Chapter 5 • Theory & Derivations

Unit 5: Linear Transformations, Kernel, Image & Change of Basis

Linear mappings between vector spaces, kernel and image subspaces, injectivity and surjectivity, the Abstract Rank-Nullity Theorem, matrix representations with respect to arbitrary bases, transition matrices, similarity transformations, and geometric operators.

§5.1 Linear Transformations, Kernel and Image

1. Definition of a Linear Transformation

Let $V$ and $W$ be vector spaces over the same scalar field $F$. A function $T: V \to W$ is a Linear Transformation (or Linear Operator if $V = W$) if for all $\vec{u}, \vec{v} \in V$ and all scalars $c \in F$:

1. Additivity: $T(\vec{u} + \vec{v}) = T(\vec{u}) + T(\vec{v})$

2. Homogeneity (Scalar Multiplication): $T(c\vec{u}) = c T(\vec{u})$

Equivalently, $T$ preserves arbitrary linear combinations:

T(c_1 \vec{v}_1 + c_2 \vec{v}_2) = c_1 T(\vec{v}_1) + c_2 T(\vec{v}_2)

Elementary Properties:

$T(\vec{0}_V) = \vec{0}_W$, since $T(\vec{0}) = T(0\vec{0}) = 0 T(\vec{0}) = \vec{0}$.
$T(-\vec{v}) = -T(\vec{v})$.
$T\left(\sum_{i=1}^k c_i \vec{v}_i\right) = \sum_{i=1}^k c_i T(\vec{v}_i)$.

2. Kernel (Null Space) and Image (Range)

The Kernel:

The Kernel (or Null Space) of $T$, denoted by $\ker(T)$ or $\text{Null}(T)$, is the set of all vectors in $V$ mapped to the zero vector of $W$:

\ker(T) = \{\vec{v} \in V : T(\vec{v}) = \vec{0}_W\}

The Image (Range):

The Image (or Range) of $T$, denoted by $\text{im}(T)$ or $T(V)$, is the set of all outputs in $W$:

\text{im}(T) = \{\vec{w} \in W : \exists \vec{v} \in V \text{ such that } T(\vec{v}) = \vec{w}\} = \{T(\vec{v}) : \vec{v} \in V\}

Theorem 5.1 (Kernel and Image are Subspaces):

$\ker(T)$ is a subspace of $V$.
$\text{im}(T)$ is a subspace of $W$.

Proof:

For $\ker(T)$: $T(\vec{0}_V) = \vec{0}_W$, so $\vec{0}_V \in \ker(T)$. If $\vec{u}, \vec{v} \in \ker(T)$ and $c, d \in F$, then $T(c\vec{u} + d\vec{v}) = c T(\vec{u}) + d T(\vec{v}) = c\vec{0}_W + d\vec{0}_W = \vec{0}_W$. Thus $c\vec{u} + d\vec{v} \in \ker(T)$.
For $\text{im}(T)$: $\vec{0}_W = T(\vec{0}_V) \in \text{im}(T)$. If $\vec{w}_1, \vec{w}_2 \in \text{im}(T)$, there exist $\vec{v}_1, \vec{v}_2 \in V$ such that $T(\vec{v}_1) = \vec{w}_1$ and $T(\vec{v}_2) = \vec{w}_2$. Then for $c, d \in F$, $c\vec{w}_1 + d\vec{w}_2 = c T(\vec{v}_1) + d T(\vec{v}_2) = T(c\vec{v}_1 + d\vec{v}_2) \in \text{im}(T)$. $\blacksquare$

3. Injectivity, Surjectivity and Isomorphism

$T$ is Injective (One-to-One) if $T(\vec{u}) = T(\vec{v}) \implies \vec{u} = \vec{v}$.
$T$ is Surjective (Onto) if $\text{im}(T) = W$.
$T$ is an Isomorphism (Bijective) if it is both injective and surjective.

