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7. Linear Algebra

Transpose of a Linear Transformation

The transpose (or dual) of a linear transformation maps the dual space of the codomain back to the dual space of the domain, preserving algebraic structure and reversing direction.

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

Definition and Existence of the Transpose Transformation
Let $U$ and $V$ be vector spaces over a field $\mathbb{F}$, and let $T: U \to V$ be a linear transformation. The transpose (or dual) of $T$ is the map $T^*: V^* \to U^*$ defined on the dual spaces $V^*$ and $U^*$ such that for every functional $\varphi \in V^*$ and every vector $u \in U$, the action of $T^*(\varphi)$ on $u$ is given by $$(T^*(\varphi))(u) = \varphi(T(u))$$

📐 Geometric Interpretation

Geometrically, while $T$ moves vectors forward from the domain space $U$ to the codomain space $V$, its transpose $T^*$ pulls linear functionals (hyperplanes or measurements) backward from the dual space $V^*$ to $U^*$. This duality preserves the inner product pairing or contraction evaluation: $\langle \varphi, T(u) \rangle = \langle T^*(\varphi), u \rangle$.

⚙️ Algebraic & Structural Meaning

Algebraically, the transpose map establishes a canonical contravariant functor between vector spaces and their duals. If $T$ is represented by a matrix $A$ relative to ordered bases, then $T^*$ is represented by the exact matrix transpose $A^T$ relative to the corresponding dual bases, intertwining null spaces and ranges via annihilators.

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 $T$ Algebraic structure over field $\mathbb{F}$ Strictly Mandatory / Necessary If $T$ is not linear, the standard transpose map $T^*$ cannot be defined consistently as a linear functional operator satisfying $(T^*(\varphi))(u) = \varphi(T(u))$. Failures break superposition.
Dual Space Domain / Codomain Formulation Algebraic vector spaces $U^*$ and $V^*$ algebraic duals Strictly Mandatory If restricted to primal spaces without an inner product isomorphism, $T^*: V^* \to U^*$ cannot be mapped directly back to $U$ unless $U$ is finite-dimensional and canonically identified with $U^{**}$.
Finite-Dimensional Basis Compatibility Finite dimensions $\dim(U) = n, \dim(V) = m$ Essential for matrix equivalence In infinite-dimensional spaces, algebraic duals are strictly larger than the primal spaces, making standard matrix transpose interpretations invalid unless continuous duals with topologies are invoked.

Useful results and shortcuts

Use these identities and results when their conditions apply:

01

Matrix Representation Rule

$$[T^*]_{\mathcal{B}^*} = ([T]_\mathcal{B})^T$$

Rule: The matrix of the transpose transformation relative to dual bases is precisely the transpose of the matrix of the original transformation relative to the primal bases.

Application: Directly tested in UP PGT and DSSSB PGT to find the matrix of $T^*$ without computing dual basis functionals explicitly.
02

Rank Preservation Property

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

Rule: A linear transformation and its transpose always have identical ranks, as row rank equals column rank for matrices.

03

Nullity and Annihilator Dimension Relation

$$\text{nullity}(T^*) = \dim(V) - \text{range}(T)$$

Rule: The nullity of $T^*$ equals the dimension of the annihilator of the range of $T$, directly related to the corank of $T$.

Application: Frequently asked in KVS PGT vector space dimension calculation problems.
04

Composition Reversal Rule

$$(S \circ T)^* = T^* \circ S^*$$

Rule: The transpose of a composite linear transformation reverses the order of the maps, identical to matrix transposition $(AB)^T = B^T A^T$.

05

Inverse Transpose Relation

$$(T^{-1})^* = (T^*)^{-1}$$

Rule: If a linear transformation $T$ is invertible, its transpose is also invertible, and the transpose of the inverse is the inverse of the transpose.

Application: Standard verification question in competitive teaching tests.
06

Rapid Options Elimination via Dimension Check

$$\text{If } T: \mathbb{R}^n \to \mathbb{R}^m, \text{ then } T^*: (\mathbb{R}^m)^* \to (\mathbb{R}^n)^*$$

Rule: Always check domain and codomain dimensions. If $T$ goes from dimension $n$ to $m$, $T^*$ goes from $m$ to $n$. Eliminate options reversing this.

Application: Saves 30 seconds on matching matrix dimension questions.

Common errors and counterexamples

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

⚠️ COMMON ERROR #01

Domain and Codomain Direction Reversal Confusion

Common Pitfall: Candidates assume that if $T: U \to V$, then $T^*$ also maps from $U$ to $V$ or from $V$ to $U$ without dual spaces.

Counterexample: If $T: \mathbb{R}^2 \to \mathbb{R}^3$, then $T^*: (\mathbb{R}^3)^* \to (\mathbb{R}^2)^{**}$, not $\mathbb{R}^3 \to \mathbb{R}^2$.

