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7. Linear Algebra

Image of linear transformation

The image (or range) of a linear transformation $T: V \to W$ is the subspace of the codomain $W$ consisting of all possible outputs $T(v)$ for vectors $v \in V$.

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Definition and statement

Fundamental Definition and Dimension Theorem (Rank-Nullity)
Let $V$ and $W$ be vector spaces over a field $\mathbb{F}$, and let $T: V \to W$ be a linear transformation. The image of $T$, denoted as $\operatorname{Im}(T)$ or $\mathcal{R}(T)$, is defined by $\operatorname{Im}(T) = \{ w \in W \mid w = T(v) \text{ for some } v \in V \}$. If $V$ is finite-dimensional, then $\dim(V) = \dim(\ker(T)) + \dim(\operatorname{Im}(T))$.

📐 Geometric Interpretation

Geometrically, the image represents the subspace in the target space $W$ that is actually reached or spanned by mapping the domain space $V$ through $T$. For example, a linear transformation from $\mathbb{R}^3$ to $\mathbb{R}^3$ with rank $2$ maps the entire 3D space onto a 2D plane passing through the origin.

⚙️ Algebraic & Structural Meaning

Algebraically, $\operatorname{Im}(T)$ is a linear subspace of the codomain $W$. Its dimension is called the rank of $T$. Computing a basis for $\operatorname{Im}(T)$ is equivalent to finding the column space of the associated transformation matrix under chosen bases.

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 the mapping $T$ Vector spaces $V, W$ over field $\mathbb{F}$ Strictly Mandatory / Necessary If $T$ is non-linear (e.g., $T(x,y) = (x^2, y)$), the set of outputs does not form a vector subspace, and the Rank-Nullity theorem fails.
Finite dimensionality of domain $V$ \dim(V) < \infty Required for standard Rank-Nullity dimension relation In infinite-dimensional spaces, dimensions cannot be directly subtracted, though $\operatorname{Im}(T)$ remains well-defined as a subspace.
Basis representation Finite spanning set Essential for matrix rank computation Without explicit basis vectors, determining the exact spanning vectors of $\operatorname{Im}(T)$ requires solving infinite parametric systems.

Useful results and shortcuts

Use these identities and results when their conditions apply:

01

Image Spanning by Column Vectors

$$ \operatorname{Im}(T) = \operatorname{Col}(A) $$

Rule: If $A$ is the standard matrix representation of $T: \mathbb{R}^n \to \mathbb{R}^m$, then $\operatorname{Im}(T)$ is spanned by the columns of $A$. Reduce $A$ to row echelon form to find a basis of non-zero pivot columns.

Application: Directly tested in UP PGT and DSSSB PGT to find basis and dimension of range spaces.
02

Rank-Nullity Quick Calculation

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

Rule: Find the nullity by solving homogeneous system $A x = 0$; subtract from domain dimension to instantly get $\dim(\operatorname{Im}(T))$ without finding basis.

Application: Saves up to 90 seconds in matrix transformation MCQs.
03

Surjectivity Criterion (Onto)

$$ \operatorname{Im}(T) = W \iff \operatorname{rank}(T) = \dim(W) $$

Rule: For $T: V \to W$, $T$ is surjective if and only if the rank of its matrix equals the dimension of the codomain $W$.

04

Dimension Bound via Codomain

$$ \dim(\operatorname{Im}(T)) \le \min(\dim(V), \dim(W)) $$

Rule: The dimension of the image cannot exceed the dimension of either the domain or the codomain. Instantly eliminate options violating this bound.

Application: Used for rapid elimination in assertion-reason type questions.
05

Matrix Rank via Determinant

$$ \det(A) \neq 0 \implies \operatorname{Im}(T) = \mathbb{R}^n $$

Rule: If $T: \mathbb{R}^n \to \mathbb{R}^n$ has a non-zero determinant, its matrix is full rank, meaning $\operatorname{Im}(T) = \mathbb{R}^n$ and $T$ is onto.

Application: Frequent arithmetic check in UP TGT Linear Algebra problems.
06

Composite Transformation Image Bound

$$ \operatorname{rank}(ST) \le \min(\operatorname{rank}(S), \operatorname{rank}(T)) $$

Rule: The rank of a composite transformation cannot exceed the individual ranks of the constituent transformations.

Common errors and counterexamples

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

⚠️ COMMON ERROR #01

Confusing Image with Codomain

Common Pitfall: Assuming that because $T: V \to W$, the image of $T$ is automatically the entire codomain $W$.

Counterexample: $T: \mathbb{R}^2 \to \mathbb{R}^3$ defined by $T(x,y) = (x, y, 0)$

Why it fails: The codomain is 3-dimensional, but $\operatorname{Im}(T)$ is only a 2-dimensional plane ($z=0$). Thus, $T$ is not surjective.

