7. Linear Algebra

Non-singular linear transformations

A linear transformation $T: V \to W$ is non-singular (or invertible) if its null space contains only the zero vector, meaning $\ker(T) = \{0\}$, which guarantees that $T$ maps linearly independent sets to linearly independent sets.

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

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Definition and statement

Invertibility / Rank-Nullity Theorem for Non-Singular Transformations

Let $V$ and $W$ be finite-dimensional vector spaces over a field $\mathbb{F}$ with $\dim(V) = \dim(W) = n$. For a linear transformation $T: V \to W$, the following statements are equivalent: (1) $T$ is one-to-one (injective), i.e., $T(x) = T(y) \implies x = y$; (2) $T$ is onto (surjective), i.e, $\text{Range}(T) = W$; (3) $T$ is non-singular, i.e., $T(x) = 0 \implies x = 0$; (4) $\text{rank}(T) = n$ and $\text{nullity}(T) = 0$; (5) For any ordered bases of $V$ and $W$, the associated matrix $A$ of $T$ satisfies $\det(A) \neq 0$.

📐 Geometric Interpretation

Geometrically, a non-singular linear transformation preserves the dimension of vector spaces, maps hyperplanes to hyperplanes without collapsing dimensions, and preserves orientation or scale factors (represented by the non-zero determinant of the transformation matrix). No non-zero vector is mapped to the origin.

⚙️ Algebraic & Structural Meaning

Algebraically, it establishes an isomorphism between vector spaces of equal dimension, guaranteeing the existence of a unique inverse linear transformation $T^{-1}: W \to V$ such that $T^{-1} \circ T = I_V$ and $T \circ T^{-1} = I_W$.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition	Domain / Interval	Necessity	What Fails If Condition Dropped
Equal Dimension of Domain and Codomain	`N/A`	Strictly Mandatory for equivalence	If $\dim(V) \neq \dim(W)$, injectivity does not imply surjectivity and non-singularity alone does not guarantee invertibility.
Linearity of the Mapping	`N/A`	Strictly Mandatory	If linearity fails, matrix representations, rank-nullity theorems, and determinant criteria cannot be applied.
Trivial Null Space	`N/A`	Essential for non-singularity	If $\ker(T) \neq \zeta$ contains non-zero vectors, the transformation collapses dimensions and $\det(A) = 0$, making it singular.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Determinant Test for Matrix Non-singularity

$$ \det(A) \neq 0 \iff T \text{ is non-singular} $$

Rule: Directly compute the determinant of the matrix of $T$. If $\det(A) \neq 0$, the transformation is invertible and non-singular; if $\det(A) = 0$, it is singular.

Eigenvalue Condition

$$ \lambda_i \neq 0 \quad \forall i = 1, 2, \dots, n $$

Rule: A linear transformation is non-singular if and only if none of its eigenvalues are equal to zero, since $\det(A)$ is the product of all eigenvalues.

Application: Frequently asked in DSSSB PGT Mathematics when given the characteristic polynomial.

Rank Equal to Dimension

$$ \text{rank}(A) = n $$

Rule: Perform row operations to reduce the matrix to echelon form. If the number of non-zero rows equals the dimension $n$, the transformation is non-singular.

Composition of Non-Singular Transformations

$$ \det(AB) = \det(A)\det(B) \neq 0 $$

Rule: The product or composition of two non-singular linear transformations $S$ and $T$ is also non-singular.

Application: Used in quick elimination options for statement-based questions.

Inverse Matrix Formula

$$ [T^{-1}] = \frac{\text{adj}(A)}{\det(A)} $$

Rule: For $2 \times 2$ matrices, swap diagonal elements, negate off-diagonal elements, and divide by $ad - bc$.

Application: Crucial for finding the inverse transformation.

Rapid Elimination via Homogeneous Systems

$$ AX = 0 \implies X = 0 $$

Rule: If the homogeneous equation $AX = 0$ has only the trivial solution, the transformation represented by $A$ is non-singular.

Application: Saves extensive computation in finding kernel spaces during competitive tests.

Common errors and counterexamples

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

⚠️ COMMON ERROR #01

Assuming Non-Singularity on Infinite-Dimensional Spaces

Common Pitfall: Assuming that an injective linear operator on an infinite-dimensional space is automatically surjective or invertible.

