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7. Linear Algebra

Basis of a vector space

A basis of a vector space is a subset that is both linearly independent and spans the entire space, forming the foundational coordinate framework of linear algebra. This topic covers basis expansion, dimension theorems, and coordinate transformations.

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

Steinitz Exchange Lemma & Basis Theorem
Let $V$ be a vector space over a field $\mathbb{F}$. A subset $B = \{v_1, v_2, \dots, v_n\}$ of $V$ is called a basis of $V$ if: (1) $B$ is linearly independent, and (2) $\operatorname{span}(B) = V$. Furthermore, any two bases of a finitely generated vector space have the exact same number of elements, defining the dimension $\dim(V) = n$.

📐 Geometric Interpretation

Geometrically, a basis represents a set of independent axes (like standard Cartesian unit vectors $\hat{i}, \hat{j}, \hat{k}$) that span the entire geometric space without redundancy, allowing every vector to be uniquely represented as a linear combination.

⚙️ Algebraic & Structural Meaning

Algebraically, a basis establishes an isomorphism between an $n$-dimensional abstract vector space $V$ and the coordinate space $\mathbb{F}^n$, transforming complex subspace intersections and transformations into matrix computations.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition Domain / Interval Necessity What Fails If Condition Dropped
Linear Independence Entire Subset $B \subset V$ Strictly Mandatory / Necessary If vectors are linearly dependent, the set contains redundancies and fails to provide unique coordinate representations.
Spanning Property Entire Space $V$ Strictly Mandatory If the subset fails to span $V$, certain vectors in the space cannot be expressed as linear combinations of the basis elements.
Finitely Generated / Finite Dimensionality Vector Space $V$ Essential for unique dimension Infinite-dimensional spaces (such as the space of polynomials $\mathbb{R}[x]$) cannot possess a finite basis.

Useful results and shortcuts

Use these identities and results when their conditions apply:

01

Dimension of Polynomial Spaces

$$\dim(P_n) = n + 1$$

Rule: The space of polynomials of degree at most $n$ has dimension $n+1$, with standard basis $\{1, x, x^2, \dots, x^n\}$.

Application: undefined
02

Dimension of Matrix Spaces

$$\dim(M_{m \times n}(\mathbb{F})) = m \times n$$

Rule: The dimension of the vector space of all $m \times n$ matrices is the product of dimensions. For symmetric matrices of order $n$, it is $\frac{n(n+1)}{2}$; for skew-symmetric, it is $\frac{n(n-1)}{2}$.

03

Row Rank Equals Column Rank

$$\operatorname{rank}(A) = \operatorname{row-rank}(A) = \operatorname{col-rank}(A)$$

Rule: To find a basis for the row/column space or range of a matrix, apply elementary row operations to reduce it to row-echelon form; the non-zero rows form the basis.

Application: Standard computational question in vector space and linear transformation sections.
04

Extension Theorem Shortcut

$$\text{Any L.I. set of } k \text{ vectors can be extended to a basis of } n \text{ dimensions ($k \le n$)}$$

Rule: If you are given $k$ linearly independent vectors in an $n$-dimensional space, append standard basis vectors and eliminate dependent ones using determinant non-zero checks.

05

Wronskian Test for Function Bases

$$W(f_1, f_2, \dots, f_n)(x) \neq 0 \implies \text{Linearly Independent}$$

Rule: For differentiable functions on an interval, if the Wronskian determinant is non-zero at least at one point, the functions form a linearly independent set.

Application: Used in differential equation solution spaces (UP PGT Mathematics).
06

Intersection and Sum Formula (Grassmann's Law)

$$\dim(W_1 + W_2) = \dim(W_1) + \dim(W_2) - \dim(W_1 \cap W_2)$$

Rule: Instantly calculate missing dimensions of subspace sums or intersections without computing explicit bases.

Application: Direct formula substitution questions appear almost every alternate year.

Common errors and counterexamples

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

⚠️ COMMON ERROR #01

Assuming Any Spanning Set is a Basis

Common Pitfall: Candidates often check only if a set spans the space and forget to verify linear independence.

Counterexample: The set $\{(1,0), (0,1), (1,1)\}$ in $\mathbb{R}^2$

Why it fails: It spans $\mathbb{R}^2$, but it is linearly dependent (contains 3 vectors in a 2-dimensional space), hence it is NOT a basis.

🛡️ Exam Defense: In an $n$-dimensional space, a basis must have exactly $n$ vectors. If $|B| > n$, it cannot be a basis (linearly dependent).
⚠️ COMMON ERROR #02

Field Dependency (Scalars Matter)

Common Pitfall: Failing to account for the underlying field $\mathbb{F}$ when determining dimension and basis.

Counterexample: $\mathbb{C}$ over $\mathbb{R}$ vs $\mathbb{C}$ over $\mathbb{C}$

Why it fails: $\dim_{\mathbb{R}}(\mathbb{C}) = 2$ (basis $\{1, i\}$), whereas $\dim_{\mathbb{C}}(\mathbb{C}) = 1$ (basis $\{1\}$)

🛡️ Exam Defense: Always read the underlying field carefully in the problem statement before stating the dimension.

Comparison

Concept / TheoremHypotheses / AssumptionsKey Formula / OutcomeGeometric MeaningExam Identification Clue
Linearly Independent Set$\sum c_i v_i = 0 \implies c_i = 0$No redundancyVectors point in independent directionsMaximal vectors without collinearity
Spanning SetEvery vector is a linear combination$\operatorname{span}(B) = V$Generates the entire geometric spaceSufficient vectors to cover the space
BasisL.I. + SpanningCardinality = $\dim(V)$Minimal spanning / maximal independent setExact coordinate frame for $V$
Grassmann's IdentitySubspaces $W_1, W_2 \subset V$$\dim(W_1+W_2) = \dim W_1 + \dim W_2 - \dim(W_1 \cap W_2)$Dimension balance of combined spacesGiven dimensions of two subspaces and intersection

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

Find the dimension of the subspace of $\mathbb{R}^3$ spanned by the vectors $v_1 = (1, -1, 0)$, $v_2 = (2, 1, 3)$, and $v_3 = (1, 2, 3)$.

(A) 1(B) 2(C) 3(D) 0
✓ Correct Answer: (B)
📚 Method A: Formal Textbook Derivation
Step 1: Form matrix $A$ with rows as vectors. Step 2: Apply row operations to find row rank. $R_2 \to R_2 - 2R_1$ and $R_3 \to R_3 - R_1$ gives row-echelon form with 2 non-zero rows. Thus, $\dim = 2$.
⚡ Method B: 25-Second Exam Shortcut
Compute the determinant of the $3 \times 3$ matrix formed by the vectors. $\det(A) = 0$, meaning vectors are linearly dependent (dim < 3). Check any $2 \times 2$ minor, e.g., $\det\begin{pmatrix} 1 & -1 \\ 2 & 1 \end{pmatrix} = 3 \neq 0$, so rank = 2.
💡 Tip: If determinant is zero, check non-zero $2 \times 2$ minors instantly to confirm rank 2.
CANONICAL PROBLEM 02 Dual-Method Breakdown

What is the dimension of the vector space of all symmetric matrices of order $n \times n$ over $\mathbb{R}$?

(A) $n^2$(B) $\frac{n(n-1)}{2}$(C) $\frac{n(n+1)}{2}$(D) $n^2 - n$
✓ Correct Answer: (C)
📚 Method A: Formal Textbook Derivation
A symmetric matrix $A = [a_{ij}]$ satisfies $a_{ij} = a_{ji}$. There are $n$ independent diagonal entries and $\frac{n^2-n}{2}$ independent entries above the main diagonal. Total = $n + \frac{n^2-n}{2} = \frac{n(n+1)}{2}$.
⚡ Method B: 25-Second Exam Shortcut
Direct formula recall: Symmetric = $\frac{n(n+1)}{2}$, Skew-symmetric = $\frac{n(n-1)}{2}$.
💡 Tip: Direct memory-based question frequently asked in UP PGT and KVS exams.

Practice questions

Choose an answer to check it and see the solution:

Q1
MCQ 1: If $V$ is a vector space of all polynomials of degree at most $n$, then the dimension of $V$ over $\mathbb{R}$ is:
A $n$
B $n-1$
C $n+1$
D $2n$
Derivation & Analysis:

The standard basis is $\{1, x, x^2, \dots, x^n\}$, which contains exactly $n+1$ elements. Thus, $\dim(P_n) = n+1$.

Q2
MCQ 2: The dimension of the vector space $\mathbb{C}$ over the field of real numbers $\mathbb{R}$ is:
A 1
B 2
C Infinite
D 4
Derivation & Analysis:

Any complex number can be uniquely written as $a + bi$ where $a, b \in \mathbb{R}$. The basis over $\mathbb{R}$ is $\{1, i\}$, hence the dimension is 2.

Q3
MCQ 3: Which of the following sets forms a basis for $\mathbb{R}^3$?
A $\{(1,0,0), (1,1,0), (1,1,1)\}$
B $\{(1,2,3), (2,4,6), (0,0,1)\}$
C $\{(1,1,1), (2,2,2), (3,3,3)\}$
D $\{(1,0,0), (0,1,0)\}$
Derivation & Analysis:

Option (A) consists of 3 vectors with a non-zero determinant $\det = 1 \neq 0$, meaning they are linearly independent and span $\mathbb{R}^3$.

Q4
MCQ 4: If $W_1$ and $W_2$ are two subspaces of a finite-dimensional vector space $V$ with $\dim(W_1) = 3$, $\dim(W_2) = 4$, and $\dim(W_1 \cap W_2) = 1$, then $\dim(W_1 + W_2)$ is:
A 7
B 6
C 5
D 3
Derivation & Analysis:

Using Grassmann's law: $\dim(W_1 + W_2) = \dim(W_1) + \dim(W_2) - \dim(W_1 \cap W_2) = 3 + 4 - 1 = 6$.

Q5
MCQ 5: The dimension of the vector space of all skew-symmetric matrices of order $n \times n$ over $\mathbb{R}$ is:
A $\frac{n(n+1)}{2}$
B $\frac{n(n-1)}{2}$
C $n^2 - n$
D $n^2$
Derivation & Analysis:

For a skew-symmetric matrix, diagonal entries are zero ($a_{ii} = 0$) and $a_{ij} = -a_{ji}$. The number of independent entries above the diagonal is $\frac{n(n-1)}{2}$.

Q6
MCQ 6: If $S$ is a subset of a vector space $V$ containing the zero vector $0$, then $S$ is:
A Always a basis
B Always linearly independent
C Always linearly dependent
D A spanning set
Derivation & Analysis:

Any set containing the zero vector is linearly dependent because $1 \cdot 0 + \sum 0 \cdot v_i = 0$ provides a non-trivial linear combination.

Q7
MCQ 7: Let $V$ be the vector space of all matrices of order $2 \times 2$ over $\mathbb{R}$. The dimension of $V$ is:
A 2
B 3
C 4
D 8
Derivation & Analysis:

A $2 \times 2$ matrix has 4 independent entries, with standard basis matrices $E_{11}, E_{12}, E_{21}, E_{22}$. Thus, dimension is 4.

Q8
MCQ 8: If $B_1$ and $B_2$ are two bases of a finite-dimensional vector space $V$, then:
A $|B_1| \neq |B_2|$
B $|B_1| = |B_2|$
C $|B_1| > |B_2|$
D $|B_1| < |B_2|$
Derivation & Analysis:

By the invariance of dimension theorem, any two bases of a finite-dimensional vector space contain the exact same number of elements.

Q9
MCQ 9: The dimension of the vector space of all diagonal matrices of order $n \times n$ is:
A $n$
B $n^2$
C $\frac{n(n+1)}{2}$
D $n^2 - n$
Derivation & Analysis:

A diagonal matrix has non-zero entries only on the main diagonal. There are $n$ independent diagonal entries, so its dimension is $n$.

Q10
MCQ 10: Which statement is INCORRECT regarding a basis $B$ of a vector space $V$?
A $B$ is a maximal linearly independent set.
B $B$ is a minimal spanning set.
C Every vector in $V$ can be expressed uniquely as a linear combination of elements of $B$.
D $B$ must be a finite set for any vector space.
Derivation & Analysis:

Infinite-dimensional vector spaces (such as polynomials $\mathbb{R}[x]$) have bases that are infinite. Thus, statement (D) is false.

Common questions

Yes! A vector space has infinitely many different bases (except for trivial spaces like $\mathbb{R}^0$ or $\mathbb{F}_p^n$ over finite fields), but every basis of the same space has the exact same number of elements (dimension).

Students frequently confuse linear independence with spanning, or forget to check the underlying field (e.g., treating $\mathbb{C}$ as a vector space over $\mathbb{C}$ instead of $\mathbb{R}$). Always verify both conditions and the scalar field.

Memorize standard dimension formulas (symmetric matrices, skew-symmetric matrices, polynomial spaces, trace-zero matrices) and use determinant shortcuts to quickly test linear independence.

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