Theorem 5.2 (Kernel Criterion for Injectivity):

A linear transformation $T: V \to W$ is injective if and only if its kernel is trivial:

\ker(T) = \{\vec{0}_V\}

Proof: $(\implies)$ If $T$ is injective, since $T(\vec{0}_V) = \vec{0}_W$, any $\vec{v} \in \ker(T)$ satisfies $T(\vec{v}) = T(\vec{0}_V) \implies \vec{v} = \vec{0}_V$. $(\impliedby)$ Suppose $\ker(T) = \{\vec{0}_V\}$. If $T(\vec{u}) = T(\vec{v})$, then by linearity $T(\vec{u} - \vec{v}) = T(\vec{u}) - T(\vec{v}) = \vec{0}_W$. This implies $\vec{u} - \vec{v} \in \ker(T) = \{\vec{0}_V\}$, so $\vec{u} - \vec{v} = \vec{0} \implies \vec{u} = \vec{v}$. Thus $T$ is injective. $\blacksquare$

§5.2 The Abstract Rank-Nullity Theorem & Matrix Representations

1. The Abstract Rank-Nullity Theorem (Dimension Theorem)

The Nullity of $T$ is $\text{nullity}(T) = \dim(\ker(T))$. The Rank of $T$ is $\text{rank}(T) = \dim(\text{im}(T))$.

Theorem 5.3 (The Rank-Nullity Theorem for Linear Maps):

Let $V$ and $W$ be vector spaces over $F$, with $V$ finite-dimensional. For any linear transformation $T: V \to W$:

\dim(\ker(T)) + \dim(\text{im}(T)) = \dim(V)

Rigorous Proof: Let $n = \dim(V)$, and let $k = \dim(\ker(T)) = \text{nullity}(T)$. Choose an ordered basis $\mathcal{B}_K = \{\vec{u}_1, \vec{u}_2, \dots, \vec{u}_k\}$ for the subspace $\ker(T)$. By the Basis Extension Theorem, extend $\mathcal{B}_K$ to an ordered basis $\mathcal{B}$ for the entire space $V$:

\mathcal{B} = \{\vec{u}_1, \dots, \vec{u}_k, \; \vec{v}_1, \dots, \vec{v}_{n-k}\}

We will prove that the set $\mathcal{S} = \{T(\vec{v}_1), T(\vec{v}_2), \dots, T(\vec{v}_{n-k})\} \subseteq W$ is a basis for $\text{im}(T)$.

1. $\mathcal{S}$ spans $\text{im}(T)$:

Let $\vec{w} \in \text{im}(T)$. Then $\vec{w} = T(\vec{x})$ for some $\vec{x} \in V$. Express $\vec{x}$ in terms of the basis $\mathcal{B}$:

\vec{x} = \sum_{i=1}^k a_i \vec{u}_i + \sum_{j=1}^{n-k} b_j \vec{v}_j

Applying the linear transformation $T$:

\vec{w} = T(\vec{x}) = \sum_{i=1}^k a_i T(\vec{u}_i) + \sum_{j=1}^{n-k} b_j T(\vec{v}_j)

Since $\vec{u}_i \in \ker(T)$, $T(\vec{u}_i) = \vec{0}_W$. Therefore:

\vec{w} = \sum_{j=1}^{n-k} b_j T(\vec{v}_j) \in \text{span}(\mathcal{S})

Thus $\mathcal{S}$ spans $\text{im}(T)$.

2. $\mathcal{S}$ is linearly independent in $W$:

Suppose there exist scalars $c_1, c_2, \dots, c_{n-k} \in F$ such that:

\sum_{j=1}^{n-k} c_j T(\vec{v}_j) = \vec{0}_W

By linearity of $T$:

T\left( \sum_{j=1}^{n-k} c_j \vec{v}_j \right) = \vec{0}_W \implies \sum_{j=1}^{n-k} c_j \vec{v}_j \in \ker(T)

Since $\mathcal{B}_K = \{\vec{u}_1, \dots, \vec{u}_k\}$ is a basis for $\ker(T)$, there exist scalars $d_1, \dots, d_k$ such that:

\sum_{j=1}^{n-k} c_j \vec{v}_j = \sum_{i=1}^k d_i \vec{u}_i \implies \sum_{i=1}^k (-d_i)\vec{u}_i + \sum_{j=1}^{n-k} c_j \vec{v}_j = \vec{0}_V

Since $\mathcal{B} = \{\vec{u}_1, \dots, \vec{u}_k, \vec{v}_1, \dots, \vec{v}_{n-k}\}$ is a basis for $V$, it is linearly independent! Therefore, all coefficients must be zero: $c_1 = c_2 = \cdots = c_{n-k} = 0$ (and $d_1 = \cdots = d_k = 0$). This proves that $\mathcal{S}$ is linearly independent.

Hence $\mathcal{S}$ is a basis for $\text{im}(T)$, which implies:

\dim(\text{im}(T)) = |\mathcal{S}| = n - k = \dim(V) - \dim(\ker(T))

Rearranging gives $\dim(\ker(T)) + \dim(\text{im}(T)) = \dim(V)$. $\blacksquare$

2. Matrix Representation of a Linear Transformation

Let $V$ have ordered basis $\mathcal{B} = \{\vec{v}_1, \dots, \vec{v}_n\}$ and $W$ have ordered basis $\mathcal{C} = \{\vec{w}_1, \dots, \vec{w}_m\}$. The Matrix of $T$ with respect to bases $\mathcal{B}$ and $\mathcal{C}$, denoted by $[T]_{\mathcal{B}}^{\mathcal{C}} \in M_{m \times n}(F)$, has column $j$ given by the coordinate vector of $T(\vec{v}_j)$ relative to $\mathcal{C}$:

[T]_{\mathcal{B}}^{\mathcal{C}} = \begin{pmatrix} [T(\vec{v}_1)]_{\mathcal{C}} & [T(\vec{v}_2)]_{\mathcal{C}} & \cdots & [T(\vec{v}_n)]_{\mathcal{C}} \end{pmatrix}

The Fundamental Commutative Property:

For every vector $\vec{x} \in V$:

[T(\vec{x})]_{\mathcal{C}} = [T]_{\mathcal{B}}^{\mathcal{C}} \, [\vec{x}]_{\mathcal{B}}

§5.3 Change of Basis, Transition Matrices & Similar Operators

1. Change of Basis and Transition Matrix

Let $\mathcal{B} = \{\vec{v}_1, \dots, \vec{v}_n\}$ and $\mathcal{B}' = \{\vec{v}_1', \dots, \vec{v}_n'\}$ be two ordered bases for vector space $V$. The Change of Basis Matrix (Transition Matrix) from $\mathcal{B}'$ to $\mathcal{B}$, denoted by $P_{\mathcal{B} \leftarrow \mathcal{B}'}$ (or simply $P$), transforms coordinates relative to $\mathcal{B}'$ into coordinates relative to $\mathcal{B}$:

[\vec{x}]_{\mathcal{B}} = P_{\mathcal{B} \leftarrow \mathcal{B}'} [\vec{x}]_{\mathcal{B}'}

The $j$-th column of $P$ is the coordinate vector of the $j$-th basis vector of $\mathcal{B}'$ expressed in the basis $\mathcal{B}$:

P_{\mathcal{B} \leftarrow \mathcal{B}'} = \begin{pmatrix} [\vec{v}_1']_{\mathcal{B}} & [\vec{v}_2']_{\mathcal{B}} & \cdots & [\vec{v}_n']_{\mathcal{B}} \end{pmatrix}

Because $\mathcal{B}'$ is a basis, $P$ is always an invertible matrix:

P_{\mathcal{B}' \leftarrow \mathcal{B}} = (P_{\mathcal{B} \leftarrow \mathcal{B}'})^{-1}

2. Transformation Under Change of Basis: Similarity

Let $T: V \to V$ be a linear operator on $V$. Let $A = [T]_{\mathcal{B}}$ be the matrix representation of $T$ relative to basis $\mathcal{B}$, and let $B = [T]_{\mathcal{B}'}$ be the representation relative to basis $\mathcal{B}'$.