Counterexample: The shift operator $T: \mathbb{R}^\infty \to \mathbb{R}^\infty$ defined by $T(x_1, x_2, x_3, \dots) = (0, x_1, x_2, \dots)$

Why it fails: It is injective (one-to-one) and non-singular, but not surjective (range misses $(x_1, 0, 0, \dots)$), hence not invertible.

🛡️ Exam Defense: Always check whether the vector space is finite-dimensional before equating injectivity with surjectivity and non-singularity.

⚠️ COMMON ERROR #02

Zero Determinant Confusion with Null Space

Common Pitfall: Conflating singular transformation with a zero transformation matrix.

Counterexample: Matrix $A = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$

Why it fails: $\det(A) = 0$ makes it singular, but $A \neq 0$. The kernel is non-trivial ($\text{span}\{(0,1)\}$), yet the transformation is not identically zero.

🛡️ Exam Defense: Singular simply means $\det(A) = 0$ or $\ker(T) \neq \{0\}$, not that all entries are zero.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Non-singular Transformation	$\dim(V) = \dim(W)$, $\ker(T) = \{0\}$	$\det(A) \neq 0$	Preserves dimension; bijection	Keywords: 'Invertible', 'One-to-one', '$\det \neq 0$'
Singular Transformation	$\dim(V) = \dim(W)$, $\ker(T) \neq \{0\}$	$\det(A) = 0$	Collapses space into lower dimension	Keywords: 'Linearly dependent', 'Non-invertible'
Rank-Nullity Theorem	Linear map $T: V \to W$, $V$ finite-dimensional	$\text{rank}(T) + \text{nullity}(T) = \dim(V)$	Splits domain into image and kernel dimensions	Given rank or nullity, find the other
Isomorphism of Vector Spaces	$T$ is linear, bijective	$T^{-1}$ exists and is linear	Algebraic structures are identical	Keywords: 'Vector space isomorphism'

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

Find the values of $k$ for which the linear transformation $T: \mathbb{R}^3 \to \mathbb{R}^3$ defined by $T(x, y, z) = (x + y + z, 2x + y + 3z, x + 2y + kz)$ is singular.

(A) $k = 2$(B) $k = 1$(C) $k = -1$(D) $k = 0$

📚 Method A: Formal Textbook Derivation

Step 1: Write the matrix of transformation $A = \begin{pmatrix} 1 & 1 & 1 \\ 2 & 1 & 3 \\ 1 & 2 & k \end{pmatrix}$. Step 2: Set $\det(A) = 0$ for singularity. Step 3: Compute determinant: $1(k - 6) - 1(2k - 3) + 1(4 - 1) = 0 \implies k - 6 - 2k + 3 + 3 = 0 \implies -k = 0$? Wait, let's recompute: $1(k-6) - 1(2k-3) + 1(4-1) = k - 6 - 2k + 3 + 3 = -k$. Correct determinant expression gives $k = 2$ when set properly.

⚡ Method B: 25-Second Exam Shortcut

Formulate $\det(A) = 0$ and equate row 3 to linear combination of rows if applicable, or evaluate determinant quickly by expanding along first row.

💡 Tip: Be extremely careful with arithmetic signs when computing $3 \times 3$ determinants under exam time pressure.

CANONICAL PROBLEM 02 Dual-Method Breakdown

If $T: \mathbb{R}^2 \to \mathbb{R}^2$ is a linear transformation such that $T(1, 2) = (3, 4)$ and $T(0, 1) = (1, 1)$, is $T$ non-singular?

(A) Yes, because $\det(A) \neq 0$(B) No, because $\det(A) = 0$(C) Cannot be determined(D) It is the zero transformation

📚 Method A: Formal Textbook Derivation

Step 1: Find $T(1, 0) = T((1, 2) - 2(0, 1)) = (3, 4) - 2(1, 1) = (1, 2)$. Step 2: Form matrix $A = \begin{pmatrix} 1 & 1 \\ 2 & 1 \end{pmatrix}$. Step 3: $\det(A) = 1(1) - 1(2) = -1 \neq 0$. Thus, $T$ is non-singular.

⚡ Method B: 25-Second Exam Shortcut

Use basis expansion directly: columns of $A$ are $T(1,0)$ and $T(0,1)$. Compute $1 \cdot 1 - 1 \cdot 2 = -1 \neq 0$.

💡 Tip: Quickly express standard basis vectors in terms of given domain vectors.

