UPSC Exam > UPSC Notes > Mathematics Optional Notes for UPSC > Linear Transformations
Linear Transformations | Mathematics Optional Notes for UPSC PDF Download
| Download, print and study this document offline |
Please wait while the PDF view is loading
Page 1
Edurev123
2. Linear Transformations
Q 2.1 Let ?? ={(?? ,?? ,?? ),(?? ,?? ,?? ),(?? ,?? ,?? )} and ?? '
={(?? ,?? ,?? ),(?? ,?? ,?? ),(-?? ,?? ,?? )} be the
two order bases of ?? ?? . Then find the matrix representing the linear transformation
?? :?? ?? ??? ?? which transforms ?? into ?? '
. Use this matrix transformation to find ?? (???)
where ??? =(?? ,?? ,?? ) .
Solution:
Insight: A change of basis matrix from ?? to ?? '
expresses the coordinates with respect to
basis ?? '
in terms of coordinates w.r.t. to ?? .
First we express elements of ?? '
in terms of ?? .
Let (?? ,?? ,?? ) any general vector be expressed as linear combination of elements of ?? .
(?? ,?? ,?? )=(1,1,0)?? +(1,0,1)?? +(0,1,1)??
? [
1 1 0
1 0 1
0 1 1
][
?? ?? ?? ]=[
?? ?? ?? ]
Using the augmented matrix to solve the linear system of equations
[
1 1 0 ? ?? 1 0 1 ? ?? 0 1 1 ? ?? ]?? 2
??? 2
-?? 1
[
1 1 0 ? ?? 0 -1 1 ? ?? -?? 0 1 1 ? ?? ]
?? 2
?-1?? 2
[
1 1 0 ? ?? 0 1 -1 ? ?? -?? 0 1 1 ? ?? ]?? 3
??? 3
-?? 2
[
1 1 0 ? ?? 0 1 -1 ? ?? -?? 0 0 2 ? ?? +?? -?? ]
?? =
?? +?? -?? 2
;?? =
?? -?? +?? 2
;?? =
?? +?? -?? 2
We use this to express elements of ?? '
as coordinates of ?? .
(2,1,1) =1?? 1
+1?? 2
+0?? 3
(1,2,1) =1?? 1
+0?? 2
+1?? 3
(-1,1,1) =
-1
2
?? 1
+
-1
2
?? 2
+
3
2
?? 3
So change of basis matrix is
[
1 1 -
1
2
1 0 -
1
2
0 1
3
2
]
(2,1,1)=(1,1,0)(
2+1-1
2
)+(1,0,1)(
2-1+1
2
)
Page 2
Edurev123
2. Linear Transformations
Q 2.1 Let ?? ={(?? ,?? ,?? ),(?? ,?? ,?? ),(?? ,?? ,?? )} and ?? '
={(?? ,?? ,?? ),(?? ,?? ,?? ),(-?? ,?? ,?? )} be the
two order bases of ?? ?? . Then find the matrix representing the linear transformation
?? :?? ?? ??? ?? which transforms ?? into ?? '
. Use this matrix transformation to find ?? (???)
where ??? =(?? ,?? ,?? ) .
Solution:
Insight: A change of basis matrix from ?? to ?? '
expresses the coordinates with respect to
basis ?? '
in terms of coordinates w.r.t. to ?? .
First we express elements of ?? '
in terms of ?? .
Let (?? ,?? ,?? ) any general vector be expressed as linear combination of elements of ?? .
(?? ,?? ,?? )=(1,1,0)?? +(1,0,1)?? +(0,1,1)??
? [
1 1 0
1 0 1
0 1 1
][
?? ?? ?? ]=[
?? ?? ?? ]
Using the augmented matrix to solve the linear system of equations
[
1 1 0 ? ?? 1 0 1 ? ?? 0 1 1 ? ?? ]?? 2
??? 2
-?? 1
[
1 1 0 ? ?? 0 -1 1 ? ?? -?? 0 1 1 ? ?? ]
?? 2
?-1?? 2
[
1 1 0 ? ?? 0 1 -1 ? ?? -?? 0 1 1 ? ?? ]?? 3
??? 3
-?? 2
[
1 1 0 ? ?? 0 1 -1 ? ?? -?? 0 0 2 ? ?? +?? -?? ]
?? =
?? +?? -?? 2
;?? =
?? -?? +?? 2
;?? =
?? +?? -?? 2
We use this to express elements of ?? '
as coordinates of ?? .
