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7. Linear Algebra

Kernel of linear transformation

The kernel (or null space) of a linear transformation is the subspace of the domain that maps to the zero vector, serving as a fundamental measure of the transformation's non-injectivity.

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

Definition of Kernel & The 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. The kernel of $T$, denoted as $\ker(T)$ or $\mathcal{N}(T)$, is defined by $\ker(T) = \{v \in V \mid T(v) = 0_W\}$. Furthermore, if $V$ is finite-dimensional, then $\dim(\ker(T)) + \dim(\text{range}(T)) = \dim(V)$, where $\dim(\ker(T))$ is the nullity and $\dim(\text{range}(T))$ is the rank of $T$.

📐 Geometric Interpretation

Geometrically, the kernel represents all vectors in the domain space $V$ that are collapsed, squashed, or projected directly onto the origin of the codomain $W$. For instance, a projection from $\mathbb{R}^3$ onto the $xy$-plane has a kernel consisting of the entire $z$-axis.

⚙️ Algebraic & Structural Meaning

Algebraically, $\ker(T)$ is always a subspace of the domain $V$. It dictates the injectivity (one-to-one nature) of $T$: $T$ is injective if and only if $\ker(T) = \{0_V\}$, providing an immediate algebraic test for invertibility when $\dim(V) = \dim(W)$.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition Domain / Interval Necessity What Fails If Condition Dropped
Linearity of Transformation ($T$) Domain/Codomain Vector spaces $V, W$ over field $F$ Strictly Mandatory / Necessary If $T$ is not linear (e.g., $T(x) = x^2$ or $T(x) = x + 1$), the set of solutions to $T(v) = 0$ does not form a vector subspace, and the Rank-Nullity Theorem completely fails.
Finite-Dimensionality of Domain ($V$) $\dim(V) < \infty$ Necessary for the numerical Rank-Nullity dimension relation In infinite-dimensional spaces (like polynomial spaces $\mathcal{P}(\mathbb{R})$ or function spaces), dimension addition fails, although the kernel remains a well-defined subspace.
Zero Preservation Property $T(0_V) = 0_W$ Essential consistency check If a purported transformation fails to map the zero vector of the domain to the zero vector of the codomain, it cannot be linear, rendering the kernel undefined or trivial.

Useful results and shortcuts

Use these identities and results when their conditions apply:

01

Injectivity via Trivial Kernel Test

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

Rule: To test if a linear operator is injective, set the general transformation expression equal to zero and solve. If only the zero vector emerges as a solution, nullity is $0$.

Application: Directly asked in UP PGT and KVS to determine whether given matrix transformations or differential operators are injective.
02

Rank-Nullity Dimension Balancing

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

Rule: Whenever asked for nullity of a matrix or transformation, compute the rank (number of non-zero rows in row-echelon form or linearly independent columns) and subtract it from the dimension of the domain.

03

Homogeneous System Equivalence

$$ \ker(T_A) = \{X \in F^n \mid AX = 0\} $$

Rule: For a matrix transformation $T_A(X) = AX$, finding the kernel is identical to solving the homogeneous linear system $AX = 0$. The nullity equals the number of free variables (dimension of null space).

04

Differential Operator Kernel (Annihilator Approach)

$$ \text{For } D: P_n(\mathbb{R}) \to P_n(\mathbb{R}), \; \ker(D) = \text{Span}\{1\} \implies \text{Nullity} = 1 $$

Rule: For differentiation operators on polynomial spaces, the kernel consists strictly of constant polynomials. The nullity equals the order of the highest derivative annihilated or $1$ for a single derivative.

Application: Frequent in DSSSB PGT Mathematics questions involving polynomial differential spaces.
05

Matrix Nullity via Row Reduction

$$ \text{Nullity}(A) = n - \text{rank}(A) \quad (A \text{ is } m \times n) $$

Rule: Do not find the basis of the kernel unless explicitly asked. Simply reduce $A$ to row echelon form, count non-zero pivot rows to get the rank, and subtract from column count $n$.

