Vector Space over R and C | Mathematics Optional Notes for UPSC PDF Download

 Page 1


Edurev123 
 
1. Vector Space over R and C 
 
Q1.1: Prove that the set ?? of the vectors (?? ?? ,?? ?? ,?? ?? ,?? ?? ) in ?? ?? which satisfy the 
equations ?? ?? +?? ?? +?? ?? ?? +?? ?? : 0 and ?? ?? ?? +?? ?? ?? -?? ?? +?? ?? =?? is a subspace of ?? ?? . 
What is the dimension of this subspace. Find one of its bases. 
(2009 : 12 Marks)  
Solution: 
ILD for vertical.reaction at ?? ; 
?? ={(?? 1
,?? 2
,?? 3
,?? 4
)??? 4
|?? 1
+?? 2
+2?? 3
+?? 4
=0,2?? 1
+3?? 2
-?? 3
+?? 4
=0}
 
Then clearly (0,0,0,0)??? and so ?? is non-empty. 
Again let ?? =(?? 1
,?? 2
,?? 3
,?? 4
)??? and ?? =(?? 1
,?? 2
,?? 3
,?? 4
)??? , Also let ?? ,?? ??? . 
Also, ?? (?? 1
'
+?? 2
+2?? 3
+?? 4
)+?? (?? 1
+?? 2
+2?? 3
+?? 4
)=0 
???? +???? =(?? ?? 1
+?? ?? 1
,?? ?? 2
+?? ?? 2
,?? ?? 3
+?? ?? 3
,?? ?? 4
+?? ?? 4
)??? 4
 
????? +???? satisfies ?? 1
+?? 2
+2?? 3
+?? 4
=0 
Similarly ???? +???? satisfies 2nd equation as well. 
? ???? +???? ??? 
Dimension of the Subspace : 
Any element of ?? is a solution to equation 
[
1 1 2 1
2 3 -1 1
][
?? 1
?? 2
?? 3
?? 4
]=[
0
0
0
0
] 
So, its dimension is same as rowspace of coefficient matrix, i.e., its rank. 
Reducing it to row reduced echelon form 
Page 2


Edurev123 
 
1. Vector Space over R and C 
 
Q1.1: Prove that the set ?? of the vectors (?? ?? ,?? ?? ,?? ?? ,?? ?? ) in ?? ?? which satisfy the 
equations ?? ?? +?? ?? +?? ?? ?? +?? ?? : 0 and ?? ?? ?? +?? ?? ?? -?? ?? +?? ?? =?? is a subspace of ?? ?? . 
What is the dimension of this subspace. Find one of its bases. 
(2009 : 12 Marks)  
Solution: 
ILD for vertical.reaction at ?? ; 
?? ={(?? 1
,?? 2
,?? 3
,?? 4
)??? 4
|?? 1
+?? 2
+2?? 3
+?? 4
=0,2?? 1
+3?? 2
-?? 3
+?? 4
=0}
 
Then clearly (0,0,0,0)??? and so ?? is non-empty. 
Again let ?? =(?? 1
,?? 2
,?? 3
,?? 4
)??? and ?? =(?? 1
,?? 2
,?? 3
,?? 4
)??? , Also let ?? ,?? ??? . 
Also, ?? (?? 1
'
+?? 2
+2?? 3
+?? 4
)+?? (?? 1
+?? 2
+2?? 3
+?? 4
)=0 
???? +???? =(?? ?? 1
+?? ?? 1
,?? ?? 2
+?? ?? 2
,?? ?? 3
+?? ?? 3
,?? ?? 4
+?? ?? 4
)??? 4
 