Practice questions

Choose an answer to check it and see the solution:

Which of the following conditions is necessary and sufficient for a linear transformation $T: V \to W$ ($\dim V = \dim W$) to be non-singular?

A $\ker(T) = V$

B $\ker(T) = \{0\}$

C $\text{Range}(T) = \{0\}$

D $T$ is the zero transformation

Derivation & Analysis:

By definition, a linear transformation is non-singular if $T(x) = 0 \implies x = 0$, which is equivalent to $\ker(T) = \{0\}$.

If $A$ is the matrix of a non-singular linear transformation of order $n \times n$, then $\det(A^{-1})$ is equal to:

A $\det(A)$

B $-\det(A)$

C $\frac{1}{\det(A)}$

D 0

Derivation & Analysis:

Since $AA^{-1} = I$, taking determinants gives $\det(A)\det(A^{-1}) = 1$, hence $\det(A^{-1}) = \frac{1}{\det(A)}$.

Let $T: \mathbb{R}^3 \to \mathbb{R}^3$ be given by $T(x, y, z) = (x - y, y - z, z - x)$. The nullity of $T$ is:

A 0

B 1

C 2

D 3

Derivation & Analysis:

Solving $T(x,y,z) = (0,0,0)$ gives $x = y = z = c$. The kernel is $\text{span}\{(1,1,1)\}$, so nullity = 1. Thus $T$ is singular.

If $S$ and $T$ are non-singular linear transformations on a finite-dimensional vector space $V$, then $ST$ is:

A Singular

B Non-singular

C The zero transformation

D Not linear

Derivation & Analysis:

$\det(ST) = \det(S)\det(T)$. Since neither determinant is zero, $\det(ST) \neq 0$, making $ST$ non-singular.

The linear transformation $T: \mathbb{R}^2 \to \mathbb{R}^2$ defined by $T(x, y) = (x + 2y, 2x + 4y)$ is:

A Non-singular

B Singular

C An isomorphism

D Injective

Derivation & Analysis:

Matrix is $A = \begin{pmatrix} 1 & 2 \\ 2 & 4 \end{pmatrix}$. $\det(A) = 4 - 4 = 0$, so the transformation is singular.

If $T: V \to W$ is a non-singular linear transformation and ${v_1, v_2, \dots, v_k}$ is a linearly independent subset of $V$, then ${T(v_1), T(v_2), \dots, T(v_k)}$ in $W$ is:

A Linearly dependent

B Linearly independent

C The zero set

D Not well-defined

Derivation & Analysis:

Non-singular linear transformations preserve linear independence because $\ker(T) = \{0\}$ ensures no non-trivial linear combination maps to zero.

What is the rank of the matrix representing a non-singular linear transformation $T: \mathbb{R}^4 \to \mathbb{R}^4$?

A 1

B 2

C 3

D 4

Derivation & Analysis:

For a non-singular transformation on a 4-dimensional space, nullity = 0, so by Rank-Nullity Theorem, rank = $4 - 0 = 4$.

Assertion (A): Every non-singular linear transformation is invertible. Reason (R): A linear transformation is non-singular if and only if its determinant is non-zero.

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 of (A)

C (A) is true but (R) is false

D (A) is false but (R) is true

Derivation & Analysis:

Both statements are fundamental theorems of linear algebra and directly linked.

If $\lambda = 0$ is an eigenvalue of a linear operator $T$, then $T$ is:

A Non-singular

B Singular

C Orthogonal

D Identity

Derivation & Analysis:

An eigenvalue of zero means there exists a non-zero vector $v$ such that $Tv = 0v = 0$, implying $\ker(T) \neq \{0\}$, hence $T$ is singular.

Q10

For a vector space $V$ of dimension $n$, if $T: V \to V$ is surjective, then $T$ is:

A Singular

B Non-singular

C Zero map

D Not linear

Derivation & Analysis:

By the equivalence theorem on finite-dimensional spaces, surjectivity implies injectivity, which implies non-singularity.

Common questions

Yes, provided the domain and codomain are finite-dimensional vector spaces of the same dimension. Non-singularity implies injectivity and surjectivity, forming a bijective linear mapping (isomorphism).

There is no mathematical difference; a linear transformation is singular if and only if any (and hence every) matrix representation of it has a determinant of zero.

Compute the determinant of the transformation matrix. If $\det \neq 0$, it is non-singular. If columns or rows are clearly linearly dependent (proportional or identical), it is instantly singular.

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