06

Rapid Option Elimination via Rank-Nullity Bounds

$$ \max(0, \dim(V) - \dim(W)) \le \text{nullity}(T) \le \dim(V) $$

Rule: Instantly eliminate options where nullity exceeds the dimension of the domain space or violates lower bounds imposed by codomain dimensions.

Application: Powerful mental check for theoretical multiple-choice questions.

Common errors and counterexamples

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

⚠️ COMMON ERROR #01

Confusing Range Dimension with Kernel Dimension

Common Pitfall: Candidates calculate the rank (number of pivot columns) and mistakenly report it as the nullity of the transformation.

Counterexample: For a $3 \times 3$ matrix of rank $2$, nullity is $3 - 2 = 1$, NOT $2$.

Why it fails: Forgetting to subtract the rank from the total dimension of the domain vector space $\dim(V)$.

🛡️ Exam Defense: Always write down $\dim(\ker) + \dim(\text{range}) = \dim(V)$ on your scratchpad before selecting the final numerical option.
⚠️ COMMON ERROR #02

Assuming Non-Zero Kernel Implies Non-Linearity

Common Pitfall: Believing that if $\ker(T)$ contains non-zero vectors, the transformation cannot be linear.

Counterexample: The zero transformation $T(v) = 0$ has $\ker(T) = V$, yet it is perfectly linear.

Why it fails: Confusing injectivity with linearity. Linearity requires preservation of vector addition and scalar multiplication, not a zero kernel.

🛡️ Exam Defense: Linearity is a structural property of the mapping rule itself; kernel size only dictates injectivity.

Comparison

Concept / TheoremHypotheses / AssumptionsKey Formula / OutcomeGeometric MeaningExam Identification Clue
Kernel of $T$ ($\ker T$)Linear transformation $T: V \to W$$\{v \in V \mid T(v) = 0_W\}$Subspace collapsed to the originKeywords: 'null space', 'mapped to zero', 'solutions to $T(x)=0$'
Rank-Nullity TheoremFinite-dimensional domain $V$$\dim(\ker T) + \dim(\text{range } T) = \dim V$Domain dimension partitioned into rank and nullityKeywords: 'find nullity given rank', 'dimension of domain'
Injective (One-to-One)Linear transformation $T: V \to W$$\ker T = \{0_V\}$No two distinct vectors collapse to the same imageKeywords: 'one-to-one', 'trivial kernel', 'monomorphism'
Matrix Null SpaceMatrix $A$ of order $m \times n$$\text{Nullity} = n - \text{rank}(A)$Solution space of homogeneous system $AX = 0$Keywords: 'homogeneous system', 'nullity of 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 $T(x, y, z) = (x + y + z, 2x + y - z, x + 2y + 3z)$. Find the dimension of the kernel of $T$ (nullity of $T$).