(2,1,1) =1?? 1
+1?? 2
+0?? 3
(1,2,1) =1?? 1
+0?? 2
+1?? 3
(-1,1,1) =
-1
2
?? 1
+
-1
2
?? 2
+
3
2
?? 3
So change of basis matrix is
[
1 1 -
1
2
1 0 -
1
2
0 1
3
2
]
(2,1,1)=(1,1,0)(
2+1-1
2
)+(1,0,1)(
2-1+1
2
)
?? ,?? ,?? =1,1,0
Similarly, (?? 2
,?? 2
,?? 2
) and (?? 3
,?? 3
,?? 3
) matrix is
=[
?? 1
?? 2
?? 3
?? 1
?? 2
?? 3
?? 1
?? 2
?? 3
]
?? (???) =?? ·??˜
'
=
[
1 1 -
1
2
1 0 -
1
2
0 1
3
2
]
[
2
3
1
]=[
9/2
3/2
9/2
]
Q 2.2 Let ?? :?? ?? ??? ?? be a linear transformation defined by
?? ((?? ?? ,?? ?? ,?? ?? ,?? ?? ))=(?? ?? +?? ?? -?? ?? -?? ?? ,?? ?? -?? ?? ,?? ?? -?? ?? )
Then find the rank and nullify of ?? . Also determine null space and range space of
?? .
(2009 : 20 Marks)
Solution:
Approach: We could use the standard basis vectors of ?? 4
or use the matrix form of the
linear transformation. Let (?? 1
,?? 2
,?? 3
,?? 4
)=?? be the standard basis of ?? 4
where ?? 1
=
(1,0,0,0) and similar to others.
?? =(?? 1
,?? 2
,?? 3
,?? 4
) spans ?? 4
so {?? (?? 1
),?? (?? 2
),?? (?? 3
),?? (?? 4
)} will span ???? (?? ) , i.e., image
space of ?? . So a linear independent subspace will be basis of range space and its
dimension rank.
?? (?? 1
)=(-1,0,-1)
?? (?? 2
)=(-1,-1,0)
?? ?? 3
)=(1,1,0)
?? ?? 4
)=(1,0,1)
To reduce it to a linearly independent subset we reduce the matrix with the vectors as
row to row reduced echelon form
[
-1 0 -1
-1 -1 0
1 1 0
1 0 1
] ?
?? 1
??? 3
?? 2
??? 4
[
1 1 0
1 0 1
-1 0 -1
-1 -1 0
]
=
?? 1
?? 2
??? 2
-?? 4
?? 4
??? 4
+?? 1
[
1 1 0
0 -1 1
0 1 -1
0 0 0
] ?
?? 2
?-1?? 2
?? 3
??? 3
-?? 2
[
1 1 0
0 1 -1
0 0 0
0 0 0
]
Page 3
Edurev123
2. Linear Transformations
Q 2.1 Let ?? ={(?? ,?? ,?? ),(?? ,?? ,?? ),(?? ,?? ,?? )} and ?? '
={(?? ,?? ,?? ),(?? ,?? ,?? ),(-?? ,?? ,?? )} be the
two order bases of ?? ?? . Then find the matrix representing the linear transformation
?? :?? ?? ??? ?? which transforms ?? into ?? '
. Use this matrix transformation to find ?? (???)
where ??? =(?? ,?? ,?? ) .
Solution:
Insight: A change of basis matrix from ?? to ?? '
expresses the coordinates with respect to
basis ?? '
in terms of coordinates w.r.t. to ?? .
First we express elements of ?? '
in terms of ?? .
Let (?? ,?? ,?? ) any general vector be expressed as linear combination of elements of ?? .
(?? ,?? ,?? )=(1,1,0)?? +(1,0,1)?? +(0,1,1)??
? [
1 1 0
1 0 1
0 1 1
][
?? ?? ?? ]=[
?? ?? ?? ]
Using the augmented matrix to solve the linear system of equations
[
1 1 0 ? ?? 1 0 1 ? ?? 0 1 1 ? ?? ]?? 2
??? 2
-?? 1
[
1 1 0 ? ?? 0 -1 1 ? ?? -?? 0 1 1 ? ?? ]
?? 2
?-1?? 2
[
1 1 0 ? ?? 0 1 -1 ? ?? -?? 0 1 1 ? ?? ]?? 3
??? 3
-?? 2
[
1 1 0 ? ?? 0 1 -1 ? ?? -?? 0 0 2 ? ?? +?? -?? ]
?? =
?? +?? -?? 2
;?? =
?? -?? +?? 2
;?? =
?? +?? -?? 2
We use this to express elements of ?? '
as coordinates of ?? .
