8.3 Linear Transformations, Kernel, Range, and the Rank-Nullity Theorem

8.3 Linear Transformations, Kernel, Range, and the Rank-Nullity Theorem

Linear transformations represent morphisms that preserve algebraic structure between vector spaces. On the GRE Subject Test in Mathematics, linear transformations are evaluated through their fundamental subspaces—kernel (null space) and range (image)—their injectivity and surjectivity characteristics, the Rank-Nullity Theorem, and composition algebra.

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Definition and Fundamental Properties

Let $V$ and $W$ be vector spaces over the same field $\mathbb{F}$. A function $T: V \to W$ is a linear transformation (or linear map) if for all $u, v \in V$ and $c \in \mathbb{F}$:

1. Additivity: $T(u + v) = T(u) + T(v)$.
2. Homogeneity: $T(c u) = c T(u)$. Equivalently, $T(c u + v) = c T(u) + T(v)$ for all $u, v \in V$ and $c \in \mathbb{F}$.

Elementary Consequences

• $T(0_V) = 0_W$. If $T(0_V) \neq 0_W$, $T$ is non-linear (e.g., affine shifts $T(x) = Ax + b$ with $b \neq 0$).
• $T(-v) = -T(v)$.
• Linear combinations are preserved: $T\left(\sum_{i=1}^k c_i v_i\right) = \sum_{i=1}^k c_i T(v_i)$.

Principle of Linear Extension

Let $\mathcal{B} = {v_1, \dots, v_n}$ be a basis of $V$, and let ${w_1, \dots, w_n}$ be arbitrary vectors in $W$. Then there exists a unique linear transformation $T: V \to W$ such that: $T(v_i) = w_i \quad \forall i \in \{1, \dots, n\}.$ A linear transformation is completely and unambiguously determined by its action on a basis.

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Kernel, Range, Injectivity, and Surjectivity

Kernel (Null Space)

The kernel of $T: V \to W$, denoted $\ker(T)$ or $\operatorname{null}(T)$, is the set of all vectors in $V$ mapped to zero: $\ker(T) = \{v \in V \mid T(v) = 0_W\}.$

• $\ker(T)$ is a subspace of the domain $V$.
• The nullity of $T$ is $\operatorname{nullity}(T) = \dim(\ker T)$.

Range (Image)

The range (or image) of $T: V \to W$, denoted $\operatorname{im}(T)$ or $\operatorname{range}(T)$, is the set of all outputs: $\operatorname{im}(T) = \{T(v) \mid v \in V\} = \{w \in W \mid \exists v \in V \text{ such that } T(v) = w\}.$

• $\operatorname{im}(T)$ is a subspace of the codomain $W$.
• If $\operatorname{span}({v_1, \dots, v_n}) = V$, then $\operatorname{im}(T) = \operatorname{span}({T(v_1), \dots, T(v_n)})$.
• The rank of $T$ is $\operatorname{rank}(T) = \dim(\operatorname{im} T)$.

Injectivity and Surjectivity Criteria

• Injective (One-to-One): $T$ is injective if and only if $\ker(T) = {0_V} \iff \operatorname{nullity}(T) = 0$.
• Surjective (Onto): $T$ is surjective if and only if $\operatorname{im}(T) = W \iff \operatorname{rank}(T) = \dim W$.
• Isomorphism (Bijective): $T$ is an isomorphism if it is both injective and surjective. Two finite-dimensional spaces over $\mathbb{F}$ are isomorphic ($V \cong W$) if and only if $\dim V = \dim W$.

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The Rank-Nullity Theorem

The Rank-Nullity Theorem (Fundamental Theorem of Linear Maps) bridges the dimensions of the kernel, image, and domain.

Theorem Statement

Let $V$ be a finite-dimensional vector space and let $T: V \to W$ be a linear transformation. Then: $\dim(\ker T) + \dim(\operatorname{im} T) = \dim V.$ $\operatorname{nullity}(T) + \operatorname{rank}(T) = \dim(\operatorname{Domain}).$

Key Proof Mechanism

Let ${u_1, \dots, u_k}$ be a basis of $\ker(T)$ (so $\operatorname{nullity}(T) = k$). By the Basis Extension Theorem, extend this to a basis ${u_1, \dots, u_k, v_1, \dots, v_r}$ of $V$, where $k + r = \dim V$. It follows that the set ${T(v_1), \dots, T(v_r)}$ is linearly independent and spans $\operatorname{im}(T)$, forming a basis of $\operatorname{im}(T)$. Thus $\operatorname{rank}(T) = r$, establishing $k + r = \dim V$.

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Endomorphisms and Dimensional Constraints

When $T: V \to V$ is an endomorphism (operator on $V$) with $\dim V = n < \infty$, the Rank-Nullity Theorem yields an equivalence: $T \text{ is injective} \iff T \text{ is surjective} \iff T \text{ is an isomorphism} \iff \ker(T) = \{0\} \iff \operatorname{rank}(T) = n.$

Dimensional Constraints Between Different Spaces

For any linear map $T: V \to W$ between finite-dimensional spaces:

1. Domain Larger than Codomain ($\dim V > \dim W$): $\operatorname{nullity}(T) = \dim V - \operatorname{rank}(T) \ge \dim V - \dim W > 0.$ Therefore, $T$ cannot be injective. There must exist a non-trivial kernel.
2. Domain Smaller than Codomain ($\dim V < \dim W$): $\operatorname{rank}(T) = \dim V - \operatorname{nullity}(T) \le \dim V < \dim W.$ Therefore, $T$ cannot be surjective. The image cannot fill $W$.