????? +???? satisfies ?? 1
+?? 2
+2?? 3
+?? 4
=0 
Similarly ???? +???? satisfies 2nd equation as well. 
? ???? +???? ??? 
Dimension of the Subspace : 
Any element of ?? is a solution to equation 
[
1 1 2 1
2 3 -1 1
][
?? 1
?? 2
?? 3
?? 4
]=[
0
0
0
0
] 
So, its dimension is same as rowspace of coefficient matrix, i.e., its rank. 
Reducing it to row reduced echelon form 
[
1 1 2 1
2 3 -1 1
] ?
?? 2
??? 2
-2?? 1
[
1 1 2 1
0 1 -5 -1
] 
As it has two non-zero rows in row reduced form. 
dim (?? )= Rank of matrix =2 
Writing the equation as matrix 
[
1 1 2 1
2 3 -1 1
][
?? 1
?? 2
?? 3
?? 4
]=[
0
0
0
0
] or ???? =0 
???? ,   ?? ={?? =(?? 1
,?? 2
,?? 3
,?? 4
)? R4 |???? =0}
???? ?? ???????? ,   ?? =(0,0,0,0)??? so ?? is non-empty. 
?????? ,   ?? =(?? 1
,?? 2
,?? 3
,?? 4
);?? =(?? 1
,?? 2
,?? 3
,?? 4
)??? and ?? ,?? ??? 
?? (???? +???? )=?? (???? )+?? (???? ) 
? ???? +???? ??? 
??? is a vector subspace. 
Clearly, ?? is the null space of the matrix 
?? =[
1 1 2 1
2 3 -1 1
] 
Reducing it to row reduced echelon form 
[
1 1 2 1
2 3 -1 1
] ?
?? ?? ??? 2
-2?? 1
[
1 1 2 1
0 1 -5 -1
]
dim (Row  space (?? )) = Number of non-zero rows 
 =2
 
By Rank-Nullity Theorem : 
dim ( Null space )
 (Nullity) 
+dim (( Pow space )
 (Rank) 
=?? =4
dim ( nuil space )=2
cim (?? )=2
 
For Finding Basis : 
dim  (null space) =2
?  No. of free variables =4-2=2
 
So, we fix 2 variables. 
Taking ?? 3
=1;?? 4
=0 first. 
Page 3


Edurev123 
 
1. Vector Space over R and C 
 
Q1.1: Prove that the set ?? of the vectors (?? ?? ,?? ?? ,?? ?? ,?? ?? ) in ?? ?? which satisfy the 
equations ?? ?? +?? ?? +?? ?? ?? +?? ?? : 0 and ?? ?? ?? +?? ?? ?? -?? ?? +?? ?? =?? is a subspace of ?? ?? . 
What is the dimension of this subspace. Find one of its bases. 
(2009 : 12 Marks)  
Solution: 
ILD for vertical.reaction at ?? ; 
?? ={(?? 1
,?? 2
,?? 3
,?? 4
)??? 4
|?? 1
+?? 2
+2?? 3
+?? 4
=0,2?? 1
+3?? 2
-?? 3
+?? 4
=0}
 
Then clearly (0,0,0,0)??? and so ?? is non-empty. 
Again let ?? =(?? 1
,?? 2
,?? 3
,?? 4
)??? and ?? =(?? 1
,?? 2
,?? 3
,?? 4
)??? , Also let ?? ,?? ??? . 
Also, ?? (?? 1
'
+?? 2
+2?? 3
+?? 4
)+?? (?? 1
+?? 2
+2?? 3
+?? 4
)=0 
???? +???? =(?? ?? 1
+?? ?? 1
,?? ?? 2
+?? ?? 2
,?? ?? 3
+?? ?? 3
,?? ?? 4
+?? ?? 4
)??? 4
 