(2,1,1) =1?? 1
+1?? 2
+0?? 3
(1,2,1) =1?? 1
+0?? 2
+1?? 3
(-1,1,1) =
-1
2
?? 1
+
-1
2
?? 2
+
3
2
?? 3
So change of basis matrix is
[
1 1 -
1
2
1 0 -
1
2
0 1
3
2
]
(2,1,1)=(1,1,0)(
2+1-1
2
)+(1,0,1)(
2-1+1
2
)
?? ,?? ,?? =1,1,0
Similarly, (?? 2
,?? 2
,?? 2
) and (?? 3
,?? 3
,?? 3
) matrix is
=[
?? 1
?? 2
?? 3
?? 1
?? 2
?? 3
?? 1
?? 2
?? 3
]
?? (???) =?? ·??˜
'
=
[
1 1 -
1
2
1 0 -
1
2
0 1
3
2
]
[
2
3
1
]=[
9/2
3/2
9/2
]
Q 2.2 Let ?? :?? ?? ??? ?? be a linear transformation defined by
?? ((?? ?? ,?? ?? ,?? ?? ,?? ?? ))=(?? ?? +?? ?? -?? ?? -?? ?? ,?? ?? -?? ?? ,?? ?? -?? ?? )
Then find the rank and nullify of ?? . Also determine null space and range space of
?? .
(2009 : 20 Marks)
Solution:
Approach: We could use the standard basis vectors of ?? 4
or use the matrix form of the
linear transformation. Let (?? 1
,?? 2
,?? 3
,?? 4
)=?? be the standard basis of ?? 4
where ?? 1
=
(1,0,0,0) and similar to others.
?? =(?? 1
,?? 2
,?? 3
,?? 4
) spans ?? 4
so {?? (?? 1
),?? (?? 2
),?? (?? 3
),?? (?? 4
)} will span ???? (?? ) , i.e., image
space of ?? . So a linear independent subspace will be basis of range space and its
dimension rank.
?? (?? 1
)=(-1,0,-1)
?? (?? 2
)=(-1,-1,0)
?? ?? 3
)=(1,1,0)
?? ?? 4
)=(1,0,1)
To reduce it to a linearly independent subset we reduce the matrix with the vectors as
row to row reduced echelon form
[
-1 0 -1
-1 -1 0
1 1 0
1 0 1
] ?
?? 1
??? 3
?? 2
??? 4
[
1 1 0
1 0 1
-1 0 -1
-1 -1 0
]
=
?? 1
?? 2
??? 2
-?? 4
?? 4
??? 4
+?? 1
[
1 1 0
0 -1 1
0 1 -1
0 0 0
] ?
?? 2
?-1?? 2
?? 3
??? 3
-?? 2
[
1 1 0
0 1 -1
0 0 0
0 0 0
]
??? '
={(1,1,0),(0,1,-1)} is a linearly independent subset which spans Im (?? ) .
? Rank (?? ) =2
and Range Space = Linear Span {(1,1,0),(0,1,-1)}
Nullify and Null Space ?? ??? 4
is in null space if ?? (?? )=0
? [
?? 3
+?? 4
-?? 1
-?? 2
?? 3
-?? 2
?? 4
-?? 1
]=[
0
0
0
]
Thus, there are two free variables ?? 1
and ?? 2
and so nullify = dim (null space) =2.
Giving arbitrary value to free variables
?? 1
=1,?? 2
=0 (?? 1
,?? 2
,?? 3
,?? 4
)
=(1,0,0,1)
?? 2
=0,?? 2
=1 (?? 1
,?? 2
,?? 3
,?? 4
)
=(0,1,1,0)
?{(1,0,0,1),(0,1,1,0} is a basis for null space of ?? .
Q 2.3. What is the null space of the differential transformation
?? ????
:?? ?? ??? ?? where
?? ?? is the space of all polynomials of degree =?? are the real numbers? What is the
null space of second derivative as a transformation of ?? ?? ? What is the null space
of ?? th derivative?
(2010 : 12 Marks)
Solution:
Let the polynomial
?????? , ?? ?? (?? )=?? 0
+?? 1
?? +?? 2
?? 2
+?..+?? ?? ?? ?? ????
?? ????
?? ?? (?? )=?? 1
+2?? 2
?? +3?? 3
?? 2
+?..+?? ?? ?? ?? ?? -1
???? ?? ?? (?? )=0??? 1
+2?? 2
?? +?…+?? ?? ?? ?? ?? -1
=0
?? h???? ?? 1
=?? 2
=?…=?? ?? =0
? Null space of the transformation
?? ????
?? ?? (?? )??? ?? (?? ) is
i.e.,
Page 4
Edurev123
2. Linear Transformations
Q 2.1 Let ?? ={(?? ,?? ,?? ),(?? ,?? ,?? ),(?? ,?? ,?? )} and ?? '
={(?? ,?? ,?? ),(?? ,?? ,?? ),(-?? ,?? ,?? )} be the
two order bases of ?? ?? . Then find the matrix representing the linear transformation
?? :?? ?? ??? ?? which transforms ?? into ?? '
. Use this matrix transformation to find ?? (???)
where ??? =(?? ,?? ,?? ) .
Solution:
Insight: A change of basis matrix from ?? to ?? '
expresses the coordinates with respect to
basis ?? '
in terms of coordinates w.r.t. to ?? .
First we express elements of ?? '
in terms of ?? .