🛡️ Exam Defense: Always check if $\operatorname{rank}(T) = \dim(W)$ before claiming $T$ is onto or $\operatorname{Im}(T) = W$.
⚠️ COMMON ERROR #02

Using Non-Pivot Columns for Basis

Common Pitfall: Taking column vectors of the original matrix $A$ corresponding to free variables as basis elements of $\operatorname{Im}(T)$.

Counterexample: Dependent columns in matrix reduction

Why it fails: Basis elements for $\operatorname{Im}(T)$ must correspond to the pivot columns of the original matrix $A$ (not the row-reduced matrix, but the original columns at pivot positions).

🛡️ Exam Defense: Identify pivot columns in row-echelon form, then select the corresponding columns from the *original* matrix.

Comparison

Concept / TheoremHypotheses / AssumptionsKey Formula / OutcomeGeometric MeaningExam Identification Clue
Image (Range)Linear transformation $T: V \to W$$\operatorname{Im}(T) = \{T(v) \mid v \in V\}$Subspace of codomain reached by domainPhrases like 'range space', 'column space'
Kernel (Null Space)Linear transformation $T: V \to W$$-\ker(T) = \{v \in V \mid T(v) = 0\}$Subspace of domain collapsed to originPhrases like 'null space', 'solution space'
Rank-Nullity TheoremFinite-dimensional domain $V$$\dim(V) = \text{nullity} + \text{rank}$Conservation of dimension across mappingQuestions asking for rank given nullity or vice versa
IsomorphismBijective linear transformation$T$ is 1-1 and onto; $\dim(V) = \dim(W)$Complete structural preservation (invertible)Keywords: 'invertible', 'nonsingular', 'vector space isomorphism'

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 $T(x, y, z) = (x + y, y + z, x + z)$. What is the dimension of the image of $T$?

(A) 0(B) 1(C) 2(D) 3
✓ Correct Answer: (D)
📚 Method A: Formal Textbook Derivation
Step 1: Write the matrix of $T$ standardly: $A = \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{pmatrix}$. Step 2: Compute determinant $\det(A) = 1(1-0) - 1(0-1) + 0 = 2 \neq 0$. Step 3: Since $\det(A) \neq 0$, the rank is 3. Step 4: $\dim(\operatorname{Im}(T)) = 3$.
⚡ Method B: 25-Second Exam Shortcut
Compute $\det(A)$. Since $\det \neq 0$, matrix is full rank. Dimension equals domain dimension 3. Mark (D) in 15 seconds.
💡 Tip: For square matrices representing transformations, a non-zero determinant instantly implies full rank and full image dimension.
CANONICAL PROBLEM 02 Dual-Method Breakdown

Consider the linear transformation $T: \mathbb{R}^4 \to \mathbb{R}^3$ whose matrix relative to standard bases is $A = \begin{pmatrix} 1 & 2 & 0 & 1 \\ 2 & 4 & 1 & 3 \\ 1 & 2 & -1 & 0 \end{pmatrix}$. Find the dimension of $\operatorname{Im}(T)$.

(A) 1(B) 2(C) 3(D) 4
✓ Correct Answer: (B)
📚 Method A: Formal Textbook Derivation
Step 1: Apply elementary row operations to find row rank. $R_2 \to R_2 - 2R_1$, $R_3 \to R_3 - R_1$. Result: $\begin{pmatrix} 1 & 2 & 0 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & -1 & -1 \end{pmatrix}$. Step 2: Further reduce to get 2 non-zero rows. Step 3: Row rank = Column rank = 2. Therefore, $\dim(\operatorname{Im}(T)) = 2$.
⚡ Method B: 25-Second Exam Shortcut
Notice column 2 is $2 \times$ column 1, and column 4 equals column 1 + column 3. Only 2 linearly independent columns exist. Rank = 2.
💡 Tip: Inspect proportional columns or linear combinations before executing full row reduction.

Practice questions

Choose an answer to check it and see the solution:

Q1
MCQ 1: If $T: V \to W$ is a linear transformation and $\dim(V) = 5$, $\dim(\ker(T)) = 2$, then $\dim(\operatorname{Im}(T))$ is:
A 2
B 3
C 5
D 7
Derivation & Analysis:

By Rank-Nullity Theorem: $\dim(V) = \dim(\ker(T)) + \dim(\operatorname{Im}(T))$. Thus $5 = 2 + \dim(\operatorname{Im}(T)) \implies \dim(\operatorname{Im}(T)) = 3$.

Q2
MCQ 2: The image of the zero transformation $T: V \to W$ defined by $T(v) = 0$ for all $v \in V$ is:
A $V$
B $W$
C $\{0_W\}$
D $\emptyset$
Derivation & Analysis:

Since $T(v) = 0$ for every vector in $V$, the set of all outputs contains only the zero vector of $W$. Thus $\operatorname{Im}(T) = \{0_W\}$.

