Lecture 11

Linear Algebra and Differential Equations

6th February 2025

Rahul Dattatraya Kitture

Quick Revision:

▶ (T/F) Zero vector is linearly independent.

▶ (T/F) Vectors (1, 2) and (3, 6) in R2 are linearly independent.

▶ (T/F) Vectors (1, 2) and (2, 1) are linearly independent in R2 .

▶ (T/F) Zero vector of vector space is in the span of every subset

of the space.

▶ A subset S = {v1 , v2 , . . . , vn } of a vector space V is called a

spanning set for V if ........

▶ (T/F) For the vector space P5 of all polynomials of degree ≤ 5,

the subset S = {x, x 2 , x 3 , x 4 , x 5 } is a spanning set for P5 .
Basis of a vector space: Definition as in book (p.196)

If V is any vector space, and if S = {v1 , v2 , . . . , vk } is a finite

set of vectors in V , then S is called a basis of V if following
two conditions hold:

▶ S is linearly independent;

▶ S spans V .
Example 1.
Consider vector space R2 over R. Then show that

S = {(1, 0), (0, 1)}

is a basis for R2 :
Step 1: S is linearly independent:

α(1, 0) + β(0, 1) = (0, 0)

=⇒ (α, β) = (0, 0)
=⇒ α = β = 0.

Step 2: S spans R2 :
Take any vector (a, b) in R2 . Then we can write

(a, b) = a(1, 0) + b(0, 1).

Hence S is a basis of R2 .
Example 2: Standard Basis

▶ We saw: {(1, 0), (0, 1)} is a basis of the vector space R2 .

It is called a natural/standard basis.

▶ In general, for Rn , the set

{(1, 0, . . . , 0), (0, 1, 0, . . . , 0), ··· (0, . . . , 1)

is a basis, called standard basis for Rn .

▶ Similarly, for vector space of 2 × 2 matrices, the set

         
1 0 0 1 0 0 0 0
, , ,
0 0 0 0 1 0 0 1

is a standard basis.

▶ Write a standard basis for vector space of all 2 × 3 real

matrices.
Example 3. Standard basis for Pn :

For the vector space Pn of all polynomials of degree ≤ n over R, a

standard basis is
1, x, x 2 , . . . , x n .
Example 4.

Consider vector space R3 over R. Is following set a basis of R3 ?

S = {(1, 0, 0), (0, 0, 1)}

Step 1: Is S is linearly independent?

α(1, 0, 0) + β(0, 0, 1) = (0, 0, 0) =⇒ (α, 0, β) = (0, 0, 0)

=⇒ α = β = 0.

Yes; S is linearly independent.

Step 2: Does S span R3 ?
NO: because in (1, 0, 0) and (0, 0, 1), the second coordinate is 0,
so in their any linear combination, the middle coordinate will be 0.
In particular, (∗, 1, ∗) can not be linear combination of vectors in S.
So S is NOT a basis of R3 .
Example 5.

If {u, v} is a basis of space V , show: {u + v, v} is also a basis.

Step 1: {u + v, v} is linearly independent: because

α(u + v) + βv = 0 =⇒ αu + (α + β)v = 0
=⇒ α + β = 0 and β = 0
=⇒ α = β = 0.

Step 2. {u + v, v} spans V :

Take any x in V .

Then x = αu + βv (since span{u, v} = V )

So x = αu + αv − αv + βv

So x = α(u + v) + (β − α)v
Exercise (Tutorial: Chapter 3: 11(b))

If previous example is clear, then you can solve following exercise

easily:

Let V be a real vector space with {u, v, w} as a basis.

Check whether

{u + v + w, v + w, w}

is also a basis of V
Property of Basis: (Theorem 4.4.1, p.198)

If S = {v1 , v2 , . . . , vn } is a basis for a vector space V , then

each vector u ∈ V has unique expression:

u = c1 v1 + c2 v2 + · · · + cn vn (ci ∈ R).

Example: For R2 , the set {(1, 0), (0, 1), (1, 1)} is NOT a basis.

Why? Consider vector (1, 1): it has non-unique expression:

(1, 1) = 1(1, 0) + 1(0, 1) + 0(1, 1)

(1, 1) = 0(1, 0) + 0(0, 1) + 1(1, 1)

Coordinates relative to a basis:

Let S = {v1 , v2 , . . . , vn } be a basis for a vector space V .

So each vector u ∈ V has unique expression:

u = c1 v1 + c2 v2 + · · · + cn vn (ci ∈ R).

The tuple
(c1 , c2 , . . . , cn )
is called coordinates of v relative to basis S.
Example 0:

▶ We saw: {u, v} = {(1, 0), (0, 1)} is a basis for R2 .

(1, 1)

(1, 0)

Q. Find coordinates of (3, 2) relative to basis {(1, 0), (0, 1)}.

Ans: We need to find c1 , c2 such that

(3, 2) = c1 (1, 0) + c2 (0, 1).

Solving it, we get: (c1 , c2 ) = (3, 2) are the coordinates of (3, 2).
Example 1:
▶ We saw: {u, v} = {(1, 0), (0, 1)} is a basis for R2 .

▶ We also saw: {u + v, v} is a basis for R2 .

▶ So {(1, 1), (0, 1)} is a basis of R2 .

(0, 1)
(1, 1)

Q. Find coordinates of (3, 2) relative to basis {(1, 1), (0, 1)}.

Ans: We need to find c1 , c2 such that
(3, 2) = c1 (1, 1) + c2 (0, 1).
Solving it, we get: (c1 , c2 ) = (3, −1) are the coordinates of (3, 2).
Example 2: Slight modification of previous:
▶ We saw: {u, v} = {(1, 0), (0, 1)} is a basis for R2 .

▶ So: {v, u} = {(0, 1), (1, 0)} is also basis for R2 .

(0, 1)

(1, 0)

Q. Find coordinates of (3, 2) relative to basis {(0, 1), (1, 0)}.

Ans: We need to find c1 , c2 such that

(3, 2) = c1 (0, 1) + c2 (1, 0).

Solving it, we get: (c1 , c2 ) = (2, 3) are the coordinates of (3, 2).
Summary of three examples:

Different bases of a space can give different coordinates for

the same vector.

First basis: second basis: third basis:

{(1, 0), (0, 1)} {(1, 1), (0, 1)} {(0, 1), (1, 0)}

Coordinates of (3, 2)

(c1 , c2 ) = (3, 2) (c1 , c2 ) = (3, −1) (c1 , c2 ) = (2, 3)

Important Result on Basis

Theorem: If V is a vector space, which has a finite spanning

set, then

any two bases of V have same size.

Example: In previous three examples, we saw three bases of R2 :

B1 = {(1, 0), (0, 1)} (standard basis)

B2 = {(1, 1), (0, 1)}

B3 = {(0, 1), (1, 0)}

Geometric way to get a basis:

One can easily get various bases for the vector space R2 :

▶ Take a non-zero vector u;

▶ Then span{u} is the line through origin which contains u.

▶ Take second vector v which is not in this line!

Then
{u, v}
is a basis for R2 !
A basis for R3 :
One can similarly get many bases for the vector space R3 :

▶ Since standard basis of R3 contains 3 vectors, so

any basis of R3 contains 3 vectors.

How to get three vectors for a basis?

(1) Take any non-zero vector u.

(2) Consider line (subspace) spanned by u; take vector v outside

this line.

(3) Consider span{u, v}.

Then span{u, v} is a plane in R3 (passing through origin).

Take a vector w which is not in the span{u, v}.

Then {u, v, w} is a basis for R3 .