: \mathbb{R}^2 \to \mathbb{R}^2$ be a reflection across the x-axis. Its matrix relative to the standard basis is:

A $\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}$

B $\begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}$

C $\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$

D $\begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}$

Derivation & Analysis:

Reflection across the x-axis maps $(1, 0) \to (1, 0)$ and $(0, 1) \to (0, -1)$. Placing these as columns gives $\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}$.

MCQ 9: If $T: V \to W$ is an isomorphism, then:

A $\dim(V) < \dim(W)$

B $\dim(V) > \dim(W)$

C $\dim(V) = \dim(W)$

D No relation between dimensions

Derivation & Analysis:

An isomorphism is bijective, which implies that finite-dimensional vector spaces $V$ and $W$ must have the exact same dimension.

Q10

MCQ 10: Let $T$ be a linear operator on $\mathbb{R}^3$ whose matrix is $\begin{bmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 5 \end{bmatrix}$. The nullity of $T$ is:

A 0

B 1

C 2

D 3

Derivation & Analysis:

The determinant of the diagonal matrix is $2 \times 3 \times 5 = 30 \neq 0$. Thus, $T$ is invertible and full rank ($3$), making its nullity $3 - 3 = 0$.

Common questions

By definition, linearity requires $T(0v) = 0 T(v) = 0w$. If a transformation maps $0$ to a non-zero vector, it fails the homogeneity property immediately and cannot be linear.

Confusing the dimensions in the Rank-Nullity theorem—specifically subtracting rank from the codomain dimension rather than the domain dimension, or failing to verify both additivity and homogeneity simultaneously.

Master standard basis column-mapping shortcuts and inspect matrix determinants. If the determinant is non-zero, nullity is instantly zero and rank equals the full dimension, saving crucial minutes.

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