Let (?? ,?? ,?? ) any general vector be expressed as linear combination of elements of ?? .
(?? ,?? ,?? )=(1,1,0)?? +(1,0,1)?? +(0,1,1)??
? [
1 1 0
1 0 1
0 1 1
][
?? ?? ?? ]=[
?? ?? ?? ]
Using the augmented matrix to solve the linear system of equations
[
1 1 0 ? ?? 1 0 1 ? ?? 0 1 1 ? ?? ]?? 2
??? 2
-?? 1
[
1 1 0 ? ?? 0 -1 1 ? ?? -?? 0 1 1 ? ?? ]
?? 2
?-1?? 2
[
1 1 0 ? ?? 0 1 -1 ? ?? -?? 0 1 1 ? ?? ]?? 3
??? 3
-?? 2
[
1 1 0 ? ?? 0 1 -1 ? ?? -?? 0 0 2 ? ?? +?? -?? ]
?? =
?? +?? -?? 2
;?? =
?? -?? +?? 2
;?? =
?? +?? -?? 2
We use this to express elements of ?? '
as coordinates of ?? .
(2,1,1) =1?? 1
+1?? 2
+0?? 3
(1,2,1) =1?? 1
+0?? 2
+1?? 3
(-1,1,1) =
-1
2
?? 1
+
-1
2
?? 2
+
3
2
?? 3
So change of basis matrix is
[
1 1 -
1
2
1 0 -
1
2
0 1
3
2
]
(2,1,1)=(1,1,0)(
2+1-1
2
)+(1,0,1)(
2-1+1
2
)
?? ,?? ,?? =1,1,0
Similarly, (?? 2
,?? 2
,?? 2
) and (?? 3
,?? 3
,?? 3
) matrix is
=[
?? 1
?? 2
?? 3
?? 1
?? 2
?? 3
?? 1
?? 2
?? 3
]
?? (???) =?? ·??˜
'
=
[
1 1 -
1
2
1 0 -
1
2
0 1
3
2
]
[
2
3
1
]=[
9/2
3/2
9/2
]
Q 2.2 Let ?? :?? ?? ??? ?? be a linear transformation defined by
?? ((?? ?? ,?? ?? ,?? ?? ,?? ?? ))=(?? ?? +?? ?? -?? ?? -?? ?? ,?? ?? -?? ?? ,?? ?? -?? ?? )
Then find the rank and nullify of ?? . Also determine null space and range space of
?? .
(2009 : 20 Marks)
Solution:
Approach: We could use the standard basis vectors of ?? 4
or use the matrix form of the
linear transformation. Let (?? 1
,?? 2
,?? 3
,?? 4
)=?? be the standard basis of ?? 4
where ?? 1
=
(1,0,0,0) and similar to others.
?? =(?? 1
,?? 2
,?? 3
,?? 4
) spans ?? 4
so {?? (?? 1
),?? (?? 2
),?? (?? 3
),?? (?? 4
)} will span ???? (?? ) , i.e., image
space of ?? . So a linear independent subspace will be basis of range space and its
dimension rank.
?? (?? 1
)=(-1,0,-1)
?? (?? 2
)=(-1,-1,0)
?? ?? 3
)=(1,1,0)
?? ?? 4
)=(1,0,1)
To reduce it to a linearly independent subset we reduce the matrix with the vectors as
row to row reduced echelon form
[
-1 0 -1
-1 -1 0
1 1 0
1 0 1
] ?
?? 1
??? 3
?? 2
??? 4
[
1 1 0
1 0 1
-1 0 -1
-1 -1 0
]
=
?? 1
?? 2
??? 2
-?? 4
?? 4
??? 4
+?? 1
[
1 1 0
0 -1 1
0 1 -1
0 0 0
] ?
?? 2
?-1?? 2
?? 3
??? 3
-?? 2
[
1 1 0
0 1 -1
0 0 0
0 0 0
]
??? '
={(1,1,0),(0,1,-1)} is a linearly independent subset which spans Im (?? ) .
? Rank (?? ) =2
and Range Space = Linear Span {(1,1,0),(0,1,-1)}
Nullify and Null Space ?? ??? 4
is in null space if ?? (?? )=0
? [
?? 3
+?? 4
-?? 1
-?? 2
?? 3
-?? 2
?? 4
-?? 1
]=[
0
0
0
]
Thus, there are two free variables ?? 1
and ?? 2
and so nullify = dim (null space) =2.
Giving arbitrary value to free variables
?? 1
=1,?? 2
=0 (?? 1
,?? 2
,?? 3
,?? 4
)
=(1,0,0,1)
?? 2
=0,?? 2
=1 (?? 1
,?? 2
,?? 3
,?? 4
)
=(0,1,1,0)
?{(1,0,0,1),(0,1,1,0} is a basis for null space of ?? .
