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7. Linear Algebra

Linear dependence

Linear dependence determines whether a set of vectors in a vector space can be expressed as a non-trivial linear combination yielding the zero vector.

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

Definition of Linear Dependence and Independence
Let $V$ be a vector space over a field $\mathbb{F}$. A subset $S = \{v_1, v_2, \dots, v_n\} \subseteq V$ is said to be linearly dependent if there exist scalars $c_1, c_2, \dots, c_n \in \mathbb{F}$, not all zero, such that $$c_1 v_1 + c_2 v_2 + \dots + c_n v_n = 0.$$ If this relation holds exclusively when $c_1 = c_2 = \dots = c_n = 0$, the set $S$ is linearly independent.

📐 Geometric Interpretation

In $\mathbb{R}^2$ or $\mathbb{R}^3$, linear dependence of two vectors means they are collinear (lie on the same line through the origin), and of three vectors means they are coplanar. If any vector in a set can be written as a linear combination of the others, the set is linearly dependent.

⚙️ Algebraic & Structural Meaning

Algebraically, linear dependence implies redundancy in a spanning set. The maximal number of linearly independent vectors defines the dimension of the subspace and corresponds directly to the rank of the associated matrix formed by these vectors.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition Domain / Interval Necessity What Fails If Condition Dropped
Field Scalar Field Membership Scalars $c_i \in \mathbb{F}$ (usually $\mathbb{R}$ or $\mathbb{C}$) Strictly Mandatory / Necessary Scalars must belong to the defined underlying field; otherwise, scaling operations break vector space axioms.
Zero Vector Inclusion Property Presence of $0 \in S$ Sufficient for Dependence Any finite set of vectors containing the zero vector is automatically linearly dependent, as a non-zero scalar can be assigned to the zero vector.
Super-dimension Cardinality Bound Number of vectors $n > \dim(V)$ Strictly Mandatory for size constraint Any set containing more vectors than the dimension of the vector space is always linearly dependent.

Useful results and shortcuts

Use these identities and results when their conditions apply:

01

Determinant Test for Square Matrices

$$ \det(A) = 0 \iff \text{Rows/Columns are Linearly Dependent} $$

Rule: If $n$ vectors in $\mathbb{R}^n$ are written as rows or columns of a square matrix, compute the determinant directly. If $\det(A) = 0$, they are linearly dependent.

Application: Frequently asked in UP PGT/TGT to test vector sets in $\mathbb{R}^3$ or $\mathbb{R}^4$.
02

Proportion Rule for Two Vectors

$$ \frac{a_1}{b_1} = \frac{a_2}{b_2} = \frac{a_3}{b_3} = \lambda $$

Rule: Two vectors $(a_1, a_2, a_3)$ and $(b_1, b_2, b_3)$ are linearly dependent if and only if their corresponding components are proportional.

03

Wronskian Test for Function Spaces

$$ W(f_1, f_2, \dots, f_n)(x) = 0 \text{ identically on } I $$

Rule: For differentiable functions on an interval, if the Wronskian determinant is identically zero, the functions are often linearly dependent.

Application: Standard question type in DSSSB and UP PGT Mathematics for polynomial and trigonometric function sets.
04

Homogeneous System Rank Rule

$$ \text{Rank}(A) < n \implies \text{Non-trivial solutions exist} $$

Rule: Form matrix $A$ with vectors as columns. Perform row reduction or check rank. If $\text{rank}(A) < \text{number of vectors}$, they are linearly dependent.

Application: Universal shortcut for vector space basis determination.
05

Zero Vector Instant Trap

$$ \text{If } 0 \in \{v_1, v_2, \dots, v_n\}, \text{ then dependent} $$

Rule: Instantly mark 'Linearly Dependent' if the zero vector is present anywhere in the given set.

06

Subset Extension Rule

$$ \text{Subset of Independent is Independent; Super-set of Dependent is Dependent} $$

Rule: If a subset is dependent, any larger set containing it is automatically dependent.

Application: Quick elimination tool for multiple-statement assertion questions.

Common errors and counterexamples

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

⚠️ COMMON ERROR #01

Wronskian Zero vs Linear Independence

Common Pitfall: Assuming that a zero Wronskian $W=0$ everywhere implies linear dependence in all cases.