(A) 0(B) 1(C) 2(D) 3
✓ Correct Answer: (B)
📚 Method A: Formal Textbook Derivation
Step 1: Write the matrix $A$ associated with $T$: $$A = \begin{pmatrix} 1 & 1 & 1 \\ 2 & 1 & -1 \\ 1 & 2 & 3 \end{pmatrix}$$. Step 2: Perform elementary row operations to find the rank. $R_2 \to R_2 - 2R_1$, $R_3 \to R_3 - R_1$: $$A \sim \begin{pmatrix} 1 & 1 & 1 \\ 0 & -1 & -3 \\ 0 & 1 & 2 \end{pmatrix}$$. Step 3: $R_3 \to R_3 + R_2$: $$A \sim \begin{pmatrix} 1 & 1 & 1 \\ 0 & -1 & -3 \\ 0 & 0 & -1 \end{pmatrix}$$. Step 4: There are 3 non-zero pivot rows, so $\text{rank}(T) = 3$. Step 5: By Rank-Nullity, $\text{Nullity}(T) = \dim(\mathbb{R}^3) - \text{rank}(T) = 3 - 3 = 0$. Wait, recalculate $R_3$: row 3 is $(1, 2, 3)$, minus row 1 $(1, 1, 1)$ gives $(0, 1, 2)$. Adding row 2 $(0, -1, -3)$ gives $(0, 0, -1)$. Let's check determinant: $\det(A) = 1(3 - (-2)) - 1(6 - (-1)) + 1(4 - 1) = 5 - 7 + 3 = 1 \neq 0$. Thus $\text{rank} = 3$, nullity $= 0$.
⚡ Method B: 25-Second Exam Shortcut
Since the determinant of the transformation matrix is non-zero ($\det A = 1 \neq 0$), the transformation is invertible and non-singular, meaning $\ker(T) = \{0_V\}$ and nullity is instantly $0$.
💡 Tip: Always check the determinant for square matrices before embarking on full row reduction.
CANONICAL PROBLEM 02 Dual-Method Breakdown

Let $T: \mathbb{R}^4 \to \mathbb{R}^3$ be a linear transformation. If the rank of $T$ is $2$, what is the nullity of $T$?

(A) 1(B) 2(C) 3(D) 4
✓ Correct Answer: (B)
📚 Method A: Formal Textbook Derivation
Step 1: Identify the domain dimension $\dim(\mathbb{R}^4) = 4$. Step 2: Apply the Rank-Nullity Theorem: $\dim(\ker T) + \text{rank}(T) = \dim(\text{domain})$. Step 3: $\text{Nullity}(T) + 2 = 4 \implies \text{Nullity}(T) = 2$.
⚡ Method B: 25-Second Exam Shortcut
Direct subtraction: $4 - 2 = 2$. Takes 5 seconds.
💡 Tip: Watch out whether the domain or codomain dimension is given in the problem statement.

Practice questions

Choose an answer to check it and see the solution:

Q1
If $T: V \to W$ is a linear transformation and $\dim(V) = 5$, $\dim(W) = 7$, what is the maximum possible value for the nullity of $T$?
A 3
B 5
C 7
D 2
Derivation & Analysis:

The kernel of $T$ is a subspace of the domain $V$. Therefore, $\dim(\ker(T)) \le \dim(V) = 5$. The maximum possible nullity is the dimension of the domain, which is $5$.

Q2
Let $T: \mathbb{R}^3 \to \mathbb{R}^3$ be defined by $T(x, y, z) = (x - y, y - z, z - x)$. The dimension of $\ker(T)$ is:
A 0
B 1
C 2
D 3
Derivation & Analysis:

Set $T(x,y,z) = (0,0,0) \implies x - y = 0, y - z = 0, z - x = 0 \implies x = y = z$. Thus, $\ker(T) = \{(t, t, t) \mid t \in \mathbb{R}\} = \text{Span}\{(1, 1, 1)\}$. The dimension of this subspace is $1$.

Q3
Which of the following conditions is equivalent to a linear transformation $T: V \to W$ being one-to-one?
A $\ker(T) = V$
B $\ker(T) = \{0_V\}$
C $\text{range}(T) = \{0_W\}$
D $\dim(\ker(T)) = \dim(V)$
Derivation & Analysis:

By definition, $T(u) = T(v) \implies T(u - v) = 0 \implies u - v \in \ker(T)$. For $T$ to be injective, $u - v = 0 \implies u = v$, which requires $\ker(T) = \{0_V\}$.

Q4
If $A$ is a $4 \times 6$ matrix of rank $3$, what is the nullity of the corresponding linear transformation?
A 1
B 2
C 3
D 4
Derivation & Analysis:

The domain dimension for a $4 \times 6$ matrix transformation is the number of columns, which is $6$. By Rank-Nullity, $\text{Nullity} = 6 - \text{rank}(A) = 6 - 3 = 3$.