Q 2.3. What is the null space of the differential transformation
?? ????
:?? ?? ??? ?? where
?? ?? is the space of all polynomials of degree =?? are the real numbers? What is the
null space of second derivative as a transformation of ?? ?? ? What is the null space
of ?? th derivative?
(2010 : 12 Marks)
Solution:
Let the polynomial
?????? , ?? ?? (?? )=?? 0
+?? 1
?? +?? 2
?? 2
+?..+?? ?? ?? ?? ????
?? ????
?? ?? (?? )=?? 1
+2?? 2
?? +3?? 3
?? 2
+?..+?? ?? ?? ?? ?? -1
???? ?? ?? (?? )=0??? 1
+2?? 2
?? +?…+?? ?? ?? ?? ?? -1
=0
?? h???? ?? 1
=?? 2
=?…=?? ?? =0
? Null space of the transformation
?? ????
?? ?? (?? )??? ?? (?? ) is
i.e.,
?? 1
=?? 2
=?..=?? ?? =0 and ?? 0
??? ??.?? .(?? 0
,?? 1
,?? 2
,…,?? ?? ) =?? 0
(1,0,0,……,0)
?????? ,
?? 2
?? ?? ?? ?? (?? ) =2?? 2
+6?? 3
+?..+?? (?? -1)?? ????
?? ?? -2
=0
?? h???? ,?? 2
=?? 3
=?..=?? ?? =0
? Null space of
?? 2
????
?? ?? (?? ) is when ?? 0
,?? 1
??? and ?? 2
=?? 3
=?.?? ?? =0
So, dimension of nuil space of ?? th
derivative is ?? .
So,
Null space =?? 0
(1,0,0,……,0)+?? 1
(0,1,0,….,0)
?? ?? ????
?? ?? =?? !?? ?? +(?? +1)!?? ?? +1
+?..+?? (?? -1)(?? -2)….(?? -?? +1)?? ?? ?? ?? -?? =0
when
?? ?? =???? +1=?=???? -1=???? =0
and ?? 0
??? ,?? 1
??? 1
…..?? ?? -?? ???
So, null space of ?? th
derivative is ?? 0
(1,0,….,0)+?? 1
(0,1,….,0)+?..+?? ?? (0,0,…..1,0,0,0)
2.4 Let ?? =(
?? ?? ?? ?? ?? ?? ) . Find the unique linear transformation ?? :?? ?? ??? ?? so that ??
is the matrix of ?? with respect to the basis.
?? ={?? ?? =(?? ,?? ,?? ),?? ?? =(?? ,?? ,?? ),?? ?? =(?? ,?? ,?? )}
of ?? ?? and ?? '
=[?? ?? =(?? ,?? ),?? ?? =(?? ,?? )} of ?? ?? . Also find ?? (?? ,?? ,?? ) .
(2010: 20 Marks)
Solution:
Given ?? is the basis of ?? 3
and ?? '
be basis of ?? ?? for transformation ?? .
Now,
?? (?? 1
) =?? (1,0,0)=4?? 1
+0?? 2
=4(1,0)+0.(1,1)=(4,0)
?? (?? 2
) =?? (1,1,0)=2w1+1?? 2
=2(1,0)+1(1,1)=(3,1)
?? (?? 3
) =?? (1,1,1)=1.?? 1
+3.?? 2
=1.(1,0)+3(1,1)=(4,3)
Page 5
Edurev123
2. Linear Transformations
Q 2.1 Let ?? ={(?? ,?? ,?? ),(?? ,?? ,?? ),(?? ,?? ,?? )} and ?? '
={(?? ,?? ,?? ),(?? ,?? ,?? ),(-?? ,?? ,?? )} be the
two order bases of ?? ?? . Then find the matrix representing the linear transformation
?? :?? ?? ??? ?? which transforms ?? into ?? '
. Use this matrix transformation to find ?? (???)
where ??? =(?? ,?? ,?? ) .
Solution:
Insight: A change of basis matrix from ?? to ?? '
expresses the coordinates with respect to
basis ?? '
in terms of coordinates w.r.t. to ?? .
First we express elements of ?? '
in terms of ?? .
Let (?? ,?? ,?? ) any general vector be expressed as linear combination of elements of ?? .
(?? ,?? ,?? )=(1,1,0)?? +(1,0,1)?? +(0,1,1)??
? [
1 1 0
1 0 1
0 1 1
][
?? ?? ?? ]=[
?? ?? ?? ]
Using the augmented matrix to solve the linear system of equations
[
1 1 0 ? ?? 1 0 1 ? ?? 0 1 1 ? ?? ]?? 2
??? 2
-?? 1
[
1 1 0 ? ?? 0 -1 1 ? ?? -?? 0 1 1 ? ?? ]
?? 2
?-1?? 2
[
1 1 0 ? ?? 0 1 -1 ? ?? -?? 0 1 1 ? ?? ]?? 3
??? 3
-?? 2
[
1 1 0 ? ?? 0 1 -1 ? ?? -?? 0 0 2 ? ?? +?? -?? ]
?? =
?? +?? -?? 2
;?? =
?? -?? +?? 2
;?? =
?? +?? -?? 2
We use this to express elements of ?? '
as coordinates of ?? .
