The document discusses linear independence and bases in vector spaces. It defines linear independence as a set of vectors where none can be written as a linear combination of the others. A basis is defined as a linearly independent set of vectors that spans the entire vector space. Several examples are provided to illustrate these concepts, including showing that the vectors (1, 2) and (4, 1) form a basis for R2. The document also discusses properties of linear independence and how it relates to bases being minimal and spanning the entire space.

1. Linear Algebra
15UMTC61
K.Anitha M.Sc., M.Phil.,
G.Nagalakshmi M.Sc., M.phil.,

2. Linear Independence
Definition and Examples
A subset of a vector space is linearly independent (L.I.) if
none of its elements is a linear combination of the others.
Otherwise, it is linearly dependent (L.D.).

3. Example : 2-D Row vectors
{ (40 15), (50 25) } is L.I.
Proof:
     40 15 50 25 0 0a b  Let
→
40 50 0
15 25 0
a b
a b
 
 
→
4 5 0
3 5 0
a b
a b
 
 
→
0
0
a
b


→
7 0 0
3 5 0
a
a b
 
 
{ (40 15), (20 7.5) } is L.D.
Proof:
     40 15 20 7.5 0 0a b Let
→
40 20 0
15 7.5 0
a b
a b
 
 
→
2 0
2 0
a b
a b
 
 
→ 2b a 

4. Example
1 0 1 0 3
0 , 2 , 2 , 1 , 3
0 0 0 1 0
S
          
                     
          
          
spans R3
The independence test equation 1 2 3 4 5
1 0 1 0 3 0
0 2 2 1 3 0
0 0 0 1 0 0
c c c c c
           
                           
           
           
gives
1 0 1 0 3
0 2 2 1 3
0 0 0 1 0
 
  
 
 
3 4
1 0 0 1 3
0 2 1 2 3
0 0 1 0 0
 
 
    
 
 
1 0 0 1 3
0 1 0 1 3/ 2
0 0 1 0 0
 
   
 
 

1
2
4 3 5
3
5
1 3
1 3/ 2
0 0
1 0
0 1
c
c
c c c
c
c
      
           
      
     
     
    
    
1
2
3 3 5
4
5
1 3
1 3/ 2
1 0
0 0
0 1
c
c
c c c
c
c
      
           
      
     
     
    
    
or →
1 0 0
0 , 2 , 1
0 0 1
      
            
      
      
is L.I. & spans R3

5. Example : P2
{ 1+x , 1x } is L.I.
Proof:
Let     2
1 1 0 0 0a x b x x x     
→
0
0
a b
a b
 
 
→
0
0
a
b


Example : R3
Let 1 2 3
3 2 4
4 9 18
5 2 4
     
            
     
     
v v v then S = { v1 ,v2 , v3 } is L.D.
Proof:
3 2 4 3 2 4
4 9 18 0 19/3 38/3
5 2 4 0 4/3 8/3
   
      
       
3 2 4
0 1 2
0 1 2
 
   
 
 
3 0 0
0 1 2
0 0 0
 
   
 
 
→ 2 32 v v 0

6. Example : Empty subset is L.I.
Example : Any subset S containing 0 is L.D.
10 ... 0 n a   v v 0 0
Alternative proof 1:
Proof:
Theorem :
Any finite subset S of a vector space V has a L.I. subset U with the same span as
S.
Proof:
• If S is L.I., setting U = S completes the proof.
• If S is not L.I.,  sk s.t. sk = jk cj sj .
By Lemma 1.1, span S1 = span S, where S1 = S  {sk} .
If S1 is L.I., the proof is complete.
• Else, repeat the extraction until L.I. is achieved. QED.
 a  R
10 ... 0 n  0 v v
Alternative proof 2: 0 is a linear combination of the empty set   S.

7. Lemma :
Any subset of a L.I. set is also L.I. ( L.I.
is preserved by the subset operation.)
Any superset of a L.D. set is also L.D. ( L.D. is
preserved by the superset operation.)
Proof: Trivial.
Subset of a L.D. set can either be L.I. or L.D. (see Example 1.13)
Superset of a L.I. set can either be L.I. or L.D. (see Example 1.15)
Example :
1 0
0 , 1
0 0
S
    
         
    
    
1 0 3
0 , 1 , 2
0 0 0
      
            
      
      
is L.D.
1 0 0
0 , 1 , 0
0 0 1
      
      
      
      
      
is L.I.
S1  S S2  S
S is L.I. S1 is L.I. --
S is L.D. -- S2 is L.D.

8. Basis
Definition : Basis
A basis of a vector space V is an ordered set of linearly
independent (non-zero) vectors that spans V.
Notation:
1 , , nβ β
Example :
2 1
,
4 1
B
   
    
   
is a basis for R2
B is L.I. :
2 1 0
4 1 0
a b
     
      
     
→
2 0
4 0
a b
a b
 
 
→
0
0
a
b


B spans R2:
2 1
4 1
x
a b
y
     
      
     
→
2
4
a b x
a b y
 
 
→
 
1
2
2
a y x
b x y
 
 
L.I. → Minimal
Span → Complete

9. Example :
The trivial space { 0 } has only one basis, the empty one  .
Note: By convention, 0 does not count as a basis vector. ( Any
set of vectors containing 0 is linearly dependent. )
Example :
The space of all finite degree polynomials has a basis with infinitely many
elements  1, x, x2, … .
Example : Solution Set of Homogeneous Systems
The solution set of
0
0
x y w
z w
  
 
1 1
1 0
,
0 1
0 1
y w y w
    
    
      
    
        
Ris
1 1
1 0
,
0 1
0 1
Span
    
    
      
    
        
( Proof of L.I. is
left as exercise )

10. Theorem :
In any finite-dimensional vector space, all of the bases have
the same number of elements.
Proof:
Let B =  β1 , …, βn  be a basis of n elements.
Any other basis D =  δ1 , …, δm  must have m  n.
1 1 1 k k n nc c c    δ β β βLet with ck  0.
By lemma 2.2, D1 =  β1 , …, βk  1 ,δ1 ,βk + 1 , …, βn  is a basis.
Next, replacing βj in D1 begets
D2 =  β1 , …, βk  1 ,δ1 ,βk + 1 , …, βj  1 ,δ2 ,βj + 1 , …, βn 
Repeating the process n times results in a basis Dn =  δ1 , …, δn  that spans V.
Which contradicts with the assumption that D is L.I.
1 1 1n n nc c   δ δ δ with at least one ck  0.If m > n, then we can write
Hence m = n.

11. Ex : Testing for linear independence
Determine whether the following set of vectors in P2 is L.I. or
L.D.
c1v1+c2v2+c3v3 = 0
i.e., c1(1+x – 2x2) + c2(2+5x – x2) + c3(x+x2) = 0+0x+0x2

c1+2c2 = 0
c1+5c2+c3 = 0
–2c1– c2+c3 = 0
Sol:
 This system has infinitely many solutions
(i.e., this system has nontrivial solutions, e.g., c1=2, c2= – 1, c3=3)
 S is (or v1, v2, v3 are) linearly dependent
1 2 0 0
1 5 1 0
2 1 1 0
 
 
 
   
1 2 0 0
10 1 0
3
0 0 0 0
 
 
 
 
 

G. E.

   2 2 2
1 2 3, , 1 2 ,2 5 ,S x x x x x x      v v v

12. Theorem : Uniqueness of basis representation for any vectors
If is a basis for a vector space V, then
everyvector in V can be written in one and only one way as a
linear combination of vectors in S
 nS vvv ,,, 21 
Pf:



basisaisS
(1) span(S) = V
(2) S is linearly independent
 span(S) = V Let v = c1v1+c2v2+…+cnvn
v = b1v1+b2v2+…+bnvn
 v + (–1)v = 0 = (c1– b1)v1 + (c2 – b2)v2 + … + (cn – bn)vn
is linearly independent with only the trivial solutionS 
(i.e., unique basis representation) c1 = b1 , c2 = b2 ,…, cn = bn
coefficients for are all zeroi v

13. Ex: Find the dimension of a vector space according to the standard basis
(1) Vector space Rn
 standard basis {e1 , e2 ,  , en}
(2) Vector space Mmn  standard basis {Eij | 1im , 1jn}
(3) Vector space Pn(x)  standard basis {1, x, x2,  , xn}
(4) Vector space P(x)  standard basis {1, x, x2, }
 dim(R
n
) = n
 dim(Mmn) = mn
 dim(Pn(x)) = n+1
 dim(P(x)) = 
※ The simplest way to find the dimension of a vector space is to count the
number of vectors in the “standard” basis for that vector space
and in Eij
1
other entries are zero
ija 



14. Pf:
Theorem : Row-equivalent matrices have the same row space
If an mn matrix A is row equivalent to an mn matrix B,
then the row space of A is equal to the row space of B
(1) Since B can be obtained from A by elementary row operations, the row vectors of
B can be expressed as linear combinations of the row vectors of A  The linear
combinations of row vectors in B must be linear combinations of row vectors in
A  any vector in RS(B) lies in RS(A)  RS(B)  RS(A)
(2) Since A can be obtained from B by elementary row operations, the row vectors of
A can be written as linear combinations of the row vectors of B  The linear
combinations of row vectors in A must be linear combinations of row vectors in
B  any vector in RS(A) lies in RS(B)  RS(A)  RS(B)
( ) = ( )RS A RS B

15. Summary of equivalent conditions for square matrices:
If A is an n×n matrix, then the following conditions are
equivalent
(1) A is invertible
(2) Ax = b has a unique solution for any n×1 matrix b
(3) Ax = 0 has only the trivial solution
(4) A is row-equivalent to In
(5) det (A)  0
(6) rank(A) = n
(7) There are n row vectors of A which are linearly independent
(8) There are n column vectors of A which are linearly independent
(The above five statements are from Slide 3.39)
(The last three statements are from the arguments on Slide 4.95)