Counterexample: $f_1(x) = x^3, f_2(x) = |x|^3$

Why it fails: Functions can be linearly independent on an interval even if their Wronskian vanishes at certain points or subintervals if analyticity conditions fail.

🛡️ Exam Defense: Remember that $W=0$ is a necessary condition for linear dependence of differentiable functions, but sufficient only under analytic conditions.
⚠️ COMMON ERROR #02

Field Dependency Confusion

Common Pitfall: Ignoring the underlying field when testing independence (e.g., vectors over $\mathbb{R}$ vs $\mathbb{C}$).

Counterexample: Set $\{1, i\}$ over field $\mathbb{R}$ vs $\mathbb{C}$

Why it fails: $\{1, i\}$ is linearly independent over $\mathbb{R}$, but linearly dependent over $\mathbb{C}$ because $1(i) + (-i)(1) = 0$.

🛡️ Exam Defense: Always check the specified scalar field in the question statement before computing.

Comparison

Concept / TheoremHypotheses / AssumptionsKey Formula / OutcomeGeometric MeaningExam Identification Clue
Linear IndependenceScalars $c_i = 0$ only for $\sum c_i v_i = 0$$\det(A) \neq 0$ or $\text{Rank} = n$Vectors span higher dimension, no redundancyBasis questions, full rank matrix
Linear DependenceNon-zero scalars exist for $\sum c_i v_i = 0$$\det(A) = 0$ or $\text{Rank} < n$Collinear, coplanar, or redundant vectorsZero determinant, proportional components
Wronskian TestFunctions $f_i$ $n-1$ times differentiable$W(f_1, \dots, f_n) = 0$Functional linear couplingPolynomial or trigonometric function sets
Dimension TheoremFinite dimensional vector space $V$$|S| = \dim(V)$ for basisExact coordinate frame for spaceFinding maximum independent vectors

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

If the vectors $(1, 2, 3)$, $(3, m, 7)$, and $(2, 3, 4)$ are linearly dependent, then the value of $m$ is:

(A) 4(B) 5(C) 6(D) 8
✓ Correct Answer: (B)
📚 Method A: Formal Textbook Derivation
Set up the determinant of the matrix formed by the three vectors and equate it to zero for linear dependence: $$\begin{vmatrix} 1 & 2 & 3 \\ 3 & m & 7 \\ 2 & 3 & 4 \end{vmatrix} = 0.$$ Expanding along the first row: $1(4m - 21) - 2(12 - 14) + 3(9 - 2m) = 0 \implies 4m - 21 - 2(-2) + 27 - 6m = 0 \implies -2m + 10 = 0 \implies m = 5$.
⚡ Method B: 25-Second Exam Shortcut
Observe row 3 and row 1: vector 3 is not directly proportional, but testing determinant via Sarrus rule or row operations $R_3 \to R_3 - 2R_1$ gives quick results.
💡 Tip: Whenever three 3-tuples are stated as linearly dependent, immediately write their determinant equal to zero.
CANONICAL PROBLEM 02 Dual-Method Breakdown

The set of functions $\{e^x, xe^x, x^2e^x\}$ on $\mathbb{R}$ is:

(A) Linearly dependent over $\mathbb{R}$(B) Linearly independent over $\mathbb{R}$(C) Linearly dependent only for $x = 0$(D) None of these
✓ Correct Answer: (B)
📚 Method A: Formal Textbook Derivation
Compute the Wronskian $W(e^x, xe^x, x^2e^x)$. $W = \begin{vmatrix} e^x & xe^x & x^2e^x \\ e^x & (x+1)e^x & (x^2+2x)e^x \\ e^x & (x+2)e^x & (x^2+4x+2)e^x \end{vmatrix} = 2e^{3x} \neq 0$ for all $x \in \mathbb{R}$. Thus, the functions are linearly independent.
⚡ Method B: 25-Second Exam Shortcut
Factor out $e^x$ from each column/row and recognize standard exponential-polynomial basis blocks.
💡 Tip: Sets of the form $x^k e^{rx}$ with distinct polynomials are always linearly independent.

Practice questions

Choose an answer to check it and see the solution:

Q1
MCQ 1: Which of the following sets of vectors in $\mathbb{R}^3$ is linearly dependent?
A $\{(1, 0, 0), (0, 1, 0), (0, 0, 1)\}$
B $\{(1, 1, 1), (0, 1, 1), (0, 0, 1)\}$
C $\{(1, 2, 3), (2, 4, 6), (1, 1, 1)\}$
D $\{(1, 0, 0), (1, 1, 0), (1, 1, 1)\}$
Derivation & Analysis:

In option (C), the second vector $(2, 4, 6) = 2(1, 2, 3)$, making the set linearly dependent.

Q2
MCQ 2: If vectors $\alpha = (1, 2, k)$, $\beta = (2, 4, 6)$ are linearly dependent, the value of $k$ is:
A 2
B 3
C 4
D 6
Derivation & Analysis:

For proportionality, $\frac{1}{2} = \frac{2}{4} = \frac{k}{6} \implies k = 3$.

Q3
MCQ 3: The dimension of the vector space of all symmetric matrices of order $n \times n$ over $\mathbb{R}$ is related to the independence of basis elements. What is this dimension?
A $n^2$
B $\frac{n(n+1)}{2}$
C $\frac{n(n-1)}{2}$
D $2n$
Derivation & Analysis:

The independent entries in a symmetric matrix occur on and above the main diagonal, numbering $\frac{n(n+1)}{2}$.

Q4
MCQ 4: Any set of vectors containing the _______ vector is always linearly dependent.
A unit
B zero
C orthogonal
D prime
Derivation & Analysis:

By definition, $c(0) = 0$ for any non-zero scalar $c$, proving linear dependence immediately.

Q5
MCQ 5: If $S_1 \subset S_2$ and $S_1$ is linearly dependent, then $S_2$ is:
A linearly independent
B linearly dependent
C orthogonal
D empty
Derivation & Analysis:

Any superset of a linearly dependent set is also linearly dependent.

Q6
MCQ 6: The functions $f_1(x) = \sin x$ and $f_2(x) = \cos x$ on $\mathbb{R}$ are:
A Linearly dependent
B Linearly independent
C Orthogonal only
D Identical
Derivation & Analysis:

Their Wronskian $W(\sin x, \cos x) = \cos^2 x - (-\sin^2 x) = 1 \neq 0$, hence they are linearly independent.

Q7
MCQ 7: In an $n$-dimensional vector space $V$, any subset containing more than $n$ vectors is:
A Linearly independent
B Linearly dependent
C A basis
D Empty
Derivation & Analysis:

By fundamental theorem of vector spaces, any set with more than $n$ vectors must be linearly dependent.

Q8
MCQ 8: Assertion (A): The vectors $(1, 1), (2, 2)$ in $\mathbb{R}^2$ are linearly dependent. Reason (R): One vector is a scalar multiple of the other.
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, (R) is false
D (A) is false, (R) is true
Derivation & Analysis:

Scalar multiples in vector sets directly imply linear dependence.

Q9
MCQ 9: If $A$ is a square matrix of order $n$ and its columns are linearly independent, then:
A $\det(A) = 0$
B $\det(A) \neq 0$
C $\text{Rank}(A) < n$
D $A$ is a zero matrix
Derivation & Analysis:

Linear independence of columns of a square matrix is equivalent to invertibility, i.e., $\det(A) \neq 0$.

Q10
MCQ 10: The set $\{(1, i), (i, -1)\}$ over the complex field $\mathbb{C}$ is:
A Linearly independent
B Linearly dependent
C Orthogonal
D Basis for $\mathbb{C}^2$
Derivation & Analysis:

Notice that $i(1, i) - 1(i, -1) = (i, -1) - (i, -1) = (0,0)$. Non-zero scalars exist, so the set is linearly dependent over $\mathbb{C}$.

Common questions

No. Any set containing the zero vector is automatically linearly dependent because we can take a non-zero scalar multiplying the zero vector to yield the zero sum.

Orthogonal vectors have a dot product of zero ($u \cdot v = 0$), whereas linearly dependent vectors are scalar multiples or combinations of each other. Orthogonal non-zero vectors are always linearly independent, but independent vectors are not necessarily orthogonal.

For vector tuples, construct a matrix and check if its determinant is zero. For function sets, compute the Wronskian or check for polynomial/exponential quotient identities.

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