Theorem 5.4 (Similarity Transformation Theorem):

The matrices $A$ and $B$ are similar, related by the transition matrix $P = P_{\mathcal{B} \leftarrow \mathcal{B}'}$:

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

Proof: For any $\vec{x} \in V$, coordinate transformation gives $[\vec{x}]_{\mathcal{B}} = P [\vec{x}]_{\mathcal{B}'}$. Applying $T$:

[T(\vec{x})]_{\mathcal{B}} = [T]_{\mathcal{B}} [\vec{x}]_{\mathcal{B}} = A P [\vec{x}]_{\mathcal{B}'}

Also, $[T(\vec{x})]_{\mathcal{B}} = P [T(\vec{x})]_{\mathcal{B}'} = P B [\vec{x}]_{\mathcal{B}'}$. Equating both expressions for all $[\vec{x}]_{\mathcal{B}'} \in F^n$:

P B = A P \implies B = P^{-1} A P

The proof is complete. $\blacksquare$

3. Canonical Geometric Linear Operators in $\mathbb{R}^2$ and $\mathbb{R}^3$

1. Rotation in $\mathbb{R}^2$ by Angle $\theta$:

R_\theta = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}

2. Reflection in $\mathbb{R}^2$ across line at angle $\theta/2$:

H_\theta = \begin{pmatrix} \cos\theta & \sin\theta \\ \sin\theta & -\cos\theta \end{pmatrix}

3. Horizontal Shear by Factor $k$:

S_k = \begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}

4. Orthogonal Projection onto Unit Vector $\vec{u}$:

P_{\vec{u}} = \vec{u}\vec{u}^T = \begin{pmatrix} u_1^2 & u_1 u_2 & u_1 u_3 \\ u_1 u_2 & u_2^2 & u_2 u_3 \\ u_1 u_3 & u_2 u_3 & u_3^2 \end{pmatrix}

TIERED UNIVERSITY HONORS PROBLEMS

Step-by-Step Solved Examination Problems

Comprehensive analytical derivations, multi-tier solutions (Foundational, Intermediate Exam, and Honors/Proof Challenge) with complete line-by-line verification.

Tier 1: Foundational Example 5.1: Matrix Representation of Derivative Operator & Rank-Nullity

Let $V = \mathbb{P}_3(\mathbb{R})$ be the space of polynomials of degree at most 3, and $W = \mathbb{P}_2(\mathbb{R})$ be the space of polynomials of degree at most 2. Let $D: V \to W$ be the differentiation operator defined by $D(p(t)) = p'(t)$.

(a) Let $\mathcal{B} = \{1, t, t^2, t^3\}$ be the standard ordered basis of $V$, and $\mathcal{C} = \{1, t, t^2\}$ be the standard ordered basis of $W$. Compute the matrix representation $[D]_{\mathcal{B}}^{\mathcal{C}}$. (b) Find bases for $\ker(D)$ and $\text{im}(D)$. (c) Verify the Rank-Nullity Theorem explicitly: $\dim(\ker(D)) + \dim(\text{im}(D)) = \dim(\mathbb{P}_3(\mathbb{R}))$.

Step 1: Compute images of basis vectors of $\mathcal{B}$:

$D(1) = 0 = 0(1) + 0(t) + 0(t^2) \implies [D(1)]_{\mathcal{C}} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$
$D(t) = 1 = 1(1) + 0(t) + 0(t^2) \implies [D(t)]_{\mathcal{C}} = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$
$D(t^2) = 2t = 0(1) + 2(t) + 0(t^2) \implies [D(t^2)]_{\mathcal{C}} = \begin{pmatrix} 0 \\ 2 \\ 0 \end{pmatrix}$
$D(t^3) = 3t^2 = 0(1) + 0(t) + 3(t^2) \implies [D(t^3)]_{\mathcal{C}} = \begin{pmatrix} 0 \\ 0 \\ 3 \end{pmatrix}$