Infinite-Dimensional Failure

The equivalence between injectivity and surjectivity fails completely when $\dim V = \infty$.

• Right-Shift Operator on $\ell^2$: $S(x_1, x_2, \dots) = (0, x_1, x_2, \dots)$ is injective ($\ker S = {0}$) but not surjective ($(1, 0, 0, \dots) \notin \operatorname{im} S$).
• Left-Shift Operator on $\ell^2$: $L(x_1, x_2, \dots) = (x_2, x_3, \dots)$ is surjective but not injective ($L(1, 0, 0, \dots) = 0$).

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Composition of Linear Maps and Rank Inequalities

For linear maps $T: V \to W$ and $S: W \to U$:

• $S \circ T: V \to U$ is linear.
• Kernel Inclusion: $\ker(T) \subseteq \ker(S \circ T)$.
• Image Inclusion: $\operatorname{im}(S \circ T) \subseteq \operatorname{im}(S)$.
• Standard Rank Bound: $\operatorname{rank}(S \circ T) \le \min(\operatorname{rank}(S), \operatorname{rank}(T)).$
• Sylvester's Rank Inequality: $\operatorname{rank}(S \circ T) \ge \operatorname{rank}(S) + \operatorname{rank}(T) - \dim W.$

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Summary of Mapping Invariants

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Step-by-Step Worked Problems

Problem 1: Rank and Nullity of a Differential Map

Let $T: P_3(\mathbb{R}) \to P_2(\mathbb{R})$ be defined by $T(p(x)) = p'(x) - 2 p''(x)$. Find $\operatorname{nullity}(T)$ and $\operatorname{rank}(T)$.

Solution:

1. Domain is $P_3(\mathbb{R})$ with $\dim(P_3(\mathbb{R})) = 4$. Codomain is $P_2(\mathbb{R})$ with dimension 3.
2. Find $\ker(T)$: Let $p(x) = ax^3 + bx^2 + cx + d$.
   • $p'(x) = 3ax^2 + 2bx + c$.
   • $p''(x) = 6ax + 2b$.
   • $T(p) = 3ax^2 + (2b - 12a)x + (c - 4b) = 0$.
3. Equating coefficients:
   • $3a = 0 \implies a = 0$.
   • $2b - 12(0) = 0 \implies b = 0$.
   • $c - 4(0) = 0 \implies c = 0$.
   • The constant term $d$ is completely unrestricted!
4. Thus $\ker(T) = {d \mid d \in \mathbb{R}} = \operatorname{span}({1})$.
5. Hence $\operatorname{nullity}(T) = 1$.
6. By the Rank-Nullity Theorem: $\operatorname{rank}(T) = \dim(P_3(\mathbb{R})) - \operatorname{nullity}(T) = 4 - 1 = 3.$ Since $\operatorname{rank}(T) = 3 = \dim(P_2(\mathbb{R}))$, $T$ is surjective!

Problem 2: Operator Invertibility

Let $T: M_{2 \times 2}(\mathbb{R}) \to M_{2 \times 2}(\mathbb{R})$ be defined by $T(X) = X + X^T$. Determine whether $T$ is an isomorphism.

Solution:

1. Compute $\ker(T)$: $T(X) = 0 \iff X + X^T = 0 \iff X^T = -X$.
2. The kernel is precisely the subspace of $2 \times 2$ skew-symmetric matrices: $\ker(T) = \left\{ \begin{pmatrix} 0 & b \\ -b & 0 \end{pmatrix} \;\middle|\; b \in \mathbb{R} \right\}.$
3. Since non-zero skew-symmetric matrices exist (e.g., $b = 1$), $\dim(\ker T) = 1 \neq 0$.
4. Because $\operatorname{nullity}(T) = 1 > 0$, $T$ is not injective.
5. Because $M_{2 \times 2}(\mathbb{R})$ is finite-dimensional, failure of injectivity immediately implies failure of surjectivity. Thus $T$ is not an isomorphism.

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GRE Exam Traps & Pitfalls

Trap 1: Applying Rank-Nullity to the Codomain Dimension A frequent blunder is setting $\operatorname{rank} + \operatorname{nullity} = \dim W$. The theorem strictly asserts $\operatorname{rank}(T) + \operatorname{nullity}(T) = \dim V$ (the dimension of the domain).

Trap 2: Assuming $S \circ T = 0$ Implies $S = 0$ or $T = 0$ In linear algebra, $S \circ T = 0$ simply means $\operatorname{im}(T) \subseteq \ker(S)$. Neither operator needs to be the zero operator.

Trap 3: Injectivity Equivalence in Infinite Dimensions On GRE questions involving function spaces ($C[0, 1]$ or $\ell^2$), never assume injectivity implies surjectivity. The shift operators and integration operators provide standard counterexamples.