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UGC CSIR NET · Mathematics

Vector Basics

Explanation

# Vector Spaces ## Concept A **vector space** over a field FF is a non-empty set VV together with two operations: - Vector Addition - Scalar Multiplication satisfying all the vector space axioms. --- ## Important Formulae ### Span
Span⁡(S)={∑i=1ncivi:ci∈F,  vi∈S} \operatorname{Span}(S)=\left\{\sum_{i=1}^{n} c_i v_i : c_i \in F,\; v_i \in S\right\}
### Linear Independence Vectors v1,v2,…,vnv_1,v_2,\ldots,v_n are linearly independent if
c1v1+c2v2+⋯+cnvn=0 c_1v_1+c_2v_2+\cdots+c_nv_n=0
implies
c1=c2=⋯=cn=0. c_1=c_2=\cdots=c_n=0.
### Dimension
dim⁡(V)=Number of vectors in a basis of V \dim(V)=\text{Number of vectors in a basis of }V
--- ## Quick Notes - Every basis is linearly independent. - Every basis spans the vector space. - All bases of a vector space have the same number of vectors. - The dimension of a vector space is unique. --- ## Solved Example Determine whether
(1,2),  (2,4) (1,2),\;(2,4)
are linearly independent. ### Solution Since
(2,4)=2(1,2), (2,4)=2(1,2),
one vector is a scalar multiple of the other. Therefore,
The vectors are linearly dependent. \boxed{\text{The vectors are linearly dependent.}}
--- ## Shortcut - If one vector is a scalar multiple of another, they are **linearly dependent**. - If the determinant is non-zero, the vectors are **linearly independent**. --- ## Common Mistakes - Assuming different vectors are automatically independent. - Forgetting to check all coefficients. - Making mistakes while calculating determinants. --- ## Formula Summary
Span⁡(S)={∑i=1ncivi} \operatorname{Span}(S)=\left\{\sum_{i=1}^{n} c_i v_i\right\}
c1v1+⋯+cnvn=0⟹c1=c2=⋯=cn=0 c_1v_1+\cdots+c_nv_n=0 \Longrightarrow c_1=c_2=\cdots=c_n=0
dim⁡(V)=Number of basis vectors \dim(V)=\text{Number of basis vectors}
--- ## Revision Tips - Basis = Independent + Span - Determinant ≠0\neq 0 ⇒ Independent - Determinant =0= 0 ⇒ Dependent - Dimension = Number of basis vectors

Example

# Vector Spaces ## Concept A **vector space** over a field FF is a non-empty set VV together with two operations: - Vector Addition - Scalar Multiplication that satisfy all the vector space axioms. --- ## Example 1 ### Question Show that R2\mathbb{R}^2 is a vector space. ### Solution Consider
V=R2={(x,y):x,y∈R}. V=\mathbb{R}^2=\{(x,y):x,y\in\mathbb{R}\}.
For any two vectors,
(2,3)+(1,4)=(3,7)∈V. (2,3)+(1,4)=(3,7)\in V.
Hence, VV is closed under addition. For any scalar,
3(2,5)=(6,15)∈V. 3(2,5)=(6,15)\in V.
Hence, VV is closed under scalar multiplication. All vector space axioms are satisfied. **Answer**
R2 is a vector space. \boxed{\mathbb{R}^2 \text{ is a vector space.}}
--- ## Example 2 ### Question Is the set
V={x∈R:x>0} V=\{x\in\mathbb{R}:x>0\}
a vector space? ### Solution Take
x=2∈V. x=2\in V.
Multiply by the scalar
−1. -1.
Then
(−1)(2)=−2. (-1)(2)=-2.
Since
−2∉V, -2\notin V,
the set is **not closed under scalar multiplication**. **Answer**
V is NOT a vector space. \boxed{V \text{ is NOT a vector space.}}
--- # Span ## Formula
Span⁡(S)={∑i=1ncivi} \operatorname{Span}(S)=\left\{\sum_{i=1}^{n}c_iv_i\right\}
--- ## Example Find the span of
S={(1,0),(0,1)}. S=\{(1,0),(0,1)\}.
### Solution Every vector
(a,b) (a,b)
can be written as
a(1,0)+b(0,1). a(1,0)+b(0,1).
Therefore,
Span⁡(S)=R2. \boxed{\operatorname{Span}(S)=\mathbb{R}^2.}
--- # Linear Independence ## Formula
c1v1+c2v2+⋯+cnvn=0 c_1v_1+c_2v_2+\cdots+c_nv_n=0
implies
c1=c2=⋯=cn=0. c_1=c_2=\cdots=c_n=0.
--- ## Example 1 Determine whether
(1,2),(2,4) (1,2),(2,4)
are linearly independent. ### Solution Observe that
(2,4)=2(1,2). (2,4)=2(1,2).
Therefore, one vector is a scalar multiple of the other. **Answer**
The vectors are linearly dependent. \boxed{\text{The vectors are linearly dependent.}}
--- ## Example 2 Determine whether
(1,0),(0,1) (1,0),(0,1)
are linearly independent. ### Solution Assume
c1(1,0)+c2(0,1)=(0,0). c_1(1,0)+c_2(0,1)=(0,0).
Then
(c1,c2)=(0,0). (c_1,c_2)=(0,0).
Hence,
The vectors are linearly independent. \boxed{\text{The vectors are linearly independent.}}
--- # Dimension ## Formula
dim⁡(V)=Number of vectors in a basis of V. \dim(V)=\text{Number of vectors in a basis of }V.
--- ## Example 1 Find the dimension of
R2. \mathbb{R}^2.
### Solution A basis is
{(1,0),(0,1)}. \{(1,0),(0,1)\}.
There are two basis vectors. Therefore,
dim⁡(R2)=2. \boxed{\dim(\mathbb{R}^2)=2.}
--- ## Example 2 Find the dimension of
R3. \mathbb{R}^3.
### Solution A basis is
{(1,0,0),(0,1,0),(0,0,1)}. \{(1,0,0),(0,1,0),(0,0,1)\}.
There are three basis vectors. Therefore,
dim⁡(R3)=3. \boxed{\dim(\mathbb{R}^3)=3.}
--- ## Exam Tips - If one vector is a scalar multiple of another, the vectors are **linearly dependent**. - If the determinant of the coefficient matrix is non-zero, the vectors are **linearly independent**. - The number of vectors in a basis equals the **dimension**. - The standard basis of Rn\mathbb{R}^n always has nn vectors.