Q5
Let $P_n(\mathbb{R})$ be the vector space of polynomials of degree at most $n$. If $D: P_3(\mathbb{R}) \to P_3(\mathbb{R})$ is the differentiation operator $D(p(x)) = p'(x)$, then the nullity of $D$ is:
A 0
B 1
C 2
D 3
Derivation & Analysis:

The kernel of the derivative operator consists of polynomials whose derivative is identically zero, i.e., constant polynomials $p(x) = c$. The space of constants has dimension $1$.

Q6
Let $T: V \to W$ be a linear transformation between two vector spaces of equal finite dimension $\dim(V) = \dim(W) = n$. Then $T$ is onto if and only if:
A $\ker(T) \neq \{0\}$
B $\ker(T) = \{0\}$
C $\text{rank}(T) = 0$
D $\text{nullity}(T) = n$
Derivation & Analysis:

By the Rank-Nullity Theorem, $\text{rank}(T) + \text{nullity}(T) = n$. For $T$ to be onto, $\text{rank}(T) = n$, which forces $\text{nullity}(T) = 0$, meaning $\ker(T) = \{0\}$.

Q7
What is the kernel of the zero transformation $T: V \to W$ defined by $T(v) = 0_W$ for all $v \in V$?
A $\{0_V\}$
B $\emptyset$
C $V$
D $W$
Derivation & Analysis:

Every vector in $V$ is mapped to $0_W$ by definition of the zero transformation. Therefore, $\ker(T) = \{v \in V \mid T(v) = 0_W\} = V$.

Q8
If $T_1: U \to V$ and $T_2: V \to W$ are linear transformations, then for their composition $T_2 \circ T_1$, which of the following holds for their kernels?
A $\ker(T_1) \subseteq \ker(T_2 \circ T_1)$
B $\ker(T_2 \circ T_1) \subseteq \ker(T_1)$
C $\ker(T_1) = \ker(T_2 \circ T_1)$
D $\ker(T_2) \subseteq \ker(T_2 \circ T_1)$
Derivation & Analysis:

If $u \in \ker(T_1)$, then $T_1(u) = 0$. Then $(T_2 \circ T_1)(u) = T_2(T_1(u)) = T_2(0) = 0$, which implies $u \in \ker(T_2 \circ T_1)$. Thus, $\ker(T_1) \subseteq \ker(T_2 \circ T_1)$.

Q9
Let $T: \mathbb{R}^2 \to \mathbb{R}^2$ be a reflection about the x-axis. What is the dimension of its kernel?
A 0
B 1
C 2
D Undefined
Derivation & Analysis:

Reflection is an invertible linear isomorphism ($T^{-1} = T$). Invertible operators are injective, so their kernel is trivial: $\ker(T) = \{(0,0)\}$ with dimension $0$.

Q10
If the nullity of a linear transformation $T: \mathbb{R}^5 \to \mathbb{R}^4$ is $2$, what is the rank of $T$?
A 1
B 2
C 3
D 4
Derivation & Analysis:

Using the Rank-Nullity Theorem: $\text{rank}(T) + \text{nullity}(T) = \dim(\text{domain}) \implies \text{rank}(T) + 2 = 5 \implies \text{rank}(T) = 3$.

Common questions

No. By definition, every linear transformation must map the zero vector of the domain to the zero vector of the codomain ($T(0_V) = 0_W$). Hence, $0_V$ always belongs to $\ker(T)$, meaning the kernel is at least the trivial subspace $\{0_V\}$ and can never be empty.

Candidates often calculate the number of pivot columns (rank) during matrix reduction and mistakenly select that number as the answer when the examiner asked for nullity. Always remember the balance equation: $\text{Rank} + \text{Nullity} = \text{Dimension of Domain}$.

Reduce the matrix to row echelon form using elementary row operations, count the number of non-zero rows (rank $r$), and subtract it directly from the total number of columns $n$ using $\text{Nullity} = n - r$.

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