Assemble the matrix $[D]_{\mathcal{B}}^{\mathcal{C}} \in M_{3 \times 4}(\mathbb{R})$:

[D]_{\mathcal{B}}^{\mathcal{C}} = \begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 3 \end{pmatrix}

Step 2: Find bases for $\ker(D)$ and $\text{im}(D)$:

Kernel $\ker(D)$:

$p(t) \in \ker(D) \iff p'(t) = 0 \iff p(t) = c$ (constant polynomials). In coordinate form: $[D]_{\mathcal{B}}^{\mathcal{C}} \vec{c} = \vec{0} \implies c_2 = c_3 = c_4 = 0$, with $c_1$ free. Basis for $\ker(D)$: $\mathcal{B}_{\ker} = \{1\}$. Thus $\dim(\ker(D)) = 1$.

Image $\text{im}(D)$:

The columns of $[D]_{\mathcal{B}}^{\mathcal{C}}$ span $\mathbb{R}^3$. Columns 2, 3, 4 are linearly independent. Corresponding polynomials in $W$: $\{1, 2t, 3t^2\}$, or equivalently standard $\{1, t, t^2\}$. Basis for $\text{im}(D)$: $\mathcal{B}_{\text{im}} = \{1, t, t^2\} = \mathbb{P}_2(\mathbb{R})$. Thus $\dim(\text{im}(D)) = 3$.

Step 3: Verification of the Rank-Nullity Theorem:

\dim(\ker(D)) + \dim(\text{im}(D)) = 1 + 3 = 4 = \dim(\mathbb{P}_3(\mathbb{R}))

The theorem is verified.

Final Answer & Physical Insight

$[D]_{\mathcal{B}}^{\mathcal{C}} = \begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 3 \end{pmatrix}$. Basis for $\ker(D)$ is $\{1\}$ ($\dim=1$); basis for $\text{im}(D)$ is $\{1, t, t^2\}$ ($\dim=3$). Rank-Nullity verified: $1 + 3 = 4$.

Tier 2: Intermediate Exam Example 5.2: Change of Basis Transition Matrix & Similarity Transformation

Let $T: \mathbb{R}^2 \to \mathbb{R}^2$ be the linear operator defined by $T(x_1, x_2) = (3x_1 + x_2, \; x_1 + 3x_2)$. Let $\mathcal{E} = \{\vec{e}_1, \vec{e}_2\} = \left\{ \begin{pmatrix} 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \end{pmatrix} \right\}$ be the standard basis, and let:

\mathcal{B} = \{\vec{v}_1, \vec{v}_2\} = \left\{ \begin{pmatrix} 1 \\ 1 \end{pmatrix}, \; \begin{pmatrix} 1 \\ -1 \end{pmatrix} \right\}

(a) Write the matrix $A = [T]_{\mathcal{E}}$ of $T$ relative to the standard basis $\mathcal{E}$. (b) Find the transition matrix $P = P_{\mathcal{E} \leftarrow \mathcal{B}}$ from $\mathcal{B}$ to $\mathcal{E}$, and compute its inverse $P^{-1}$. (c) Compute the matrix $B = [T]_{\mathcal{B}}$ relative to the basis $\mathcal{B}$ directly from the definition. (d) Verify the similarity relation $B = P^{-1} A P$, and explain why $B$ is diagonal.

Step 1: Standard matrix $A = [T]_{\mathcal{E}}$:

$T(\vec{e}_1) = T(1, 0) = (3, 1)^T$
$T(\vec{e}_2) = T(0, 1) = (1, 3)^T$

A = [T]_{\mathcal{E}} = \begin{pmatrix} 3 & 1 \\ 1 & 3 \end{pmatrix}

Step 2: Transition matrix $P$ and $P^{-1}$: The columns of $P$ are the coordinates of the vectors of $\mathcal{B}$ in $\mathcal{E}$:

P = \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}

Determinant: $\det(P) = 1(-1) - 1(1) = -2$. Inverse matrix formula $P^{-1} = \frac{1}{\det(P)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$:

P^{-1} = -\frac{1}{2} \begin{pmatrix} -1 & -1 \\ -1 & 1 \end{pmatrix} = \frac{1}{2} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}

Step 3: Direct computation of $B = [T]_{\mathcal{B}}$:

Evaluate $T(\vec{v}_1)$:

T\begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 3(1) + 1 \\ 1 + 3(1) \end{pmatrix} = \begin{pmatrix} 4 \\ 4 \end{pmatrix} = 4 \begin{pmatrix} 1 \\ 1 \end{pmatrix} + 0 \begin{pmatrix} 1 \\ -1 \end{pmatrix} = 4\vec{v}_1 \implies [T(\vec{v}_1)]_{\mathcal{B}} = \begin{pmatrix} 4 \\ 0 \end{pmatrix}

Evaluate $T(\vec{v}_2)$:

T\begin{pmatrix} 1 \\ -1 \end{pmatrix} = \begin{pmatrix} 3(1) - 1 \\ 1 - 3 \end{pmatrix} = \begin{pmatrix} 2 \\ -2 \end{pmatrix} = 0 \begin{pmatrix} 1 \\ 1 \end{pmatrix} + 2 \begin{pmatrix} 1 \\ -1 \end{pmatrix} = 2\vec{v}_2 \implies [T(\vec{v}_2)]_{\mathcal{B}} = \begin{pmatrix} 0 \\ 2 \end{pmatrix}

Therefore:

B = [T]_{\mathcal{B}} = \begin{pmatrix} 4 & 0 \\ 0 & 2 \end{pmatrix}

Step 4: Verify similarity transformation $P^{-1} A P$:

A P = \begin{pmatrix} 3 & 1 \\ 1 & 3 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} = \begin{pmatrix} 3(1)+1(1) & 3(1)+1(-1) \\ 1(1)+3(1) & 1(1)+3(-1) \end{pmatrix} = \begin{pmatrix} 4 & 2 \\ 4 & -2 \end{pmatrix}

P^{-1} (A P) = \frac{1}{2} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} 4 & 2 \\ 4 & -2 \end{pmatrix} = \frac{1}{2} \begin{pmatrix} 4+4 & 2-2 \\ 4-4 & 2+2 \end{pmatrix} = \frac{1}{2} \begin{pmatrix} 8 & 0 \\ 0 & 4 \end{pmatrix} = \begin{pmatrix} 4 & 0 \\ 0 & 2 \end{pmatrix} = B

$B$ is diagonal because the basis vectors $\vec{v}_1, \vec{v}_2$ are eigenvectors of $T$ with corresponding eigenvalues $\lambda_1 = 4$ and $\lambda_2 = 2$!

Final Answer & Physical Insight

$A = \begin{pmatrix} 3 & 1 \\ 1 & 3 \end{pmatrix}$, $P = \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}$, $P^{-1} = \frac{1}{2}\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}$. $B = [T]_{\mathcal{B}} = \begin{pmatrix} 4 & 0 \\ 0 & 2 \end{pmatrix}$. Verified: $P^{-1}AP = B$. $B$ is diagonal because $\mathcal{B}$ is an eigenbasis.

Tier 3: Honors / Proof Challenge Example 5.3: Complete Proof of Abstract Rank-Nullity & Injective/Surjective Duality

Let $V$ and $W$ be finite-dimensional vector spaces over field $F$ with $\dim(V) = \dim(W) = n$, and let $T: V \to W$ be a linear transformation.

(a) Prove that the following statements are logically equivalent: (i) $T$ is injective ($\ker(T) = \{\vec{0}\}$). (ii) $T$ is surjective ($\text{im}(T) = W$). (iii) $T$ is an isomorphism. (b) Provide a counterexample showing that this equivalence fails when $V$ is infinite-dimensional. (c) Let $S: U \to V$ and $T: V \to W$ be linear transformations between finite-dimensional vector spaces. Prove that:

\text{nullity}(T \circ S) \le \text{nullity}(T) + \text{nullity}(S)

Part (a): Proof of Equivalence for $\dim(V) = \dim(W) = n$: By the Rank-Nullity Theorem:

\dim(\ker(T)) + \dim(\text{im}(T)) = \dim(V) = n

$(i) \implies (ii)$: If $T$ is injective, then $\ker(T) = \{\vec{0}\}$, so $\dim(\ker(T)) = 0$.

Then $0 + \dim(\text{im}(T)) = n \implies \dim(\text{im}(T)) = n$. Since $\text{im}(T)$ is an $n$-dimensional subspace of the $n$-dimensional space $W$, we must have $\text{im}(T) = W$. Thus $T$ is surjective.

$(ii) \implies (i)$: If $T$ is surjective, then $\text{im}(T) = W$, so $\dim(\text{im}(T)) = \dim(W) = n$.

Then $\dim(\ker(T)) + n = n \implies \dim(\ker(T)) = 0$. Therefore $\ker(T) = \{\vec{0}\}$, proving $T$ is injective.

$(i) \text{ and } (ii) \iff (iii)$: An isomorphism is by definition both injective and surjective. Since $(i) \iff (ii)$, either condition immediately implies the other, establishing (iii). $\blacksquare$

Part (b): Infinite-Dimensional Counterexample: Let $V = \mathbb{R}[t]$ (the space of all real polynomials, infinite-dimensional).

Injective but NOT Surjective:

Consider the shift/multiplication operator $T(p(t)) = t \cdot p(t)$. If $t \cdot p(t) = 0$, then $p(t) = 0$, so $\ker(T) = \{0\}$ ($T$ is injective). However, non-zero constant polynomials (e.g. $p(t) = 1$) have no pre-image under $T$. Hence $\text{im}(T) \ne V$ ($T$ is not surjective!).

Surjective but NOT Injective:

Consider the derivative operator $D(p(t)) = p'(t)$. Every polynomial has an antiderivative, so $\text{im}(D) = V$ ($D$ is surjective). However, $D(c) = 0$ for all constant polynomials, so $\ker(D) = \text{span}\{1\} \ne \{0\}$ ($D$ is not injective!).

Part (c): Proof of $\text{nullity}(T \circ S) \le \text{nullity}(T) + \text{nullity}(S)$: Notice that $\vec{u} \in \ker(T \circ S) \iff T(S(\vec{u})) = \vec{0} \iff S(\vec{u}) \in \ker(T)$. Restrict the operator $S$ to the subspace $K = \ker(T \circ S) \subseteq U$. The map $S|_K : K \to \ker(T)$ has:

Kernel: $\ker(S|_K) = \{\vec{u} \in K : S(\vec{u}) = \vec{0}\} = \ker(S) \cap K = \ker(S)$.
Image: $S(K) \subseteq \ker(T)$.

By the Rank-Nullity Theorem applied to $S|_K$:

\dim(K) = \dim(\ker(S|_K)) + \dim(S(K))

\text{nullity}(T \circ S) = \text{nullity}(S) + \dim(S(K))

Since $S(K) \subseteq \ker(T)$, its dimension is bounded by $\dim(\ker(T)) = \text{nullity}(T)$:

\dim(S(K)) \le \text{nullity}(T)

Substituting this inequality:

\text{nullity}(T \circ S) \le \text{nullity}(S) + \text{nullity}(T)

The inequality is proved. $\blacksquare$

Final Answer & Physical Insight

In finite dimensions with $\dim(V)=\dim(W)$, injectivity $\iff$ surjectivity $\iff$ isomorphism. In infinite dimensions, $p(t) \mapsto t p(t)$ is injective but not surjective, while $p(t) \mapsto p'(t)$ is surjective but not injective. $\text{nullity}(T \circ S) \le \text{nullity}(T) + \text{nullity}(S)$ proved via Rank-Nullity on $S|_{\ker(TS)}$.