(2,1,1) =1?? 1
+1?? 2
+0?? 3
(1,2,1) =1?? 1
+0?? 2
+1?? 3
(-1,1,1) =
-1
2
?? 1
+
-1
2
?? 2
+
3
2
?? 3
So change of basis matrix is
[
1 1 -
1
2
1 0 -
1
2
0 1
3
2
]
(2,1,1)=(1,1,0)(
2+1-1
2
)+(1,0,1)(
2-1+1
2
)
?? ,?? ,?? =1,1,0
Similarly, (?? 2
,?? 2
,?? 2
) and (?? 3
,?? 3
,?? 3
) matrix is
=[
?? 1
?? 2
?? 3
?? 1
?? 2
?? 3
?? 1
?? 2
?? 3
]
?? (???) =?? ·??˜
'
=
[
1 1 -
1
2
1 0 -
1
2
0 1
3
2
]
[
2
3
1
]=[
9/2
3/2
9/2
]
Q 2.2 Let ?? :?? ?? ??? ?? be a linear transformation defined by
?? ((?? ?? ,?? ?? ,?? ?? ,?? ?? ))=(?? ?? +?? ?? -?? ?? -?? ?? ,?? ?? -?? ?? ,?? ?? -?? ?? )
Then find the rank and nullify of ?? . Also determine null space and range space of
?? .
(2009 : 20 Marks)
Solution:
Approach: We could use the standard basis vectors of ?? 4
or use the matrix form of the
linear transformation. Let (?? 1
,?? 2
,?? 3
,?? 4
)=?? be the standard basis of ?? 4
where ?? 1
=
(1,0,0,0) and similar to others.
?? =(?? 1
,?? 2
,?? 3
,?? 4
) spans ?? 4
so {?? (?? 1
),?? (?? 2
),?? (?? 3
),?? (?? 4
)} will span ???? (?? ) , i.e., image
space of ?? . So a linear independent subspace will be basis of range space and its
dimension rank.
?? (?? 1
)=(-1,0,-1)
?? (?? 2
)=(-1,-1,0)
?? ?? 3
)=(1,1,0)
?? ?? 4
)=(1,0,1)
To reduce it to a linearly independent subset we reduce the matrix with the vectors as
row to row reduced echelon form
[
-1 0 -1
-1 -1 0
1 1 0
1 0 1
] ?
?? 1
??? 3
?? 2
??? 4
[
1 1 0
1 0 1
-1 0 -1
-1 -1 0
]
=
?? 1
?? 2
??? 2
-?? 4
?? 4
??? 4
+?? 1
[
1 1 0
0 -1 1
0 1 -1
0 0 0
] ?
?? 2
?-1?? 2
?? 3
??? 3
-?? 2
[
1 1 0
0 1 -1
0 0 0
0 0 0
]
??? '
={(1,1,0),(0,1,-1)} is a linearly independent subset which spans Im (?? ) .
? Rank (?? ) =2
and Range Space = Linear Span {(1,1,0),(0,1,-1)}
Nullify and Null Space ?? ??? 4
is in null space if ?? (?? )=0
? [
?? 3
+?? 4
-?? 1
-?? 2
?? 3
-?? 2
?? 4
-?? 1
]=[
0
0
0
]
Thus, there are two free variables ?? 1
and ?? 2
and so nullify = dim (null space) =2.
Giving arbitrary value to free variables
?? 1
=1,?? 2
=0 (?? 1
,?? 2
,?? 3
,?? 4
)
=(1,0,0,1)
?? 2
=0,?? 2
=1 (?? 1
,?? 2
,?? 3
,?? 4
)
=(0,1,1,0)
?{(1,0,0,1),(0,1,1,0} is a basis for null space of ?? .
Q 2.3. What is the null space of the differential transformation
?? ????
:?? ?? ??? ?? where
?? ?? is the space of all polynomials of degree =?? are the real numbers? What is the
null space of second derivative as a transformation of ?? ?? ? What is the null space
of ?? th derivative?
(2010 : 12 Marks)
Solution:
Let the polynomial
?????? , ?? ?? (?? )=?? 0
+?? 1
?? +?? 2
?? 2
+?..+?? ?? ?? ?? ????
?? ????
?? ?? (?? )=?? 1
+2?? 2
?? +3?? 3
?? 2
+?..+?? ?? ?? ?? ?? -1
???? ?? ?? (?? )=0??? 1
+2?? 2
?? +?…+?? ?? ?? ?? ?? -1
=0
?? h???? ?? 1
=?? 2
=?…=?? ?? =0
? Null space of the transformation
?? ????