Q3
MCQ 3: Let $T: \mathbb{R}^3 \to \mathbb{R}^2$ be a linear transformation. What is the maximum possible dimension of $\operatorname{Im}(T)$?
A 0
B 1
C 2
D 3
Derivation & Analysis:

The image subspace lies inside the codomain $\mathbb{R}^2$, so its dimension cannot exceed $\dim(\mathbb{R}^2) = 2$. Also $\dim(\operatorname{Im}(T)) \le \dim(\mathbb{R}^3) = 3$. Hence maximum dimension is $\min(3, 2) = 2$.

Q4
MCQ 4: A linear transformation $T: V \to W$ is one-to-one (injective) if and only if:
A $\operatorname{Im}(T) = W$
B $\ker(T) = \{0\}$
C $\dim(\operatorname{Im}(T)) = \dim(V)$
D Both (B) and (C)
Derivation & Analysis:

A linear transformation is injective if and only if its kernel is trivial ($\ker(T) = \{0\}$). By Rank-Nullity, nullity = 0 implies $\dim(\operatorname{Im}(T)) = \dim(V)$. Thus both (B) and (C) are correct.

Q5
MCQ 5: If $A$ is an $m \times n$ matrix with real entries, then the dimension of the column space of $A$ is equal to:
A Row rank of $A$
B $n - \text{nullity}(A)$
C Dimension of $\operatorname{Im}(T_A)$
D All of the above
Derivation & Analysis:

Row rank equals column rank, column rank equals the dimension of the image of the associated linear transformation $T_A$, and by Rank-Nullity this equals $n - \text{nullity}(A)$. Thus all statements are correct.

Q6
MCQ 6: Let $T: P_2(\mathbb{R}) \to P_3(\mathbb{R})$ be the linear transformation defined by $T(p(x)) = x p(x)$. The dimension of $\operatorname{Im}(T)$ is:
A 1
B 2
C 3
D 4
Derivation & Analysis:

The domain $P_2(\mathbb{R})$ has dimension 3 (basis $\{1, x, x^2\}$). Since multiplication by $x$ is an injective linear map ($x p(x) = 0 \implies p(x) = 0$), nullity = 0. By Rank-Nullity, $\dim(\operatorname{Im}(T)) = \dim(P_2(\mathbb{R})) - 0 = 3$.

Q7
MCQ 7: If $T: V \to W$ is onto (surjective) and $\dim(V) = \dim(W) < \infty$, then $T$ must be:
A Zero transformation
B Injective (one-to-one)
C Not invertible
D Singular
Derivation & Analysis:

If $T$ is onto, $\operatorname{Im}(T) = W$, so $\dim(\operatorname{Im}(T)) = \dim(W) = \dim(V)$. By Rank-Nullity, nullity is 0, so $\ker(T) = \{0\}$, making $T$ injective.

Q8
MCQ 8: Assertion (A): The image of a finite-dimensional vector space under a linear transformation is a vector space. Reason (R): The image subspace inherits vector addition and scalar multiplication from the codomain.
A Both (A) and (R) are true and (R) is the correct explanation of (A)
B Both (A) and (R) are true but (R) is not the correct explanation
C (A) is true but (R) is false
D (A) is false but (R) is true
Derivation & Analysis:

The image of a linear transformation is always a subspace of the codomain, satisfying closure under addition and scalar multiplication inherited from the codomain.

Q9
MCQ 9: Let $T: \mathbb{R}^2 \to \mathbb{R}^2$ be reflection across the $x$-axis. The image of $T$ is:
A Only the origin
B The $y$-axis
C The $x$-axis
D The entire $\mathbb{R}^2$
Derivation & Analysis:

Reflection across the $x$-axis maps every point $(x,y)$ to $(x, -y)$. Every vector in $\mathbb{R}^2$ can be reached in this manner, so the image is the entire space $\mathbb{R}^2$.

Q10
MCQ 10: If $T: \mathbb{R}^4 \to \mathbb{R}^4$ is a linear transformation with $\dim(\operatorname{Im}(T)) = 4$, which of the following is false?
A $T$ is invertible
B $\ker(T) = \{0\}$
C $T$ is singular
D $T$ is surjective
Derivation & Analysis:

If $\dim(\operatorname{Im}(T)) = 4$ for a 4D space, then $T$ is full rank, nonsingular, bijective, and invertible. Therefore, stating that $T$ is singular is false.

Common questions

The codomain $W$ is the target space specified in the definition $T: V \to W$. The image $\operatorname{Im}(T)$ is only the subset of $W$ actually hit by elements of $V$. While the codomain is given, the image is computed and is always a subspace of the codomain.

Candidates often confuse the row space with the column space or fail to test whether the transformation is surjective when asked about the range. Always use pivot columns of the original matrix to describe the basis of the image.

Check the determinant if it is a square matrix ($\det \neq 0 \implies$ full rank). For rectangular matrices, immediately inspect for dependent rows/columns or use elementary row operations to count non-zero pivot rows.

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