Why it fails: The transpose acts on linear functionals, reversing the direction of mapping between dual spaces.

🛡️ Exam Defense: Always remember: Primal goes $U \to V$, Transpose dual goes $V^* \to U^*$.
⚠️ COMMON ERROR #02

Assuming Transpose Equals Adjoint in Non-Inner Product Spaces

Common Pitfall: Confusing the algebraic transpose $T^*$ with the Hermitian adjoint $T^\dagger$ or metric adjoint over general vector spaces.

Counterexample: Over vector spaces without a specified inner product, $T^*$ is defined purely algebraically via dual spaces, whereas adjoint requires an inner product structure.

Why it fails: Using inner product properties on spaces that only have a field structure leads to incorrect algebraic conclusions.

🛡️ Exam Defense: Check if an inner product is explicitly given; if not, use pure dual-space definition $(T^*\varphi)(u) = \varphi(Tu)$.

Comparison

Concept / TheoremHypotheses / AssumptionsKey Formula / OutcomeGeometric MeaningExam Identification Clue
Linear Transformation $T$Vector spaces $U, V$, $T(au+bv) = aTu + bTv$$T: U \to V$Maps vectors forward from domain to codomainGiven a mapping rule between spaces
Transpose Transformation $T^*$Dual spaces $U^*, V^*$$T^*: V^* \to U^*, \; (T^*\varphi)u = \varphi(Tu)$Pulls linear functionals backwardAsked about dual map or matrix transpose
Matrix Transpose $A^T$Matrix $A$ of order $m \times n$$[A^T]_{ij} = A_{ji}$Reflects matrix elements across main diagonalComputation of matrix corresponding to $T^*$
Adjoint Transformation $T^*$Inner product spaces $U, V$$\langle Tu, v \rangle = \langle u, T^*v \rangle$Reflects operation across inner productKeyword 'inner product' or 'Hilbert space'

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

Let $T: \mathbb{R}^2 \to \mathbb{R}^2$ be the linear transformation defined by $T(x, y) = (x + 2y, 3x - y)$. Find the matrix of the transpose transformation $T^*$ relative to the standard dual bases.

(A) $\begin{pmatrix} 1 & 3 \\ 2 & -1 \end{pmatrix}$(B) $\begin{pmatrix} 1 & 2 \\ 3 & -1 \end{pmatrix}$(C) $\begin{pmatrix} -1 & 2 \\ 3 & 1 \end{pmatrix}$(D) $\begin{pmatrix} 1 & -2 \\ -3 & -1 \end{pmatrix}$
✓ Correct Answer: (A)
📚 Method A: Formal Textbook Derivation
Step 1: Write down the standard matrix representation of $T$, which is $A = \begin{pmatrix} 1 & 2 \\ 3 & -1 \end{pmatrix}$. Step 2: Use the theorem that the matrix of $T^*$ is the transpose of $A$. Step 3: Compute $A^T = \begin{pmatrix} 1 & 3 \\ 2 & -1 \end{pmatrix}$.
⚡ Method B: 25-Second Exam Shortcut
Instantly write the coefficients of $x$ and $y$ as columns for $T$, then transpose rows and columns to get $A^T$.
💡 Tip: Do not compute dual functionals from scratch; matrix transposition is algebraically equivalent and takes 10 seconds.
CANONICAL PROBLEM 02 Dual-Method Breakdown

If $T: V \to W$ is a linear transformation between finite-dimensional vector spaces, which of the following is always true regarding $\text{rank}(T)$ and $\text{rank}(T^*)$?

(A) $\text{rank}(T^*) > \text{rank}(T)$(B) $\text{rank}(T^*) < \text{rank}(T)$(C) $\text{rank}(T^*) = \text{rank}(T)$(D) $\text{rank}(T^*) = \text{nullity}(T)$
✓ Correct Answer: (C)
📚 Method A: Formal Textbook Derivation
By definition, the rank of a matrix is equal to the rank of its transpose ($\text{rank}(A) = \text{rank}(A^T)$). Since the matrix of $T^*$ is the transpose of the matrix of $T$, their ranks must be equal.
⚡ Method B: 25-Second Exam Shortcut
Recall the fundamental matrix property: row rank equals column rank, hence rank of transpose equals rank of original map.
💡 Tip: undefined

Practice questions

Choose an answer to check it and see the solution:

Q1
MCQ 1: If $T: U \to V$ is a linear transformation, then its transpose $T^*$ maps:
A $U \to V$
B $V \to U$
C $U^* \to V^*$
D $V^* \to U^*$
Derivation & Analysis:

By definition, the transpose map $T^*$ goes from the dual of the codomain to the dual of the domain, i.e., $T^*: V^* \to U^*$.

Q2
MCQ 2: Let $T: \mathbb{R}^3 \to \mathbb{R}^2$ be a linear transformation. What is the dimension of the domain of $T^*$?
A 1
B 2
C 3
D 5
Derivation & Analysis:

The domain of $T^*$ is $V^*$, where $V$ is the codomain of $T$. Here $\dim(V) = 2$, so $\dim(V^*) = 2$.

Q3
MCQ 3: For any two linear transformations $S$ and $T$ for which the composition $S \circ T$ is defined, $(S \circ T)^*$ is equal to:
A $S^* \circ T^*$
B $T^* \circ S^*$
C $S^* \circ T$
D $T \circ S^*$
Derivation & Analysis:

Reversal law for transpose of compositions: $(S \circ T)^* = T^* \circ S^*$, exactly analogous to matrix multiplication $(AB)^T = B^T A^T$.

Q4
MCQ 4: If $I$ is the identity transformation on vector space $V$, then its transpose $I^*$ is:
A Zero transformation
B $\text{Identity transformation on } V^*$
C Negative identity
D Undefined
Derivation & Analysis:

For the identity map $I: V \to V$, $(I^* \varphi)(v) = \varphi(I(v)) = \varphi(v)$, so $I^*$ is the identity map on $V^*$.

Q5
MCQ 5: The rank of the transpose linear transformation $T^*$ satisfies:
A $\text{rank}(T^*) \neq \text{rank}(T)$
B $\text{rank}(T^*) = \text{rank}(T)$
C $\text{rank}(T^*) \cdot \text{rank}(T) = 1$
D None of these
Derivation & Analysis:

Since matrix rank equals transpose matrix rank, $\text{rank}(T^*) = \text{rank}(T)$.

Q6
MCQ 6: If $T: U \to V$ is an isomorphism, then $T^*: V^* \to U^*$ is:
A Not linear
B An isomorphism
C Zero map
D Singular always
Derivation & Analysis:

If $T$ is invertible, $T^{-1}$ exists, and $(T^{-1})^* = (T^*)^{-1}$, proving $T^*$ is an isomorphism.

Q7
MCQ 7: Let $A$ be the matrix of $T$ relative to basis $\mathcal{B}$. The matrix of $T^*$ relative to the dual basis $\mathcal{B}^*$ is:
A A
B $A^{-1}$
C $A^T$
D -A
Derivation & Analysis:

Standard matrix theorem states $[T^*]_{\mathcal{B}^*} = ([T]_\mathcal{B})^T = A^T$.

Q8
MCQ 8: Assertion (A): $\text{nullity}(T^*) = \dim(V) - \text{rank}(T)$. Reason (R): The dimension of the annihilator of the range of $T$ equals $\dim(V) - \text{rank}(T)$.
A Both A and R are true and R is the correct explanation
B Both A and R are true but R is not explanation
C A is true, R is false
D A is false, R is true
Derivation & Analysis:

The kernel of $T^*$ is precisely the annihilator of $\text{range}(T)$. Its dimension is $\dim(V) - \dim(\text{range}(T))$. Both assertion and reason are correct.

Q9
MCQ 9: If $T: \mathbb{R}^2 \to \mathbb{R}^2$ has matrix $\begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix}$, then the matrix of $T^*$ is:
A $\begin{pmatrix} 2 & 1 \\ 5 & 3 \end{pmatrix}$
B $\begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix}$
C $\begin{pmatrix} 3 & -5 \\ -1 & 2 \end{pmatrix}$
D $\begin{pmatrix} -3 & 5 \\ 1 & -2 \end{pmatrix}$
Derivation & Analysis:

Taking the transpose of $\begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix}$ yields $\begin{pmatrix} 2 & 1 \\ 5 & 3 \end{pmatrix}$.

Q10
MCQ 10: Which of the following is false for a linear transformation $T: U \to V$ and its transpose $T^*$:
A (S+T)^* = S^* + T^*
B (kT)^* = k T^*
C $(T^*)^* = T \text{ (under natural embedding)}$
D $\text{rank}(T^*) \neq \text{rank}(T)$
Derivation & Analysis:

Statement (D) claims ranks are unequal, which is false since $\text{rank}(T^*) = \text{rank}(T)$. Thus (D) is the correct option to choose as false.

Common questions

Because linear functionals in the dual space act on the codomain $V$. To pull them back via $T$, a functional $\varphi \in V^*$ must take elements of $V$ (which are images $T(u)$) to scalars. Hence, composing $\varphi$ with $T$ yields a functional in $U^*$, naturally forcing the map $T^*$ to go from $V^*$ to $U^*$

Not quite. The transpose $T^*$ is defined purely algebraically for any vector spaces and their duals without requiring an inner product. The adjoint operator requires inner product spaces to translate between primal and dual spaces via the Riesz Representation Theorem.

Avoid constructing dual basis functionals explicitly. Simply write down the standard matrix $A$ of the linear transformation, and immediately write its matrix transpose $A^T$. Match it directly with the given options.

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