?? ?? (?? )??? ?? (?? ) is
i.e.,
?? 1
=?? 2
=?..=?? ?? =0 and ?? 0
??? ??.?? .(?? 0
,?? 1
,?? 2
,…,?? ?? ) =?? 0
(1,0,0,……,0)
?????? ,
?? 2
?? ?? ?? ?? (?? ) =2?? 2
+6?? 3
+?..+?? (?? -1)?? ????
?? ?? -2
=0
?? h???? ,?? 2
=?? 3
=?..=?? ?? =0
? Null space of
?? 2
????
?? ?? (?? ) is when ?? 0
,?? 1
??? and ?? 2
=?? 3
=?.?? ?? =0
So, dimension of nuil space of ?? th
derivative is ?? .
So,
Null space =?? 0
(1,0,0,……,0)+?? 1
(0,1,0,….,0)
?? ?? ????
?? ?? =?? !?? ?? +(?? +1)!?? ?? +1
+?..+?? (?? -1)(?? -2)….(?? -?? +1)?? ?? ?? ?? -?? =0
when
?? ?? =???? +1=?=???? -1=???? =0
and ?? 0
??? ,?? 1
??? 1
…..?? ?? -?? ???
So, null space of ?? th
derivative is ?? 0
(1,0,….,0)+?? 1
(0,1,….,0)+?..+?? ?? (0,0,…..1,0,0,0)
2.4 Let ?? =(
?? ?? ?? ?? ?? ?? ) . Find the unique linear transformation ?? :?? ?? ??? ?? so that ??
is the matrix of ?? with respect to the basis.
?? ={?? ?? =(?? ,?? ,?? ),?? ?? =(?? ,?? ,?? ),?? ?? =(?? ,?? ,?? )}
of ?? ?? and ?? '
=[?? ?? =(?? ,?? ),?? ?? =(?? ,?? )} of ?? ?? . Also find ?? (?? ,?? ,?? ) .
(2010: 20 Marks)
Solution:
Given ?? is the basis of ?? 3
and ?? '
be basis of ?? ?? for transformation ?? .
Now,
?? (?? 1
) =?? (1,0,0)=4?? 1
+0?? 2
=4(1,0)+0.(1,1)=(4,0)
?? (?? 2
) =?? (1,1,0)=2w1+1?? 2
=2(1,0)+1(1,1)=(3,1)
?? (?? 3
) =?? (1,1,1)=1.?? 1
+3.?? 2
=1.(1,0)+3(1,1)=(4,3)
Now, let
(?? ,?? ,?? )=?? (1,0,0)+?? (1,1,0)+?? (1,1,1)=?? ?? 1
+?? ?? 2
+?? ?? 3
(?? ,?? ,?? )=(?? +?? +?? ,?? +?? ,?? )
Comparing LHS & RHS, we get
?? =?? (?? ,?? ,?? ) =(?? -?? )·?? 1
+(?? -?? )·?? 2
+?? ·?? 3
? ?? (?? ,?? ,?? ) =?? ((?? -?? )·?? 1
+(?? -?? )·?? 2
+?? ·?? 3
)
=(?? -?? )?? (?? 1
)+(?? -?? )?? (?? 2
)+?? ·?? (?? 3
)
=(?? -?? )(4,0)+(?? -?? )(3,1)+?? (4,3)
=(4(?? -?? ),0)+(3(?? -?? ),(?? -?? )+(4?? ,3?? )
=(4?? -4?? +3?? -3?? +4?? ,0+?? -?? +3?? )
=(4?? -?? +?? ,?? +2?? )
? The transformation ?? is ?? (?? ,?? ,?? )=(4?? -?? +?? ,?? +2?? )
2.5 Let ?? be a linear transformation from a vector space V over reals into ?? such
that ?? -?? ?? =?? . Show that ?? is invertible.
(2010 : 10 Marks)
Solution:
Given, ?? be the linear transformation, such that
?? -?? 2
=?? ? ?? (?? -?? ) =?? ? ?? (?? -?? ) =?? ? |?? ||(?? -?? )| =|?? |, i.e., |?? |×|?? -?? |=1
|?? | ?0
??? is invertible.
Q 2.6 Find the nullity and a basis of the null space of the linear transformation
?? :?? (?? )??? (?? ) given by the matrix
?? =[
?? ?? -?? -?? ?? ?? ?? ?? ?? ?? ?? ?? ?? ?? -?? ?? ]
(2011 : 10 Marks)
Solution :
Let ?? ={?? 1
,?? 2
,?? 3
,?? 4
} be the usual basis of ?? 4
.