????? +???? satisfies ?? 1
+?? 2
+2?? 3
+?? 4
=0 
Similarly ???? +???? satisfies 2nd equation as well. 
? ???? +???? ??? 
Dimension of the Subspace : 
Any element of ?? is a solution to equation 
[
1 1 2 1
2 3 -1 1
][
?? 1
?? 2
?? 3
?? 4
]=[
0
0
0
0
] 
So, its dimension is same as rowspace of coefficient matrix, i.e., its rank. 
Reducing it to row reduced echelon form 
[
1 1 2 1
2 3 -1 1
] ?
?? 2
??? 2
-2?? 1
[
1 1 2 1
0 1 -5 -1
] 
As it has two non-zero rows in row reduced form. 
dim (?? )= Rank of matrix =2 
Writing the equation as matrix 
[
1 1 2 1
2 3 -1 1
][
?? 1
?? 2
?? 3
?? 4
]=[
0
0
0
0
] or ???? =0 
???? ,   ?? ={?? =(?? 1
,?? 2
,?? 3
,?? 4
)? R4 |???? =0}
???? ?? ???????? ,   ?? =(0,0,0,0)??? so ?? is non-empty. 
?????? ,   ?? =(?? 1
,?? 2
,?? 3
,?? 4
);?? =(?? 1
,?? 2
,?? 3
,?? 4
)??? and ?? ,?? ??? 
?? (???? +???? )=?? (???? )+?? (???? ) 
? ???? +???? ??? 
??? is a vector subspace. 
Clearly, ?? is the null space of the matrix 
?? =[
1 1 2 1
2 3 -1 1
] 
Reducing it to row reduced echelon form 
[
1 1 2 1
2 3 -1 1
] ?
?? ?? ??? 2
-2?? 1
[
1 1 2 1
0 1 -5 -1
]
dim (Row  space (?? )) = Number of non-zero rows 
 =2
 
By Rank-Nullity Theorem : 
dim ( Null space )
 (Nullity) 
+dim (( Pow space )
 (Rank) 
=?? =4
dim ( nuil space )=2
cim (?? )=2
 
For Finding Basis : 
dim  (null space) =2
?  No. of free variables =4-2=2
 
So, we fix 2 variables. 
Taking ?? 3
=1;?? 4
=0 first. 
?? 1
+?? 2
=-2
2?? 1
+3?? 2
=1
}
?? 1
=-7
?? 2
=5
 ?? =(-7,5,1,0) 
Taking ?? 3
=0,?? 4
=1 
?? 1
+?? 2
=-1
2?? 1
+3?? 2
=-1
}
?? 1
=-2
?? 2
=1
 ?? =(-2,1,0,1) 
?(-7,5,1,0) and (-2,1,0,1) are two elements of ?? . And since they are linearly 
independent (because of choice of 3rd and 4th element) they form a basis. 
Q 1.2 Prove that set ?? of all ?? ×?? real symmetric matrices form a linear subspace 
of the space of all ?? ×?? real matrices. What is the dimension of this subspace? 
Find at least one of the bases for ?? . 
(2009 : 20 Marks) 
Solution: 
Approach: Use definition of subspaces for first part. For the 2nd impose conditions due 
to symmetricity on the matrix. 
Let ?? be cubset of all 3×3 symmetric matrix. 
Then ?? 3
??? so ?? is not empty. Again, let ?? ,?? ??? . 
? 
?? =?? ?? ?? =?? ?? (definition of symmetric) 
 
and ?? ,?? ??? . 
Then 
(???? +???? )
?
 =(???? )
?
+(???? )
?
 =?? ?? ?? +?? ?? '
=???? +????
 
????? +???? ??? . 
So, ?? is a vector subspace of the space of all 3×3 real matrices over ?? . 
Dimension : Let 
?? =[
?? ?? ?? ?? ?? ?? ?? h ?? ]??? 
? 
Page 4


Edurev123 
 
1. Vector Space over R and C 
 
Q1.1: Prove that the set ?? of the vectors (?? ?? ,?? ?? ,?? ?? ,?? ?? ) in ?? ?? which satisfy the 
equations ?? ?? +?? ?? +?? ?? ?? +?? ?? : 0 and ?? ?? ?? +?? ?? ?? -?? ?? +?? ?? =?? is a subspace of ?? ?? . 
What is the dimension of this subspace. Find one of its bases. 
(2009 : 12 Marks)  
Solution: 
ILD for vertical.reaction at ?? ; 
?? ={(?? 1
,?? 2
,?? 3
,?? 4
)??? 4
|?? 1
+?? 2
+2?? 3
+?? 4
=0,2?? 1
+3?? 2
-?? 3
+?? 4
=0}
 
Then clearly (0,0,0,0)??? and so ?? is non-empty. 
Again let ?? =(?? 1
,?? 2
,?? 3
,?? 4
)??? and ?? =(?? 1
,?? 2
,?? 3
,?? 4
)??? , Also let ?? ,?? ??? . 
Also, ?? (?? 1
'
+?? 2
+2?? 3
+?? 4
)+?? (?? 1
+?? 2
+2?? 3
+?? 4
)=0 
???? +???? =(?? ?? 1
+?? ?? 1
,?? ?? 2
+?? ?? 2
,?? ?? 3
+?? ?? 3
,?? ?? 4
+?? ?? 4
)??? 4
 
????? +???? satisfies ?? 1
+?? 2
+2?? 3
+?? 4
=0 
Similarly ???? +???? satisfies 2nd equation as well. 
? ???? +???? ??? 
Dimension of the Subspace : 
Any element of ?? is a solution to equation 
[
1 1 2 1
2 3 -1 1
][
?? 1
?? 2
?? 3
?? 4
]=[
0
0
0
0
] 
So, its dimension is same as rowspace of coefficient matrix, i.e., its rank. 
Reducing it to row reduced echelon form 
[
1 1 2 1
2 3 -1 1
] ?
?? 2
??? 2
-2?? 1
[
1 1 2 1
0 1 -5 -1
] 
As it has two non-zero rows in row reduced form. 
dim (?? )= Rank of matrix =2 
Writing the equation as matrix 
[
1 1 2 1
2 3 -1 1
][
?? 1
?? 2
?? 3
?? 4
]=[
0
0
0
0
] or ???? =0 
???? ,   ?? ={?? =(?? 1
,?? 2
,?? 3
,?? 4
)? R4 |???? =0}
???? ?? ???????? ,   ?? =(0,0,0,0)??? so ?? is non-empty. 
?????? ,   ?? =(?? 1
,?? 2
,?? 3
,?? 4
);?? =(?? 1
,?? 2
,?? 3
,?? 4
)??? and ?? ,?? ??? 
?? (???? +???? )=?? (???? )+?? (???? ) 
? ???? +???? ??? 
??? is a vector subspace. 
Clearly, ?? is the null space of the matrix 
?? =[
1 1 2 1
2 3 -1 1
] 
Reducing it to row reduced echelon form 
[
1 1 2 1
2 3 -1 1
] ?
?? ?? ??? 2
-2?? 1
[
1 1 2 1
0 1 -5 -1
]
dim (Row  space (?? )) = Number of non-zero rows 
 =2
 
By Rank-Nullity Theorem : 
dim ( Null space )
 (Nullity) 
+dim (( Pow space )
 (Rank) 
=?? =4
dim ( nuil space )=2
cim (?? )=2
 
For Finding Basis : 
dim  (null space) =2
?  No. of free variables =4-2=2
 
So, we fix 2 variables. 
Taking ?? 3
=1;?? 4
=0 first. 
?? 1
+?? 2
=-2
2?? 1
+3?? 2
=1
}
?? 1
=-7
?? 2
=5
 ?? =(-7,5,1,0) 
Taking ?? 3
=0,?? 4
=1 
?? 1
+?? 2
=-1
2?? 1
+3?? 2
=-1
}
?? 1
=-2
?? 2
=1
 ?? =(-2,1,0,1) 
?(-7,5,1,0) and (-2,1,0,1) are two elements of ?? . And since they are linearly 
independent (because of choice of 3rd and 4th element) they form a basis. 
Q 1.2 Prove that set ?? of all ?? ×?? real symmetric matrices form a linear subspace 
of the space of all ?? ×?? real matrices. What is the dimension of this subspace? 
Find at least one of the bases for ?? . 
(2009 : 20 Marks) 
Solution: 
Approach: Use definition of subspaces for first part. For the 2nd impose conditions due 
to symmetricity on the matrix. 
Let ?? be cubset of all 3×3 symmetric matrix. 
Then ?? 3
??? so ?? is not empty. Again, let ?? ,?? ??? . 
? 
?? =?? ?? ?? =?? ?? (definition of symmetric) 
 
and ?? ,?? ??? . 
Then 
(???? +???? )
?
 =(???? )
?
+(???? )
?
 =?? ?? ?? +?? ?? '
=???? +????
 
????? +???? ??? . 
So, ?? is a vector subspace of the space of all 3×3 real matrices over ?? . 
Dimension : Let 
?? =[
?? ?? ?? ?? ?? ?? ?? h ?? ]??? 
? 
?? ?? =?? ?[
?? ?? ?? ?? ?? ?? ?? h ?? ]=[
?? ?? ?? ?? ?? h
?? ?? ?? ] 
?? =?? ,?? =?? ,h=?? 
?? =[
?? ?? ?? ?? ?? ?? ?? ?? ?? ] 
Thus, any general element has 6 variables (instead of 9 for a 3×3 real matrix). So, 
dimension of ?? is 6. 
Basis: Putting each of the variables as 1 and rest 0 gives us a basis, i.e., 
?? ={[
1 0 0
0 0 0
0 0 0
],[
0 0 0
0 1 0
0 0 0
][
0 0 0
0 0 0
0 0 1
],[
0 1 0
1 0 0
0 0 0
],[
0 0 1
0 0 0
1 0 0
],[
0 0 0
0 0 1
0 1 0
]} 
Q 1.3 In the n-space ?? ?? , determine whether or not the set {?? ?? -?? ?? ,?? ?? -
?? ?? ,….,?? ?? -?? -?? ?? ,?? ?? -?? ?? } is linearly independent. 
(2010 : 10 Marks) 
Solution: 
Given the set is {?? 1
-?? 2
,?? 2
-?? 3
,….,?? ?? -1
-?? ?? ,?? ?? -?? 1
}. 
Let ?? 1
(?? 1
-?? 2
)+?? 2
(?? 2
-?? 3
)+?..+?? ?? -1
(?? ?? -1
-?? ?? )+?? ?? (?? ?? -?? 1
)=0 
??? 1
(?? 1
-?? ?? )+?? 2
(?? 2
-?? 1
)+?? 3
(?? 3
-?? 2
)+?..+?? ?? (?? ?? -?? ?? -1
)=0 
As ?? 1
,?? 2
,….,?? ?? from basis of ?? ?? , so they are linearly independent. 
? 
?? 1
-?? ?? =0 (1)
?? 2
-?? 1
=0 (2)
?? 3
-?? 2
=0 (3)
?? ?? -?? ?? -1
 =0 (?? )
 
? from eqn. (1) to ( ?? ) it can be deducted that 
?? 1
=?? 2
=?? 3
=?..?? ?? -1
=?? ?? 
and not need to be zero. 
? The given set is linearly dependent. 
Q 1.4 Show that the subspaces of R
?? spanned by two sets of vectors 
{(?? ,?? ,-?? ),(?? ,?? ,?? ) and {(?? ,?? ,-?? ),(?? ,?? ,?? )} are identical. Also find the dimension of 
this subspace. 
Page 5


Edurev123 
 
1. Vector Space over R and C 
 
Q1.1: Prove that the set ?? of the vectors (?? ?? ,?? ?? ,?? ?? ,?? ?? ) in ?? ?? which satisfy the 
equations ?? ?? +?? ?? +?? ?? ?? +?? ?? : 0 and ?? ?? ?? +?? ?? ?? -?? ?? +?? ?? =?? is a subspace of ?? ?? . 
What is the dimension of this subspace. Find one of its bases. 
(2009 : 12 Marks)  
Solution: 
ILD for vertical.reaction at ?? ; 
?? ={(?? 1
,?? 2
,?? 3
,?? 4
)??? 4
|?? 1
+?? 2
+2?? 3
+?? 4
=0,2?? 1
+3?? 2
-?? 3
+?? 4
=0}
 
Then clearly (0,0,0,0)??? and so ?? is non-empty. 
Again let ?? =(?? 1
,?? 2
,?? 3
,?? 4
)??? and ?? =(?? 1
,?? 2
,?? 3
,?? 4
)??? , Also let ?? ,?? ??? . 
Also, ?? (?? 1
'
+?? 2
+2?? 3
+?? 4
)+?? (?? 1
+?? 2
+2?? 3
+?? 4
)=0 
???? +???? =(?? ?? 1
+?? ?? 1
,?? ?? 2
+?? ?? 2
,?? ?? 3
+?? ?? 3
,?? ?? 4
+?? ?? 4
)??? 4
 
????? +???? satisfies ?? 1
+?? 2
+2?? 3
+?? 4
=0 
Similarly ???? +???? satisfies 2nd equation as well. 
? ???? +???? ??? 
Dimension of the Subspace : 
Any element of ?? is a solution to equation 
[
1 1 2 1
2 3 -1 1
][
?? 1
?? 2
?? 3
?? 4
]=[
0
0
0
0
] 
So, its dimension is same as rowspace of coefficient matrix, i.e., its rank. 
Reducing it to row reduced echelon form 
[
1 1 2 1
2 3 -1 1
] ?
?? 2
??? 2
-2?? 1
[
1 1 2 1
0 1 -5 -1
] 
As it has two non-zero rows in row reduced form. 
dim (?? )= Rank of matrix =2 
Writing the equation as matrix 
[
1 1 2 1
2 3 -1 1
][
?? 1
?? 2
?? 3
?? 4
]=[
0
0
0
0
] or ???? =0 
???? ,   ?? ={?? =(?? 1
,?? 2
,?? 3
,?? 4
)? R4 |???? =0}
???? ?? ???????? ,   ?? =(0,0,0,0)??? so ?? is non-empty. 
?????? ,   ?? =(?? 1
,?? 2
,?? 3
,?? 4
);?? =(?? 1
,?? 2
,?? 3
,?? 4
)??? and ?? ,?? ??? 
?? (???? +???? )=?? (???? )+?? (???? ) 
? ???? +???? ??? 
??? is a vector subspace. 
Clearly, ?? is the null space of the matrix 
?? =[
1 1 2 1
2 3 -1 1
] 
Reducing it to row reduced echelon form 
[
1 1 2 1
2 3 -1 1
] ?
?? ?? ??? 2
-2?? 1
[
1 1 2 1
0 1 -5 -1
]
dim (Row  space (?? )) = Number of non-zero rows 
 =2
 
By Rank-Nullity Theorem : 
dim ( Null space )
 (Nullity) 
+dim (( Pow space )
 (Rank) 
=?? =4
dim ( nuil space )=2
cim (?? )=2
 
For Finding Basis : 
dim  (null space) =2
?  No. of free variables =4-2=2
 
So, we fix 2 variables. 
Taking ?? 3
=1;?? 4
=0 first. 
?? 1
+?? 2
=-2
2?? 1
+3?? 2
=1
}
?? 1
=-7
?? 2
=5
 ?? =(-7,5,1,0) 
Taking ?? 3
=0,?? 4
=1 
?? 1
+?? 2
=-1
2?? 1
+3?? 2
=-1
}
?? 1
=-2
?? 2
=1
 ?? =(-2,1,0,1) 
?(-7,5,1,0) and (-2,1,0,1) are two elements of ?? . And since they are linearly 
independent (because of choice of 3rd and 4th element) they form a basis. 
Q 1.2 Prove that set ?? of all ?? ×?? real symmetric matrices form a linear subspace 
of the space of all ?? ×?? real matrices. What is the dimension of this subspace? 
Find at least one of the bases for ?? . 
(2009 : 20 Marks) 
Solution: 
Approach: Use definition of subspaces for first part. For the 2nd impose conditions due 
to symmetricity on the matrix. 
Let ?? be cubset of all 3×3 symmetric matrix. 
Then ?? 3
??? so ?? is not empty. Again, let ?? ,?? ??? . 
? 
?? =?? ?? ?? =?? ?? (definition of symmetric) 
 
and ?? ,?? ??? . 
Then 
(???? +???? )
?
 =(???? )
?
+(???? )
?
 =?? ?? ?? +?? ?? '
=???? +????
 
????? +???? ??? . 
So, ?? is a vector subspace of the space of all 3×3 real matrices over ?? . 
Dimension : Let 
?? =[
?? ?? ?? ?? ?? ?? ?? h ?? ]??? 
? 
?? ?? =?? ?[
?? ?? ?? ?? ?? ?? ?? h ?? ]=[
?? ?? ?? ?? ?? h
?? ?? ?? ] 
?? =?? ,?? =?? ,h=?? 
?? =[
?? ?? ?? ?? ?? ?? ?? ?? ?? ] 
Thus, any general element has 6 variables (instead of 9 for a 3×3 real matrix). So, 
dimension of ?? is 6. 
Basis: Putting each of the variables as 1 and rest 0 gives us a basis, i.e., 
?? ={[
1 0 0
0 0 0
0 0 0
],[
0 0 0
0 1 0
0 0 0
][
0 0 0
0 0 0
0 0 1
],[
0 1 0
1 0 0
0 0 0
],[
0 0 1
0 0 0
1 0 0
],[
0 0 0
0 0 1
0 1 0
]} 
Q 1.3 In the n-space ?? ?? , determine whether or not the set {?? ?? -?? ?? ,?? ?? -
?? ?? ,….,?? ?? -?? -?? ?? ,?? ?? -?? ?? } is linearly independent. 
(2010 : 10 Marks) 
Solution: 
Given the set is {?? 1
-?? 2
,?? 2
-?? 3
,….,?? ?? -1
-?? ?? ,?? ?? -?? 1
}. 
Let ?? 1
(?? 1
-?? 2
)+?? 2
(?? 2
-?? 3
)+?..+?? ?? -1
(?? ?? -1
-?? ?? )+?? ?? (?? ?? -?? 1
)=0 
??? 1
(?? 1
-?? ?? )+?? 2
(?? 2
-?? 1
)+?? 3
(?? 3
-?? 2
)+?..+?? ?? (?? ?? -?? ?? -1
)=0 
As ?? 1
,?? 2
,….,?? ?? from basis of ?? ?? , so they are linearly independent. 
? 
?? 1
-?? ?? =0 (1)
?? 2
-?? 1
=0 (2)
?? 3
-?? 2
=0 (3)
?? ?? -?? ?? -1
 =0 (?? )
 
? from eqn. (1) to ( ?? ) it can be deducted that 
?? 1
=?? 2
=?? 3
=?..?? ?? -1
=?? ?? 
and not need to be zero. 
? The given set is linearly dependent. 
Q 1.4 Show that the subspaces of R
?? spanned by two sets of vectors 
{(?? ,?? ,-?? ),(?? ,?? ,?? ) and {(?? ,?? ,-?? ),(?? ,?? ,?? )} are identical. Also find the dimension of 
this subspace. 
(2011 : 10 Marks) 
Solution: 
Let ?? 1
 be the subspace generated by the vectors (1,1,-1),(1,0,1) . 
Consider a matrix ?? , whose rows are the given vectors (1,1,-1),(1,0,1) and reduce it to 
row reduced Echelon form. 
? ?? =[
1 1 -1
1 0 1
] 
~[
1 1 -1
0 -1 2
] by ?? 2
??? 2
-?? 1
(??) 
Again ?? 2
 be the subspace generated by the vectors (1,2,-3),(5,2,1) . 
Consider a matrix ?? , whose rows are the given vectors (1,2,-3) and (5,2,1) and reduce 
it to row reduced Echelon form. 
i.e., 
?? =[
1 2 -3
5 2 1
]
 ~[
1 2 -3
0 -8 16
] by ?? 2
-?? 2
-5?? 1
 ~[
1 2 -3
0 -1 2
]?? 2
?
1
8
?? 2
 
 
From (i) and (ii), we find that the non-zero rows in the row-reduced Echelon forms of 
matrices ?? and ?? are the same. 
? Row space of ?? = Row space of ?? 
? ?? 1
=?? 2
 
Again consider the vectors (1,1,-1) and (1,0,1) . 
 Let ?? 1
(1,1,-1)+?? 2
(1,0,1) =(0,0,0),?? 1
,?? 2
,?? 3
??? ? (?? 1
,?? 1
,-?? 1
)+(?? 2
,0,?? 2
) =(0,0,0)
? (?? 1
+?? 2
,?? 1
,-?? 1
+?? 2
) =(0,0,0)
? ?? 1
+?? 2
=0,?? 1
=0,-?? 1
+?? 2
=0
? ?? 1
=0=?? 2
 
? The vectors (1,1,-1) and (1,0,1) are linearly independent. 
? The vectors form basis of ?? 1
=?? 2
.