Read More
|
387 videos|203 docs
|
FAQs on Linear Transformations - Mathematics Optional Notes for UPSC
| 1. How do linear transformations work in the context of UPSC exams? |  |
Ans. Linear transformations in the context of UPSC exams refer to the process of transforming one set of coordinates into another set by using a matrix operation. This concept is often tested in the mathematics section of the UPSC exam.
| 2. Can you provide an example of a linear transformation question that may appear in the UPSC exam? | |
Ans. An example of a linear transformation question in the UPSC exam could be to find the image of a given vector under a specific transformation matrix, or to determine whether a given matrix represents a valid linear transformation.
| 3. How important is it to understand linear transformations for UPSC preparation? | |
Ans. Understanding linear transformations is crucial for UPSC preparation, as it forms the basis of various mathematical concepts that are frequently tested in the exam. It is essential to be familiar with the properties and operations of linear transformations to solve complex problems.
| 4. What are some tips for mastering linear transformations for the UPSC exam? | |
Ans. Some tips for mastering linear transformations for the UPSC exam include practicing with a variety of sample questions, understanding the properties of linear transformations thoroughly, and familiarizing yourself with different types of transformation matrices.
| 5. Are there any specific resources or study materials that can help in improving knowledge of linear transformations for the UPSC exam? | |
Ans. Yes, there are several resources available online, such as textbooks, video tutorials, and practice questions, that can help improve your understanding of linear transformations for the UPSC exam. It is recommended to utilize these resources to enhance your preparation.
Related Exams
About this Document
|
Dec 18, 2024 Last updated |
Document Description: Linear Transformations for UPSC 2024 is part of Mathematics Optional Notes for UPSC preparation.
The notes and questions for Linear Transformations have been prepared according to the UPSC exam syllabus. Information about Linear Transformations covers topics
like and Linear Transformations Example, for UPSC 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for Linear Transformations.
Introduction of Linear Transformations in English is available as part of our Mathematics Optional Notes for UPSC
for UPSC & Linear Transformations in Hindi for Mathematics Optional Notes for UPSC course.
Download more important topics related with notes, lectures and mock test series for UPSC
Exam by signing up for free. UPSC: Linear Transformations | Mathematics Optional Notes for UPSC
Description
Full syllabus notes, lecture & questions for Linear Transformations | Mathematics Optional Notes for UPSC - UPSC | Plus excerises question with solution to help you revise complete syllabus for Mathematics Optional Notes for UPSC | Best notes, free PDF download
Information about Linear Transformations
In this doc you can find the meaning of Linear Transformations defined & explained in the simplest way possible. Besides explaining types of
Linear Transformations theory, EduRev gives you an ample number of questions to practice Linear Transformations tests, examples and also practice UPSC
tests
|
Explore Courses for UPSC exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
Related Searches
Linear Transformations | Mathematics Optional Notes for UPSC
,shortcuts and tricks
,Objective type Questions
,Linear Transformations | Mathematics Optional Notes for UPSC
,ppt
,Extra Questions
,past year papers
,Semester Notes
,Viva Questions
,Linear Transformations | Mathematics Optional Notes for UPSC
,Important questions
,Sample Paper
,Exam
,mock tests for examination
,Free
,Summary
,practice quizzes
,video lectures
,MCQs
,Previous Year Questions with Solutions
,study material
;
Additional Information about Linear Transformations for UPSC Preparation
Linear Transformations Free PDF Download
The Linear Transformations is an invaluable resource that delves deep into the core of the UPSC exam.
These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective.
With the help of these notes, you can grasp complex subjects quickly, revise important points easily,
and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner,
allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material,
or toppers' notes, this PDF has got you covered. Download the Linear Transformations now and kickstart your journey towards success in the UPSC exam.
Importance of Linear Transformations
The importance of Linear Transformations cannot be overstated, especially for UPSC aspirants.
This document holds the key to success in the UPSC exam.
It offers a detailed understanding of the concept, providing invaluable insights into the topic.
By knowing the concepts well in advance, students can plan their preparation effectively.
Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.
Linear Transformations Notes
Linear Transformations Notes offer in-depth insights into the specific topic to help you master it with ease.
This comprehensive document covers all aspects related to Linear Transformations.
It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation.
Practice papers and question papers enable you to assess your progress effectively.
Additionally, the paper analysis provides valuable tips for tackling the exam strategically.
Access to Toppers' notes gives you an edge in understanding complex concepts.
Whether you're a beginner or aiming for advanced proficiency, Linear Transformations Notes on EduRev are your ultimate resource for success.
Linear Transformations UPSC Questions
The "Linear Transformations UPSC Questions" guide is a valuable resource for all aspiring students preparing for the
UPSC exam. It focuses on providing a wide range of practice questions to help students gauge
their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation.
The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level.
Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.
Study Linear Transformations on the App
Students of UPSC can study Linear Transformations alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the Linear Transformations,
students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of Linear Transformations is prepared as per the latest UPSC syllabus.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup