Mathematics-II (MATH F112)universe.bits-pilani.ac.in/uploads/L1_Chapter 5.pdfMathematics-II (MATH F112) Linear Algebra Jitender Kumar Department of Mathematics Birla Institute of Technology

Mathematics-II (MATH F112)Linear Algebra

Jitender Kumar

Department of MathematicsBirla Institute of Technology and Science Pilani

Pilani-333031

Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 1 / 122

Chapter: 5 (Linear Transformations)

1 Introduction to Linear Transformations2 The Dimension Theorem3 One-to-One and Onto Linear Transformations4 Isomorphism5 Coordinatization (4.7)6 The Matrix of a Linear Transformation


Section 5.1: Linear Transformations

Let V and W be real vector spaces.



Let V and W be real vector spaces. A mapL : V → W is called a Linear map or Lineartransformation (LT)



Let V and W be real vector spaces. A mapL : V → W is called a Linear map or Lineartransformation (LT) if and only if both of thefollowing are true:




L(u + v) = L(u) + L(v) for all u, v ∈ V




L(u + v) = L(u) + L(v) for all u, v ∈ V

L(cu) = cL(u) for all c ∈ R and all u ∈ V


Example 1: For A ∈ Mmn, consider the mapping

L : Mmn → Mnm given by

L(A) = AT .




L(A) = AT .

Check whether L is a LT.




L(A) = AT .


Solution: Let A, B ∈ Mmn and c ∈ R. Note that




L(A) = AT .



L(A + B) = (A + B)T = AT + BT = L(A) + L(B)




L(A) = AT .



L(A + B) = (A + B)T = AT + BT = L(A) + L(B)

L(cA) = (cA)T = cAT = cL(A).




L(A) = AT .



L(A + B) = (A + B)T = AT + BT = L(A) + L(B)

L(cA) = (cA)T = cAT = cL(A).

Hence, L is a LT.


Example 2: For each [x, y] ∈ R2, consider a map

L : R2 → R

3 given by

L([x, y]) = [x, y, xy].




L : R2 → R

3 given by

L([x, y]) = [x, y, xy].


Solution: For c = 2 ∈ R and [1, 2] ∈ R2 consider



L : R2 → R

3 given by

L([x, y]) = [x, y, xy].



L(2([1, 2])) = L([2, 4])



L : R2 → R

3 given by

L([x, y]) = [x, y, xy].



L(2([1, 2])) = L([2, 4])

= [2, 4, 8] 6= 2L([1, 2])



L : R2 → R

3 given by

L([x, y]) = [x, y, xy].



L(2([1, 2])) = L([2, 4])

= [2, 4, 8] 6= 2L([1, 2])

Thus, L(c([x, y])) 6= cL([x, y]) ∀c ∈ R and [x, y] ∈ R2



L : R2 → R

3 given by

L([x, y]) = [x, y, xy].



L(2([1, 2])) = L([2, 4])

= [2, 4, 8] 6= 2L([1, 2])

Thus, L(c([x, y])) 6= cL([x, y]) ∀c ∈ R and [x, y] ∈ R2

Hence, L is not a LT.


Exercise: Check which of the following maps are LT.


Exercise: Check which of the following maps are LT.1 L : P2 → R

3 given by L(a + bx + cx2) = [a, b, c].



3 given by L(a + bx + cx2) = [a, b, c].2 L : R

3 → R2 given by L([x, y, z]) = [x − y, y + z].





3 L : R2 → R

2 given by L([a, b]) = [a,−b].





3 L : R2 → R

2 given by L([a, b]) = [a,−b].4 L : R → Φ given by L(x) = sinx.





3 L : R2 → R

2 given by L([a, b]) = [a,−b].4 L : R → Φ given by L(x) = sinx.5 L : R → R given by L(x) = x2.


Linear Operator: Let V be a vector space.


Linear Operator: Let V be a vector space. A linearoperator on V is a LT whose domain and codomainare both V.


Linear Operator: Let V be a vector space. A linearoperator on V is a LT whose domain and codomainare both V.

Example 3: The mapping L : R3 → R

3 given byL([x, y, z]) = [x, y,−z] is a linear operator.


Theorem 1: Let V and W be vector spaces, and letL : V → W be a LT.


Theorem 1: Let V and W be vector spaces, and letL : V → W be a LT. Suppose 0V be the zero vector inV and 0W be the zero vector in W. Then



1 L(0V) = 0W



1 L(0V) = 0W

2 L(−v) = −L(v), for all v ∈ V



1 L(0V) = 0W

2 L(−v) = −L(v), for all v ∈ V3 For n ≥ 2 and a1, a2, . . . , an ∈ R,

If v = a1v1 + a2v2 + · · · + anvn, then



1 L(0V) = 0W


If v = a1v1 + a2v2 + · · · + anvn, thenL(v) = L(a1v1 + a2v2 + · · · + anvn)



1 L(0V) = 0W


If v = a1v1 + a2v2 + · · · + anvn, thenL(v) = L(a1v1 + a2v2 + · · · + anvn)

= a1L(v1) + a2L(v2) + · · · + anL(vn).


Example 4: Let L : R3 → R

3 be a linear operatorsuch that L([1, 0, 0]) = [−2, 1, 0],L([0, 1, 0]) = [3,−2, 1], and L([0, 0, 1]) = [0,−1, 3].

Find L([−3, 2, 4]).




Find L([−3, 2, 4]).Find L([x, y, z]) for all [x, y, z] in R

3.





3.

Solution:





3.

Solution:

[−3, 2, 4] = −3[1, 0, 0] + 2[0, 1, 0] + 4[0, 0, 1]





3.

Solution:

[−3, 2, 4] = −3[1, 0, 0] + 2[0, 1, 0] + 4[0, 0, 1]

L([−3, 2, 4]) = L(−3[1, 0, 0] + 2[0, 1, 0] + 4[0, 0, 1])





3.

Solution:

[−3, 2, 4] = −3[1, 0, 0] + 2[0, 1, 0] + 4[0, 0, 1]

L([−3, 2, 4]) = L(−3[1, 0, 0] + 2[0, 1, 0] + 4[0, 0, 1])

= −3L([1, 0, 0]) + 2L([0, 1, 0]) + 4L([0, 0, 1])





3.

Solution:

[−3, 2, 4] = −3[1, 0, 0] + 2[0, 1, 0] + 4[0, 0, 1]

L([−3, 2, 4]) = L(−3[1, 0, 0] + 2[0, 1, 0] + 4[0, 0, 1])

= −3L([1, 0, 0]) + 2L([0, 1, 0]) + 4L([0, 0, 1])

= −3[−2, 1, 0] + 2[3,−2, 1] + 4[0,−1, 3]





3.

Solution:

[−3, 2, 4] = −3[1, 0, 0] + 2[0, 1, 0] + 4[0, 0, 1]

L([−3, 2, 4]) = L(−3[1, 0, 0] + 2[0, 1, 0] + 4[0, 0, 1])

= −3L([1, 0, 0]) + 2L([0, 1, 0]) + 4L([0, 0, 1])

= −3[−2, 1, 0] + 2[3,−2, 1] + 4[0,−1, 3]

= [12, 11, 14]


Similarly,

L([x, y, z]) = L(x[1, 0, 0] + y[0, 1, 0] + z[0, 0, 1])


Similarly,

L([x, y, z]) = L(x[1, 0, 0] + y[0, 1, 0] + z[0, 0, 1])

L([x, y, z]) = x[−2, 1, 0] + y[3,−2, 1] + z[0,−1, 3]


Similarly,

L([x, y, z]) = L(x[1, 0, 0] + y[0, 1, 0] + z[0, 0, 1])

L([x, y, z]) = x[−2, 1, 0] + y[3,−2, 1] + z[0,−1, 3]

L([x, y, z]) = [−2x + 3y, x − 2y − z, y + 3z]


Similarly,

L([x, y, z]) = L(x[1, 0, 0] + y[0, 1, 0] + z[0, 0, 1])

L([x, y, z]) = x[−2, 1, 0] + y[3,−2, 1] + z[0,−1, 3]

L([x, y, z]) = [−2x + 3y, x − 2y − z, y + 3z]

Note that

L

xyz

=

−2 3 01 −2 −10 1 3

xyz


Exercise: Suppose L : R2 → R

2 is a linear operatorand L([1, 1]) = [3, 0] and L([−1, 1]) = [0, 1]. FindL([x, y]) for all [x, y] ∈ R

2.




2.

Answer: L([x, y]) =[

3x+3y2

, −x+y2

]

.




2.


3x+3y2

, −x+y2

]

.

Remark: Let V and W be vector spaces, and letL : V → W be a LT. Also, let {v1, v2, . . . , vn} be abasis for V.




2.


3x+3y2

, −x+y2

]

.

Remark: Let V and W be vector spaces, and letL : V → W be a LT. Also, let {v1, v2, . . . , vn} be abasis for V. If v ∈ V, L(v) is completely determinedby {L(v1), L(v2), . . . , L(vn)}.


Composition of Linear transformations


Composition of Linear transformations

Theorem 2: Let V1, V2 and V3 be vector spaces andlet L1 : V1 → V2 and L2 : V2 → V3 be lineartransformations. Then L2 ◦ L1 : V1 → V3 given by(L2 ◦ L1)(v) = L2(L1(v)), for all v ∈ V1, is a LT.


Example 5: Let L1 : P2 → P2 and L2 : P2 → P2 belinear operators defined as L1(ax2 + bx + c) = 2ax + band L2(ax2 + bx + c) = 2ax2 + bx, respectively.


Example 5: Let L1 : P2 → P2 and L2 : P2 → P2 belinear operators defined as L1(ax2 + bx + c) = 2ax + band L2(ax2 + bx + c) = 2ax2 + bx, respectively.Compute L2 ◦ L1 and L1 ◦ L2.



Answer:L2 ◦ L1(ax2 + bx + c) = 2ax.




L1 ◦ L2(ax2 + bx + c) = 4ax + b.




L1 ◦ L2(ax2 + bx + c) = 4ax + b.

Clearly , L2 ◦ L1 6= L1 ◦ L2.


Section 5.3: The Dimension Theorem

Kernel of a linear transformation: Let L : V → Wbe a LT.



Kernel of a linear transformation: Let L : V → Wbe a LT. The kernel of L, denoted by ker(L),



Kernel of a linear transformation: Let L : V → Wbe a LT. The kernel of L, denoted by ker(L), is thesubset of all vectors in V that map to 0W , i.e.



Kernel of a linear transformation: Let L : V → Wbe a LT. The kernel of L, denoted by ker(L), is thesubset of all vectors in V that map to 0W , i.e.

ker(L) = {v ∈ V | L(v) = 0W}.



2 be a LT given byL([x, y, z]) = [0, y].



2 be a LT given byL([x, y, z]) = [0, y]. Find ker(L).




Solution:

ker(L) = {[x, y, z] ∈ R3 | L([x, y, z]) = 0R2}




Solution:

ker(L) = {[x, y, z] ∈ R3 | L([x, y, z]) = 0R2}

= {[x, y, z] ∈ R3 | [0, y] = [0, 0]}




Solution:

ker(L) = {[x, y, z] ∈ R3 | L([x, y, z]) = 0R2}

= {[x, y, z] ∈ R3 | [0, y] = [0, 0]}

= {[x, y, z] ∈ R3 | y = 0}




Solution:

ker(L) = {[x, y, z] ∈ R3 | L([x, y, z]) = 0R2}

= {[x, y, z] ∈ R3 | [0, y] = [0, 0]}

= {[x, y, z] ∈ R3 | y = 0}

= {[x, 0, z] | x, z ∈ R}




Solution:

ker(L) = {[x, y, z] ∈ R3 | L([x, y, z]) = 0R2}

= {[x, y, z] ∈ R3 | [0, y] = [0, 0]}

= {[x, y, z] ∈ R3 | y = 0}

= {[x, 0, z] | x, z ∈ R}

In this Example, Note that

ker(L) = {[x, 0, z] | x, z ∈ R}




Solution:

ker(L) = {[x, y, z] ∈ R3 | L([x, y, z]) = 0R2}

= {[x, y, z] ∈ R3 | [0, y] = [0, 0]}

= {[x, y, z] ∈ R3 | y = 0}

= {[x, 0, z] | x, z ∈ R}

In this Example, Note that

ker(L) = {[x, 0, z] | x, z ∈ R}

is a subspace of the vector space R3.


Result: If L : V → W is a LT, then ker(L) is asubspace of V.


Range of a linear transformation:



Definition: Let L : V → W be a LT.



Definition: Let L : V → W be a LT. The range of L,denoted by range(L), is the subset of all vectors inW that are image of some vector in V, i.e.




range(L) = {L(v) | v ∈ V}

Thus a vector w ∈ range(L)




range(L) = {L(v) | v ∈ V}

Thus a vector w ∈ range(L) implies there existssome vector v ∈ V such that L(v) = w.


Result: If L : V → W is a LT, then range(L) is asubspace of W.



Exercise: Let L : R3 → R

3 be a LT given by

L

xyz

=

5 1 −1−3 0 11 −1 −1

xyz




3 be a LT given by

L

xyz

=

5 1 −1−3 0 11 −1 −1

xyz

Is [1,−2, 3]T ∈ ker(L)?




3 be a LT given by

L

xyz

=

5 1 −1−3 0 11 −1 −1

xyz

Is [1,−2, 3]T ∈ ker(L)?Is [2,−1, 4]T ∈ range(L)?


Hint: Note that L([1,−2, 3]T) = [0, 0, 0]T


Hint: Note that L([1,−2, 3]T) = [0, 0, 0]T implies[1,−2, 3]T ∈ ker(L).



Note that to check [2,−1, 4]T ∈ range(L) is same asto check whether given system of linear equations

5x + y − z = 2

−3x + z = −1

x − y − z = 4

is consistent or not.




5x + y − z = 2

−3x + z = −1

x − y − z = 4


Since above system of equations is inconsistent(show it!),




5x + y − z = 2

−3x + z = −1

x − y − z = 4


Since above system of equations is inconsistent(show it!), [2,−1, 4]T /∈ range(L).



2 be a LT given by

L([x, y, z]) = [0, y] for all [x, y, z] ∈ R3



2 be a LT given by


Find range(L).



2 be a LT given by


Find range(L).Find the dimension of ker(L) and range(L).



2 be a LT given by



Solution:

range(L) ={

L([x, y, z]) | [x, y, z] ∈ R3}



2 be a LT given by



Solution:

range(L) ={

L([x, y, z]) | [x, y, z] ∈ R3}

= {[0, y] | y ∈ R}


range(L) = {y[0, 1] | y ∈ R}


range(L) = {y[0, 1] | y ∈ R}

Note that range(L) = span{[0, 1]}.


range(L) = {y[0, 1] | y ∈ R}

Note that range(L) = span{[0, 1]}. Since {[0, 1]} is LIsubset of R

2 (Why?).


range(L) = {y[0, 1] | y ∈ R}


2 (Why?). Thus, the set B = {[0, 1]} is abasis of range(L) so that dim(range(L)) = 1


range(L) = {y[0, 1] | y ∈ R}



ker(L) = {[x, 0, z] | x, z ∈ R}


range(L) = {y[0, 1] | y ∈ R}



ker(L) = {[x, 0, z] | x, z ∈ R} (see Example 6)


range(L) = {y[0, 1] | y ∈ R}




= {x[1, 0, 0] + z[0, 0, 1] | x, z ∈ R}


range(L) = {y[0, 1] | y ∈ R}




= {x[1, 0, 0] + z[0, 0, 1] | x, z ∈ R}

= span{[1, 0, 0], [0, 0, 1]}


range(L) = {y[0, 1] | y ∈ R}




= {x[1, 0, 0] + z[0, 0, 1] | x, z ∈ R}

= span{[1, 0, 0], [0, 0, 1]}

Now, the set {[1, 0, 0], [0, 0, 1]} of vectors is LI subsetof R

3 (verify!).


range(L) = {y[0, 1] | y ∈ R}




= {x[1, 0, 0] + z[0, 0, 1] | x, z ∈ R}

= span{[1, 0, 0], [0, 0, 1]}

Now, the set {[1, 0, 0], [0, 0, 1]} of vectors is LI subsetof R

3 (verify!). Hence, the set {[1, 0, 0], [0, 0, 1]} formsa basis of ker(L) and dim(ker(L)) = 2.



2 be a LT given by

L([x, y, z]) = [x − 2y, y + z].



2 be a LT given by

L([x, y, z]) = [x − 2y, y + z].

Find ker(L) and range(L).



2 be a LT given by

L([x, y, z]) = [x − 2y, y + z].

Find ker(L) and range(L). Also, find basis for ker(L)and range(L).



2 be a LT given by

L([x, y, z]) = [x − 2y, y + z].


Solution:

ker(L) ={

[x, y, z] ∈ R3 | L([x, y, z]) = 0R2

}



2 be a LT given by

L([x, y, z]) = [x − 2y, y + z].


Solution:

ker(L) ={

[x, y, z] ∈ R3 | L([x, y, z]) = 0R2

}

={

[x, y, z] ∈ R3 | [x − 2y, y + z] = [0, 0]

}



2 be a LT given by

L([x, y, z]) = [x − 2y, y + z].


Solution:

ker(L) ={

[x, y, z] ∈ R3 | L([x, y, z]) = 0R2

}

={

[x, y, z] ∈ R3 | [x − 2y, y + z] = [0, 0]

}

={

[x, y, z] ∈ R3 | x = 2y, z = −y

}



2 be a LT given by

L([x, y, z]) = [x − 2y, y + z].


Solution:

ker(L) ={

[x, y, z] ∈ R3 | L([x, y, z]) = 0R2

}

={

[x, y, z] ∈ R3 | [x − 2y, y + z] = [0, 0]

}

={

[x, y, z] ∈ R3 | x = 2y, z = −y

}

= {[2y, y,−y] | y ∈ R}



2 be a LT given by

L([x, y, z]) = [x − 2y, y + z].


Solution:

ker(L) ={

[x, y, z] ∈ R3 | L([x, y, z]) = 0R2

}

={

[x, y, z] ∈ R3 | [x − 2y, y + z] = [0, 0]

}

={

[x, y, z] ∈ R3 | x = 2y, z = −y

}

= {[2y, y,−y] | y ∈ R}

= span{[2, 1,−1]}


Since the set B = {[2, 1,−1]} is LI. Therefore,B = {[2, 1,−1]} is a basis of ker(L). Now

range(L) ={

L([x, y, z]) | [x, y, z] ∈ R3}



range(L) ={

L([x, y, z]) | [x, y, z] ∈ R3}

= {[x − 2y, y + z] | x, y, z ∈ R}



range(L) ={

L([x, y, z]) | [x, y, z] ∈ R3}

= {[x − 2y, y + z] | x, y, z ∈ R}

= {x[1, 0] + y[−2, 1] + z[0, 1] | x, y, z ∈ R}



range(L) ={

L([x, y, z]) | [x, y, z] ∈ R3}

= {[x − 2y, y + z] | x, y, z ∈ R}

= {x[1, 0] + y[−2, 1] + z[0, 1] | x, y, z ∈ R}

= span{[1, 0], [−2, 1], [0, 1]}



range(L) ={

L([x, y, z]) | [x, y, z] ∈ R3}

= {[x − 2y, y + z] | x, y, z ∈ R}

= {x[1, 0] + y[−2, 1] + z[0, 1] | x, y, z ∈ R}

= span{[1, 0], [−2, 1], [0, 1]}

= span{[1, 0], [0, 1]}



range(L) ={

L([x, y, z]) | [x, y, z] ∈ R3}

= {[x − 2y, y + z] | x, y, z ∈ R}

= {x[1, 0] + y[−2, 1] + z[0, 1] | x, y, z ∈ R}

= span{[1, 0], [−2, 1], [0, 1]}

= span{[1, 0], [0, 1]}

Since the set {[1, 0], [0, 1]} is LI. Thus,

{[1, 0], [0, 1]}

is a basis for range(L).Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 23 / 122

Exercise: Given a map

L : P3 → P2 given by

L(ax3 + bx2 + cx + d) = 3ax2 + 2bx + c.

1 Show that L is a linear transformation.




L(ax3 + bx2 + cx + d) = 3ax2 + 2bx + c.

1 Show that L is a linear transformation.2 Find ker(L) and range(L).




L(ax3 + bx2 + cx + d) = 3ax2 + 2bx + c.

1 Show that L is a linear transformation.2 Find ker(L) and range(L).

Answer:

ker(L) ={

0x3 + 0x2 + 0x + d | d ∈ R}

range(L) = P2.



L : R4 → P2 given by

L([a, b, c, d]) = a + (b + c)x + (b − d)x2.

1 Find ker(L) and range(L).




L([a, b, c, d]) = a + (b + c)x + (b − d)x2.

1 Find ker(L) and range(L).2 Find a basis for ker(L) and range(L).




L([a, b, c, d]) = a + (b + c)x + (b − d)x2.


Answer:

ker(L) = {[0, b,−b, b] | b ∈ R} and B = {[0, 1,−1, 1]}




L([a, b, c, d]) = a + (b + c)x + (b − d)x2.


Answer:

ker(L) = {[0, b,−b, b] | b ∈ R} and B = {[0, 1,−1, 1]}

range(L) = span{1, x + x2, x, x2} and B = {1, x, x2}.



4 be a LT given by

L([x, y, z]) = [x, y − z, x − y + z, x + y − z].



4 be a LT given by

L([x, y, z]) = [x, y − z, x − y + z, x + y − z].

Find a basis for ker(L) and range(L).



4 be a LT given by

L([x, y, z]) = [x, y − z, x − y + z, x + y − z].

Find a basis for ker(L) and range(L).

Answer:

{[0, 1, 1]} is a basis of ker(L).{[1, 0, 1, 1], [0, 1,−1, 1]} is a basis for range(L).


Alternative approach for finding a basis for ker(L)(Kernel Method) Let L : R

n → Rm be a LT.



n → Rm be a LT.

Step 1: Express L(X) = AX for some m × n matrixA.



n → Rm be a LT.

Step 1: Express L(X) = AX for some m × n matrixA. In Example 9, note that L(X) = AX where



n → Rm be a LT.

Step 1: Express L(X) = AX for some m × n matrixA. In Example 9, note that L(X) = AX where

X =

xyz

and A =

1 0 00 1 −11 −1 11 1 −1


Step 2: Find matrix B, the RREF of A.



B = RREF(A) =

1 0 00 1 −10 0 00 0 0



B = RREF(A) =

1 0 00 1 −10 0 00 0 0

Step 3: Solve the system BX = 0 to find ker(L)such that ker(L) = span(S) for some S ⊆ R

n.



B = RREF(A) =

1 0 00 1 −10 0 00 0 0


n. Thesystem corresponding to B is x = 0, y = z.

ker(L) = {X ∈ Rn|L(X) = AX = 0}



B = RREF(A) =

1 0 00 1 −10 0 00 0 0



ker(L) = {X ∈ Rn|L(X) = AX = 0}

= {X ∈ Rn | BX = 0}



B = RREF(A) =

1 0 00 1 −10 0 00 0 0



ker(L) = {X ∈ Rn|L(X) = AX = 0}

= {X ∈ Rn | BX = 0}

= {[0, y, y] | y ∈ R}


ker(L) = span{[0, 1, 1]}


ker(L) = span{[0, 1, 1]}

ker(L) = span(S), where S = {[0, 1, 1]}


ker(L) = span{[0, 1, 1]}


Step 4: Find a LI subset of S which forms a basis forker(L).


ker(L) = span{[0, 1, 1]}


Step 4: Find a LI subset of S which forms a basis forker(L). Since the set {[0, 1, 1]} is a LI so it is a basisof ker(L).


Alternative approach to find a basis for range (L)(Range Method)

Step 1: Find RREF of A.




RREF(A) =

1 0 00 1 −10 0 00 0 0




RREF(A) =

1 0 00 1 −10 0 00 0 0

Step 2: Column vectors in A corresponding to pivotcolumns of RREF(A) forms a basis for range(L).




RREF(A) =

1 0 00 1 −10 0 00 0 0

Step 2: Column vectors in A corresponding to pivotcolumns of RREF(A) forms a basis for range(L).Note that, Columns I and II have leading entry.




RREF(A) =

1 0 00 1 −10 0 00 0 0

Step 2: Column vectors in A corresponding to pivotcolumns of RREF(A) forms a basis for range(L).Note that, Columns I and II have leading entry. Thus,the corresponding column vector of A i.e.{[1, 0, 1, 1], [0, 1,−1, 1]} is a basis of range (L).


The Dimension Theorem:


The Dimension Theorem: If L : V → W is a LT andV is finite dimensional, then range(L) is finitedimensional, and

dim(ker(L)) + dim(range(L)) = dim(V).


The Dimension Theorem: If L : V → W is a LT andV is finite dimensional, then range(L) is finitedimensional, and

dim(ker(L)) + dim(range(L)) = dim(V).

Sometimes dim(ker(L)) and dim(range(L)) is alsoknown as nullity (L) and rank (L), respectively.


Example 10: Consider a LT L : P2 → P3 given by

L(a + bx + cx2) = x(a + bx + cx2).



L(a + bx + cx2) = x(a + bx + cx2).

Find dim(ker(L)) and dim(range(L)).



L(a + bx + cx2) = x(a + bx + cx2).


Solution:

ker(L) = {a + bx + cx2 ∈ P2 | L(a + bx + cx2) = 0P3}



L(a + bx + cx2) = x(a + bx + cx2).


Solution:


ker(L) = {0P2}



L(a + bx + cx2) = x(a + bx + cx2).


Solution:


ker(L) = {0P2} implies dim(ker(L)) = 0.



L(a + bx + cx2) = x(a + bx + cx2).


Solution:



Since dimP2 = 3 by dimension theorem, we have



L(a + bx + cx2) = x(a + bx + cx2).


Solution:



Since dimP2 = 3 by dimension theorem, we have

dim(range(L)) = 3 − 0 = 3.


Exercise: Consider a LT L : M33 → R given by

L(A) = trace(A)



L(A) = trace(A)(sum of the diagonal entries ofA).




Find ker(L), dim(ker(L)), range(L) anddim(range(L)).





Answer:

ker(L) =

a b cd e fg h −a − e

3×3

| a, b, c, d, e, f, g, h ∈ R

Note that





Answer:

ker(L) =


3×3

| a, b, c, d, e, f, g, h ∈ R

Note that dim(ker(L)) = 8 (show it).





Answer:

ker(L) =


3×3

| a, b, c, d, e, f, g, h ∈ R

Note that dim(ker(L)) = 8 (show it). Sincerange(L) = R





Answer:

ker(L) =


3×3

| a, b, c, d, e, f, g, h ∈ R

Note that dim(ker(L)) = 8 (show it). Sincerange(L) = R so that dim(range(L)) = 1.


Exercise: Let W be the vector space of all 2 × 2symmetric matrices. Define a LT L : W → P2 by

L

([

a bb c

])

= (a − b) + (b − c)x + (c − a)x2



L

([

a bb c

])

= (a − b) + (b − c)x + (c − a)x2




L

([

a bb c

])

= (a − b) + (b − c)x + (c − a)x2


Answer: dim(ker(L)) = 1 and dim(range(L)) = 2.


Exercise: Let {e1, e2, e3, e4} be standard basis for R4

and L : R4 → R

3 be a LT given by



and L : R4 → R

3 be a LT given by

L(e1) = [1, 1, 1], L(e2) = [1,−1, 1]

L(e3) = [1, 0, 0], L(e4) = [1, 0, 1]



and L : R4 → R

3 be a LT given by

L(e1) = [1, 1, 1], L(e2) = [1,−1, 1]

L(e3) = [1, 0, 0], L(e4) = [1, 0, 1]

Find ker(L) and dim(ker(L)).Find range(L) and dim(range(L)).



and L : R4 → R

3 be a LT given by

L(e1) = [1, 1, 1], L(e2) = [1,−1, 1]

L(e3) = [1, 0, 0], L(e4) = [1, 0, 1]




Exercise: For each p ∈ P2, consider L : P2 → P4

given by L(p) = x2p.

Find ker(L) and dim(ker(L)).











Section 5.4

Definition: A linear transformation L : V → Wone-to-one if and only if


Section 5.4

Definition: A linear transformation L : V → Wone-to-one if and only if for all v1, v2 ∈ V,L(v1) = L(v2) implies v1 = v2,


Section 5.4

Definition: A linear transformation L : V → Wone-to-one if and only if for all v1, v2 ∈ V,L(v1) = L(v2) implies v1 = v2, or if v1 6= v2 impliesL(v1) 6= L(v2).


Section 5.4


L is onto if and only if,


Section 5.4


L is onto if and only if, for each w ∈ W, there issome v ∈ V such that L(v) = w.


Example 11: Consider a LT


L(p) = p′.




L(p) = p′.

Check if L is one-to-one and onto.




L(p) = p′.


Solution: Consider p1 = x + 2 and p2 = x + 4.




L(p) = p′.


Solution: Consider p1 = x + 2 and p2 = x + 4.Since, L(p1) = L(p2) = 1 implies




L(p) = p′.


Solution: Consider p1 = x + 2 and p2 = x + 4.Since, L(p1) = L(p2) = 1 implies L is not one-to-one.




L(p) = p′.



Let q be an arbitrary element in P2 i.e.q = a + bx + cx2.




L(p) = p′.



Let q be an arbitrary element in P2 i.e.q = a + bx + cx2. Note that a + bx + cx2 = p′, wherep = ax +

(

b2

)

x2 +(

c3

)

x3 so that L(p) = q.




L(p) = p′.



Let q be an arbitrary element in P2 i.e.q = a + bx + cx2. Note that a + bx + cx2 = p′, wherep = ax +

(

b2

)

x2 +(

c3

)

x3 so that L(p) = q. Hence, Lis onto.


Exercise: Which of the following transformations areone-to-one? onto?

1 L : R2 → R

3 given by L([x, y]) = [2x, x− y, 0].2 L : R

3 → R4 given by L([x, y, z]) = [y, z,−y, 0].

3 L : M22 → M22 given by L(A) = AT .



1 L : R2 → R



3 L : M22 → M22 given by L(A) = AT .

Answer:1 one-to-one but not onto.



1 L : R2 → R



3 L : M22 → M22 given by L(A) = AT .

Answer:1 one-to-one but not onto.2 neither one-to-one nor onto



1 L : R2 → R



3 L : M22 → M22 given by L(A) = AT .

Answer:1 one-to-one but not onto.2 neither one-to-one nor onto3 one-to-one and onto.


Theorem 3: Let V and W be vector spaces, and letL : V → W be a LT. Then


Theorem 3: Let V and W be vector spaces, and letL : V → W be a LT. Then L is one-to-one if and onlyif ker(L) = {0V} (i.e., dimker(L) = 0).



Theorem 4: Let V and W be vector spaces, and letL : V → W be a LT.



Theorem 4: Let V and W be vector spaces, and letL : V → W be a LT. If W is finite dimensional, then Lis onto



Theorem 4: Let V and W be vector spaces, and letL : V → W be a LT. If W is finite dimensional, then Lis onto if and only if dim(range(L)) = dim(W).


Example 12: Consider a LT L : M22 → M23 givenby

L

([

a bc d

])

=

[

a − b 0 c − dc + d a + b 0

]



L

([

a bc d

])

=

[

a − b 0 c − dc + d a + b 0

]

Is L one-to-one and onto?



L

([

a bc d

])

=

[

a − b 0 c − dc + d a + b 0

]


Solution: Let[

a bc d

]

∈ ker(L).



L

([

a bc d

])

=

[

a − b 0 c − dc + d a + b 0

]


Solution: Let[

a bc d

]

∈ ker(L). Then

L

([

a bc d

])

=

[

a − b 0 c − dc + d a + b 0

]

=

[

0 0 00 0 0

]

.



L

([

a bc d

])

=

[

a − b 0 c − dc + d a + b 0

]


Solution: Let[

a bc d

]

∈ ker(L). Then

L

([

a bc d

])

=

[

a − b 0 c − dc + d a + b 0

]

=

[

0 0 00 0 0

]

.

We have a − b = c − d = c + d = a + b = 0 impliesa = b = c = d = 0.


Hence, ker(L) contains only the zero matrix (the zerovector of M22).


Hence, ker(L) contains only the zero matrix (the zerovector of M22). Thus, L is one-to-one.



Note that



Note that

dim(range(L)) = dim(M22) − dim(ker(L))

= 4

6= dim(M23).



Note that


= 4

6= dim(M23).

Hence, L is not onto.



Note that


= 4

6= dim(M23).


Try to find a basis of range(L).


Example 13: Consider a LT L : R3 → R

3 given by

L

xyz

=

−7 4 −216 −7 24 −3 2

xyz



3 given by

L

xyz

=

−7 4 −216 −7 24 −3 2

xyz




3 given by

L

xyz

=

−7 4 −216 −7 24 −3 2

xyz


Solution: The RREF of matrix A =

−7 4 −216 −7 24 −3 2

is

1 0 −2

5

0 1 −6

5

0 0 0

.


From range method,


From range method, dim(range(L)) = 2


From range method, dim(range(L)) = 2 and byDimension theorem, dim(ker(L)) = 1.


From range method, dim(range(L)) = 2 and byDimension theorem, dim(ker(L)) = 1. Hence, L isneither one-to-one nor onto.


Example 14: Let A be a fixed n × n matrix, andconsider a LT L : Mnn → Mnn given byL(B) = AB − BA.


Example 14: Let A be a fixed n × n matrix, andconsider a LT L : Mnn → Mnn given byL(B) = AB − BA. Is L one-to-one and onto?

Solution: L(In) =



Solution: L(In) = AIn − InA = 0n×n.



Solution: L(In) = AIn − InA = 0n×n. Hence,In ∈ ker(L)



Solution: L(In) = AIn − InA = 0n×n. Hence,In ∈ ker(L) and so, L is not one-to-one.



Solution: L(In) = AIn − InA = 0n×n. Hence,In ∈ ker(L) and so, L is not one-to-one. ByDimension theorem, we see that




dim(range(L)) = n2 − dim(ker(L))





6= n2

6= dimMnn





6= n2

6= dimMnn



Example 15: Consider a LT L : P → P given byL(p(x)) = xp(x).


Example 15: Consider a LT L : P → P given byL(p(x)) = xp(x). Is L one-to-one and onto?



Solution:

ker(L) = {p(x)|L(p(x)) = 0P}



Solution:

ker(L) = {p(x)|L(p(x)) = 0P}

implies ker(L) = {0P}.



Solution:

ker(L) = {p(x)|L(p(x)) = 0P}

implies ker(L) = {0P}. Hence, L is one-to-one.



Solution:

ker(L) = {p(x)|L(p(x)) = 0P}

implies ker(L) = {0P}. Hence, L is one-to-one. Notethat the nonzero constant polynomials is not inrange(L),



Solution:

ker(L) = {p(x)|L(p(x)) = 0P}

implies ker(L) = {0P}. Hence, L is one-to-one. Notethat the nonzero constant polynomials is not inrange(L), L is not onto.



Solution:

ker(L) = {p(x)|L(p(x)) = 0P}

implies ker(L) = {0P}. Hence, L is one-to-one. Notethat the nonzero constant polynomials is not inrange(L), L is not onto.

Question : Can we apply Dimension theorem here?


Exercise: Consider a LT L : M23 → M22 given by

L

([

a b cd e f

])

=

[

a + b a + cd + e d + f

]



L

([

a b cd e f

])

=

[

a + b a + cd + e d + f

]




L

([

a b cd e f

])

=

[

a + b a + cd + e d + f

]


Answer: L is onto but not one-to-one.



L

([

a b cd e f

])

=

[

a + b a + cd + e d + f

]



Exercise: Consider a LT L : P2 → P2 given by

L(ax2 + bx + c) = (a + b)x2 + (b + c)x + (a + c).



L

([

a b cd e f

])

=

[

a + b a + cd + e d + f

]




L(ax2 + bx + c) = (a + b)x2 + (b + c)x + (a + c). Is Lone-to-one and onto?



L

([

a b cd e f

])

=

[

a + b a + cd + e d + f

]




L(ax2 + bx + c) = (a + b)x2 + (b + c)x + (a + c). Is Lone-to-one and onto?

Answer: L is one-to-one and onto.


Exercise: Consider a LT L : R4 → R

3 given by

L

x1

x2

x3

x4

=

−5 3 1 18−2 1 1 6−7 3 4 19

x1

x2

x3

x4



3 given by

L

x1

x2

x3

x4

=

−5 3 1 18−2 1 1 6−7 3 4 19

x1

x2

x3

x4




3 given by

L

x1

x2

x3

x4

=

−5 3 1 18−2 1 1 6−7 3 4 19

x1

x2

x3

x4


Answer: L is not one-to-one but onto.



3 given by

L

x1

x2

x3

x4

=

−5 3 1 18−2 1 1 6−7 3 4 19

x1

x2

x3

x4


Answer: L is not one-to-one but onto.


Section 5.5


Section 5.5

Invertible linear transformation:


Section 5.5

Invertible linear transformation: Let L : V → W bea LT. Then L is an invertible LT if and only if


Section 5.5

Invertible linear transformation: Let L : V → W bea LT. Then L is an invertible LT if and only if there isa function M : W → V such that (M ◦ L)(v) = v, forall v ∈ V, and (L ◦ M)(w) = w, for all w ∈ W.


Section 5.5

Invertible linear transformation: Let L : V → W bea LT. Then L is an invertible LT if and only if there isa function M : W → V such that (M ◦ L)(v) = v, forall v ∈ V, and (L ◦ M)(w) = w, for all w ∈ W.

Such a function M , denoted by L−1, is called aninverse of L.


Isomorphism: A LT L : V → W that is bothone-to-one and onto is called as isomorphism fromV to W.



Example 16: Show that L : Pn → Pn given byL(p) = p + p ′ is an isomorphism.




Solution: First we need to show that L is a linearoperator.





L(p + q) = (p + q) + (p + q)′





L(p + q) = (p + q) + (p + q)′

= p + p′ + q + q′





L(p + q) = (p + q) + (p + q)′

= p + p′ + q + q′

= L(p) + L(q) for all p, q ∈ Pn.


Similarly, (show that) L(c p) = cL(p) for all real c andp ∈ Pn.


Similarly, (show that) L(c p) = cL(p) for all real c andp ∈ Pn. Hence, L is a linear operator.



kerL = {p ∈ Pn | L(p) = 0Pn}



kerL = {p ∈ Pn | L(p) = 0Pn}

= {p ∈ Pn | p + p′ = 0Pn}



kerL = {p ∈ Pn | L(p) = 0Pn}

= {p ∈ Pn | p + p′ = 0Pn}

= {0Pn}



kerL = {p ∈ Pn | L(p) = 0Pn}

= {p ∈ Pn | p + p′ = 0Pn}

= {0Pn}

implies L is one-to-one.



kerL = {p ∈ Pn | L(p) = 0Pn}

= {p ∈ Pn | p + p′ = 0Pn}

= {0Pn}


By Dimension theorem,

dim range(L) = dimPn = n + 1

so that range(L) = Pn.



kerL = {p ∈ Pn | L(p) = 0Pn}

= {p ∈ Pn | p + p′ = 0Pn}

= {0Pn}




so that range(L) = Pn. Thus, L is onto.



kerL = {p ∈ Pn | L(p) = 0Pn}

= {p ∈ Pn | p + p′ = 0Pn}

= {0Pn}




so that range(L) = Pn. Thus, L is onto. Hence, L isan isomorphism.


Example 17: Show that the linear operator

L : R3 → R

3 such that

L(e1) = e1 + e2, L(e2) = e2 + e3, L(e3) = e1 + e2 + e3

is an isomorphism.



L : R3 → R

3 such that

L(e1) = e1 + e2, L(e2) = e2 + e3, L(e3) = e1 + e2 + e3

is an isomorphism.

Hint: First find L([x, y, z]) for all [x, y, z] ∈ R3.



L : R3 → R

3 such that

L(e1) = e1 + e2, L(e2) = e2 + e3, L(e3) = e1 + e2 + e3

is an isomorphism.

Hint: First find L([x, y, z]) for all [x, y, z] ∈ R3. Note

thatL([x, y, z]) = [x + z, x + y + z, y + z].



L : R3 → R

3 such that

L(e1) = e1 + e2, L(e2) = e2 + e3, L(e3) = e1 + e2 + e3

is an isomorphism.



and ker(L) = {0R3}.



L : R3 → R

3 such that

L(e1) = e1 + e2, L(e2) = e2 + e3, L(e3) = e1 + e2 + e3

is an isomorphism.



and ker(L) = {0R3}. Thus, L is one-to-one.



L : R3 → R

3 such that

L(e1) = e1 + e2, L(e2) = e2 + e3, L(e3) = e1 + e2 + e3

is an isomorphism.



and ker(L) = {0R3}. Thus, L is one-to-one. Usedimension theorem and Theorem 4 to conclude L isonto. Hence, L is an isomorphism.


Result: Let L : V → W be a linear transformation,where V and W be finite dimensional vector spacessuch that dim(V) = dim(W). Then L is one-to-one ifand only if L is onto.


Exercise: Show that the linear operator L : P2 → P2

given by L(a+ bx+ cx2) = (b+ c)+(a+ c)x+(a+ b)x2

is an isomorphism.


Exercise: Show that the linear operator L : P2 → P2

given by L(a+ bx+ cx2) = (b+ c)+(a+ c)x+(a+ b)x2

is an isomorphism.

Exercise: Show that the linear transformationL : Mmn → Mnm given by L(A) = AT is anisomorphism.


Theorem: A LT L : V → W is an isomorphism if andonly if L is an invertible LT.


Theorem: A LT L : V → W is an isomorphism if andonly if L is an invertible LT. Moreover, if L isinvertible, then L−1 is also a LT.



Example 18: Let L : R3 → P2 be a LT given by

L([x, y, z]) = x + (x + y − z)t + (x + y + z)t2.




L([x, y, z]) = x + (x + y − z)t + (x + y + z)t2.

Is L invertible?




L([x, y, z]) = x + (x + y − z)t + (x + y + z)t2.

Is L invertible? If yes, find L−1.




L([x, y, z]) = x + (x + y − z)t + (x + y + z)t2.

Is L invertible? If yes, find L−1.

Solution: First show that L is both one-to-one andonto. Hence, invertible.


Let L−1 : P2 → R3 be defined by



L−1(a + bt + ct2) = [x, y, z]



L−1(a + bt + ct2) = [x, y, z]

⇒ L([x, y, z]) = a + bt + ct2



L−1(a + bt + ct2) = [x, y, z]

⇒ L([x, y, z]) = a + bt + ct2

⇒ x + (x + y − z)t + (x + y + z)t2 = a + bt + ct2



L−1(a + bt + ct2) = [x, y, z]

⇒ L([x, y, z]) = a + bt + ct2

⇒ x + (x + y − z)t + (x + y + z)t2 = a + bt + ct2

⇒ x = a, x + y − z = b, x + y + z = c



L−1(a + bt + ct2) = [x, y, z]

⇒ L([x, y, z]) = a + bt + ct2

⇒ x + (x + y − z)t + (x + y + z)t2 = a + bt + ct2

⇒ x = a, x + y − z = b, x + y + z = c

⇒ x = a, y =b + c − 2a

2, z =

c − b

2.



L−1(a + bt + ct2) = [x, y, z]

⇒ L([x, y, z]) = a + bt + ct2

⇒ x + (x + y − z)t + (x + y + z)t2 = a + bt + ct2

⇒ x = a, x + y − z = b, x + y + z = c

⇒ x = a, y =b + c − 2a

2, z =

c − b

2.

Hence, L−1(a + bx + cx2) =[

a, b+c−2a2

, c−b2

]

.


Exercise: Let L : P2 → P2 be a LT given byL(a + bx + cx2) = (b + c) + (a + c)x + (a + b)x2.


Exercise: Let L : P2 → P2 be a LT given byL(a + bx + cx2) = (b + c) + (a + c)x + (a + b)x2. Is Linvertible?


Exercise: Let L : P2 → P2 be a LT given byL(a + bx + cx2) = (b + c) + (a + c)x + (a + b)x2. Is Linvertible? If yes, find L−1.


Exercise: Let L : P2 → P2 be a LT given byL(a + bx + cx2) = (b + c) + (a + c)x + (a + b)x2. Is Linvertible? If yes, find L−1.

Answer:

L−1(a+bx+cx2) =1

2(b+c−a)+

1

2(a+c−b)x+

1

2(a+b−c)x2.



3 be a LT given byL(e1) = e1 + e2, L(e2) = e2 + e3, L(e3) = e1 + e2 + e3.



3 be a LT given byL(e1) = e1 + e2, L(e2) = e2 + e3, L(e3) = e1 + e2 + e3.Is L invertible?



3 be a LT given byL(e1) = e1 + e2, L(e2) = e2 + e3, L(e3) = e1 + e2 + e3.Is L invertible? If yes, find L−1.



3 be a LT given byL(e1) = e1 + e2, L(e2) = e2 + e3, L(e3) = e1 + e2 + e3.Is L invertible? If yes, find L−1.

Answer: L−1([x, y, z]) = [y − z, y − x, x − y + z].


Isomorphic vector spaces: Let V and W be vectorspaces. Then V is isomorphic to W, denoted byV ∼= W, if and only if there exists an isomorphismL : V → W.


Isomorphic vector spaces: Let V and W be vectorspaces. Then V is isomorphic to W, denoted byV ∼= W, if and only if there exists an isomorphismL : V → W.

Theorem 5: Suppose V ∼= W and V and W are finitedimensional. Then V is isomorphic to W if and only ifdim(V) = dim(W).


Exercise: Show that Rn and Pn are not isomorphic.



Solution: Since, dim(Rn) = n 6= n + 1 = dim(Pn),



Solution: Since, dim(Rn) = n 6= n + 1 = dim(Pn),R

n and Pn are not isomorphic.



Solution: Since, dim(Rn) = n 6= n + 1 = dim(Pn),R

n and Pn are not isomorphic.

Exercise: Let W be the vector space of allsymmetric 2 × 2 matrices. Show that W isisomorphic to R

3.


Exercise: Show that the subspace

W = {p ∈ P3 | p(0) = 0}

is isomorphic to P2.


Section 4.7

Ordered Basis: An ordered basis for vector spaceV is an ordered n-tuple of vectors (v1, v2, . . . , vn)such that the set {v1, v2, . . . , vn} is a basis for V.


Section 4.7

Ordered Basis: An ordered basis for vector spaceV is an ordered n-tuple of vectors (v1, v2, . . . , vn)such that the set {v1, v2, . . . , vn} is a basis for V.

(e1, e2) and (e2, e1) are two ordered bases for R2.


Coordinatization: Let B = (v1, v2, . . . , vn) be anordered basis for a vector space V.


Coordinatization: Let B = (v1, v2, . . . , vn) be anordered basis for a vector space V. Suppose thatw ∈ V such that

w = a1v1 + · · · + anvn


Coordinatization: Let B = (v1, v2, . . . , vn) be anordered basis for a vector space V. Suppose thatw ∈ V such that

w = a1v1 + · · · + anvn

Then [w]B, the coordinatization or coordinates of wwith respect to B is the n-vector [a1, a2, . . . , an].


Example 19: Let B = ([4, 2], [1, 3]) be an orderedbasis for R

2.



2. Note that

[4, 2] = 1[4, 2] + 0[1, 3].



2. Note that

[4, 2] = 1[4, 2] + 0[1, 3].

Hence, [4, 2]B = [1, 0].



2. Note that

[4, 2] = 1[4, 2] + 0[1, 3].

Hence, [4, 2]B = [1, 0]. Similarly,

[11, 13] = 2[4, 2] + 3[1, 3].



2. Note that

[4, 2] = 1[4, 2] + 0[1, 3].

Hence, [4, 2]B = [1, 0]. Similarly,

[11, 13] = 2[4, 2] + 3[1, 3].

Hence, [11, 13]B = [2, 3].


Example 20: Let

B = ([−4, 5,−1, 0,−1], [1,−3, 2, 2, 5], [1,−2, 1, 1, 3])

be an ordered basis of the subspace V of R5.


Example 20: Let

B = ([−4, 5,−1, 0,−1], [1,−3, 2, 2, 5], [1,−2, 1, 1, 3])


Compute [−23, 30,−7,−1,−7]B, [1, 2, 3, 4, 5]B.


Example 20: Let

B = ([−4, 5,−1, 0,−1], [1,−3, 2, 2, 5], [1,−2, 1, 1, 3])


Compute [−23, 30,−7,−1,−7]B, [1, 2, 3, 4, 5]B.

Solution: To find [−23, 30,−7,−1,−7]B,


Example 20: Let

B = ([−4, 5,−1, 0,−1], [1,−3, 2, 2, 5], [1,−2, 1, 1, 3])


Compute [−23, 30,−7,−1,−7]B, [1, 2, 3, 4, 5]B.

Solution: To find [−23, 30,−7,−1,−7]B, we need tosolve the following equation[−23, 30,−7,−1,−7] =a[−4, 5,−1, 0,−1] + b[1,−3, 2, 2, 5] + c[1,−2, 1, 1, 3]


or equivalently


or equivalently

−4a + b + c = −23

5a − 3b − 2c = 30

−a + 2b + c = −7

2b + c = −1

−a + 5b + 3c = −7


or equivalently

−4a + b + c = −23

5a − 3b − 2c = 30

−a + 2b + c = −7

2b + c = −1

−a + 5b + 3c = −7

To solve this system, note that the RREF of theaugmented matrix


−4 1 1 −235 −3 −2 30

−1 2 1 −70 2 1 −1

−1 5 3 −7


−4 1 1 −235 −3 −2 30

−1 2 1 −70 2 1 −1

−1 5 3 −7

is


−4 1 1 −235 −3 −2 30

−1 2 1 −70 2 1 −1

−1 5 3 −7

is

1 0 0 60 1 0 −20 0 1 30 0 0 00 0 0 0


−4 1 1 −235 −3 −2 30

−1 2 1 −70 2 1 −1

−1 5 3 −7

is

1 0 0 60 1 0 −20 0 1 30 0 0 00 0 0 0

Hence, the unique solution for the system is

a = 6, b = −2, c = 3

implies


−4 1 1 −235 −3 −2 30

−1 2 1 −70 2 1 −1

−1 5 3 −7

is

1 0 0 60 1 0 −20 0 1 30 0 0 00 0 0 0

Hence, the unique solution for the system is

a = 6, b = −2, c = 3

implies

[−23, 30,−7,−1,−7]B = [6,−2, 3].


To find [1, 2, 3, 4, 5]B, we need solve the followingsystem



−4a + b + c = 1

5a − 3b − 2c = 2

−a + 2b + c = 3

2b + c = 4

−a + 5b + 3c = 5



−4a + b + c = 1

5a − 3b − 2c = 2

−a + 2b + c = 3

2b + c = 4

−a + 5b + 3c = 5

To solve this system, note that the RREF of


−4 1 1 15 −3 −2 2

−1 2 1 30 2 1 4

−1 5 3 5


−4 1 1 15 −3 −2 2

−1 2 1 30 2 1 4

−1 5 3 5

is

1 0 0 00 1 0 00 0 1 00 0 0 10 0 0 0


−4 1 1 15 −3 −2 2

−1 2 1 30 2 1 4

−1 5 3 5

is

1 0 0 00 1 0 00 0 1 00 0 0 10 0 0 0

This system has no solution,


−4 1 1 15 −3 −2 2

−1 2 1 30 2 1 4

−1 5 3 5

is

1 0 0 00 1 0 00 0 1 00 0 0 10 0 0 0

This system has no solution, implies that the vector[1, 2, 3, 4, 5] is not in span(B) = V.


Coordinatization Method:Let V be a nontrivial subspace of R

n, letB = (v1, v2, . . . , vk) be an ordered basis for V, and letv ∈ R

n.




n. To compute [v]B, we perform the followingsteps:





Form an augmented matrix [A|v]





Form an augmented matrix [A|v] by using thevectors in B as the columns of A, in order,





Form an augmented matrix [A|v] by using thevectors in B as the columns of A, in order, andusing v as a column on the right.





Form an augmented matrix [A|v] by using thevectors in B as the columns of A, in order, andusing v as a column on the right.Find RREF([A|v]),





Form an augmented matrix [A|v] by using thevectors in B as the columns of A, in order, andusing v as a column on the right.Find RREF([A|v]), say [C|w] = RREF([A|v]).





Form an augmented matrix [A|v] by using thevectors in B as the columns of A, in order, andusing v as a column on the right.Find RREF([A|v]), say [C|w] = RREF([A|v]).If there is a row of [C|w] that contains all zeroson the left and has a nonzero entry on the right,





Form an augmented matrix [A|v] by using thevectors in B as the columns of A, in order, andusing v as a column on the right.Find RREF([A|v]), say [C|w] = RREF([A|v]).If there is a row of [C|w] that contains all zeroson the left and has a nonzero entry on the right,then v /∈ span(B) = V,





Form an augmented matrix [A|v] by using thevectors in B as the columns of A, in order, andusing v as a column on the right.Find RREF([A|v]), say [C|w] = RREF([A|v]).If there is a row of [C|w] that contains all zeroson the left and has a nonzero entry on the right,then v /∈ span(B) = V, i.e., coordinatization isnot possible.





Form an augmented matrix [A|v] by using thevectors in B as the columns of A, in order, andusing v as a column on the right.Find RREF([A|v]), say [C|w] = RREF([A|v]).If there is a row of [C|w] that contains all zeroson the left and has a nonzero entry on the right,then v /∈ span(B) = V, i.e., coordinatization isnot possible. Otherwise, v ∈ span(B) = V.


Eliminate all rows consisting entirely of zeros in[C|w] to obtain [Ik|y].


Eliminate all rows consisting entirely of zeros in[C|w] to obtain [Ik|y]. Then, [v]B = y,


Eliminate all rows consisting entirely of zeros in[C|w] to obtain [Ik|y]. Then, [v]B = y, the lastcolumn of [Ik|y].


Example 21: Let

B =

([

1 −20 1

]

,

[

2 −11 0

]

,

[

1 −13 1

])

be an ordered basis of the subspace W of M22.


Example 21: Let

B =

([

1 −20 1

]

,

[

2 −11 0

]

,

[

1 −13 1

])

be an ordered basis of the subspace W of M22.

Compute [v]B if exists, where v =

[

−3 −20 3

]

.


Solution: Consider

[A|v] =

1 2 1 −3−2 −1 −1 −2

0 1 3 01 0 1 3


Solution: Consider

[A|v] =

1 2 1 −3−2 −1 −1 −2

0 1 3 01 0 1 3

Note that the row reduced echelon form is

RREF[A|v] =

1 0 0 20 1 0 −30 0 1 10 0 0 0


The row reduced matrix contains no rows with allzero entries on the left and a nonzero entry on theright, so [v]B exists,


The row reduced matrix contains no rows with allzero entries on the left and a nonzero entry on theright, so [v]B exists, and

[v]B = [2,−3, 1].


Fundamental properties of Coordinatization: LetB = (v1, v2, . . . , vk) be an ordered basis for a vectorspace V. Suppose w1, w2, . . . , wk ∈ V anda1, a2, . . . , ak are scalars. Then



[w1 + w2]B = [w1]B + [w2]B



[w1 + w2]B = [w1]B + [w2]B[a1w1]B = a1[w1]B



[w1 + w2]B = [w1]B + [w2]B[a1w1]B = a1[w1]B[a1w1 + a2w2 + . . . akwk]B= a1[w1]B + a2[w2]B + · · · + ak[wk]B.


Exercise: Let

B = (3x2 − x + 2, x2 + 2x − 3, 2x2 + 3x − 1)

be an ordered basis of the subspace W of P2.


Exercise: Let

B = (3x2 − x + 2, x2 + 2x − 3, 2x2 + 3x − 1)

be an ordered basis of the subspace W of P2.Compute [v]B if exists, where v = 13x2 − 5x + 20.


Exercise: Let

B = (3x2 − x + 2, x2 + 2x − 3, 2x2 + 3x − 1)

be an ordered basis of the subspace W of P2.Compute [v]B if exists, where v = 13x2 − 5x + 20.

Answer: [v]B = [4,−5, 3].


Exercise: Let

B = ([−4, 5,−1, 0,−1], [1,−3, 2, 2, 5], [1,−2, 1, 1, 3])

be an ordered basis of the subspace W of R5.


Exercise: Let

B = ([−4, 5,−1, 0,−1], [1,−3, 2, 2, 5], [1,−2, 1, 1, 3])


Consider x = [1, 0,−1, 0, 4], y = [0, 1,−1, 0, 3] andz = [0, 0, 0, 1, 5].


Exercise: Let

B = ([−4, 5,−1, 0,−1], [1,−3, 2, 2, 5], [1,−2, 1, 1, 3])


Consider x = [1, 0,−1, 0, 4], y = [0, 1,−1, 0, 3] andz = [0, 0, 0, 1, 5]. Compute [2x − 7y + 3z]B.


Exercise: Let

B = ([−4, 5,−1, 0,−1], [1,−3, 2, 2, 5], [1,−2, 1, 1, 3])


Consider x = [1, 0,−1, 0, 4], y = [0, 1,−1, 0, 3] andz = [0, 0, 0, 1, 5]. Compute [2x − 7y + 3z]B.

Answer: [2x − 7y + 3z]B = [−2, 9,−15].


Example 22: LetC = ([−4, 5,−1, 0,−1], [1,−3, 2, 2, 5], [1,−2, 1, 1, 3]) bean ordered basis of the subspace W of R

5.



5. Usingsimplified span method on C, compute an orderedbasis B = (x, y, z) for W.



5. Usingsimplified span method on C, compute an orderedbasis B = (x, y, z) for W. Also, compute[x]C, [y]C, [z]C.



5. Usingsimplified span method on C, compute an orderedbasis B = (x, y, z) for W. Also, compute[x]C, [y]C, [z]C.

Solution: We have the following augmented matrix

[

A x y z]

=

−4 1 1 1 0 05 −3 −2 0 1 0

−1 2 1 −1 −1 00 2 1 0 0 1

−1 5 3 4 3 5


Row reduce echelon form of the above matrix is



1 0 0 1 1 10 1 0 −5 −4 −30 0 1 10 8 70 0 0 0 0 00 0 0 0 0 0



1 0 0 1 1 10 1 0 −5 −4 −30 0 1 10 8 70 0 0 0 0 00 0 0 0 0 0

Clearly, [x]C = [1,−5, 10], [y]C = [1,−4, 8] and[z]C = [1,−3, 7].



1 0 0 1 1 10 1 0 −5 −4 −30 0 1 10 8 70 0 0 0 0 00 0 0 0 0 0

Clearly, [x]C = [1,−5, 10], [y]C = [1,−4, 8] and[z]C = [1,−3, 7]. Here, the matrix

P =

1 1 1−5 −4 −310 8 7



1 0 0 1 1 10 1 0 −5 −4 −30 0 1 10 8 70 0 0 0 0 00 0 0 0 0 0

Clearly, [x]C = [1,−5, 10], [y]C = [1,−4, 8] and[z]C = [1,−3, 7]. Here, the matrix

P =

1 1 1−5 −4 −310 8 7

is called the transition matrix

from B-coordinates to C-coordinates.Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 79 / 122

Transition Matrix:


Transition Matrix: Suppose that V is a nontrivialn-dimensional vector space with ordered bases Band C.


Transition Matrix: Suppose that V is a nontrivialn-dimensional vector space with ordered bases Band C. Let P be the n × n matrix whose ith column,for 1 ≤ i ≤ n, equals [bi]C, where bi is the ith basisvector in B.


Transition Matrix: Suppose that V is a nontrivialn-dimensional vector space with ordered bases Band C. Let P be the n × n matrix whose ith column,for 1 ≤ i ≤ n, equals [bi]C, where bi is the ith basisvector in B. Then P is called the transition matrixfrom B-coordinates to C-coordinates


Transition Matrix: Suppose that V is a nontrivialn-dimensional vector space with ordered bases Band C. Let P be the n × n matrix whose ith column,for 1 ≤ i ≤ n, equals [bi]C, where bi is the ith basisvector in B. Then P is called the transition matrixfrom B-coordinates to C-coordinates (or transitionmatrix from B to C).


Transition Matrix Method:


Transition Matrix Method: To find the transitionmatrix P from B to C,


Transition Matrix Method: To find the transitionmatrix P from B to C, we apply row reduction on

1st 2nd kth 1st 2nd kth

vector vector ·· vector vector vector ·· vectorin in in in in inC C C B B B


Transition Matrix Method: To find the transitionmatrix P from B to C, we apply row reduction on

1st 2nd kth 1st 2nd kth

vector vector ·· vector vector vector ·· vectorin in in in in inC C C B B B

to produce[

Ik Prows of zeroes

]


Example 23: For the ordered bases

B =

([

7 30 0

]

,

[

1 20 −1

]

,

[

1 −10 1

])



B =

([

7 30 0

]

,

[

1 20 −1

]

,

[

1 −10 1

])

and

C =

([

22 70 2

]

,

[

12 40 1

]

,

[

33 120 2

])

of U2 (the set of 2 × 2 upper triangular matrices),



B =

([

7 30 0

]

,

[

1 20 −1

]

,

[

1 −10 1

])

and

C =

([

22 70 2

]

,

[

12 40 1

]

,

[

33 120 2

])

of U2 (the set of 2 × 2 upper triangular matrices), findthe transition matrix P from B to C.


Solution: Apply row reduction on



22 12 33 7 1 17 4 12 3 2 −10 0 0 0 0 02 1 2 0 −1 1



22 12 33 7 1 17 4 12 3 2 −10 0 0 0 0 02 1 2 0 −1 1

we get

1 0 0 1 −2 10 1 0 −4 1 10 0 1 1 1 −10 0 0 0 0 0


The transition matrix P from B to C is


The transition matrix P from B to C is

1 −2 1−4 1 11 1 −1

.


Example 24: LetC = (a, b, c) = ([1, 0, 1], [1, 1, 0], [0, 0, 1]) andB = (x, y, z) be ordered bases of R

3.



3. Let

P =

1 1 22 1 1−1 −1 1

be the transition matrix from B to C.



3. Let

P =

1 1 22 1 1−1 −1 1

be the transition matrix from B to C. Find the basisB.



3. Let

P =

1 1 22 1 1−1 −1 1


Solution:

x = 1 · a + 2 · b − 1 · c = [3, 2, 0]


3. Let

P =

1 1 22 1 1−1 −1 1


Solution:

x = 1 · a + 2 · b − 1 · c = [3, 2, 0]


3. Let

P =

1 1 22 1 1−1 −1 1


Solution:

x = 1 · a + 2 · b − 1 · c = [3, 2, 0]

y = 1 · a + 1 · b − 1 · c = [2, 1, 0]

z = 2 · a + 1 · b + 1 · c = [3, 1, 3].Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 85 / 122

Hence,B = ([3, 2, 0], [2, 1, 0], [3, 1, 3]).


Hence,B = ([3, 2, 0], [2, 1, 0], [3, 1, 3]).

Change of Coordinates Using the TransitionMatrix

Theorem: Suppose that V is a nontrivialn-dimensional vector space with ordered bases Band C.


Hence,B = ([3, 2, 0], [2, 1, 0], [3, 1, 3]).


Theorem: Suppose that V is a nontrivialn-dimensional vector space with ordered bases Band C. Let P be an n × n matrix.


Hence,B = ([3, 2, 0], [2, 1, 0], [3, 1, 3]).


Theorem: Suppose that V is a nontrivialn-dimensional vector space with ordered bases Band C. Let P be an n × n matrix. Then P is thetransition matrix from B to C if and only if for everyv ∈ V, P [v]B = [v]C.



B =

([

7 30 0

]

,

[

1 20 −1

]

,

[

1 −10 1

])



B =

([

7 30 0

]

,

[

1 20 −1

]

,

[

1 −10 1

])

and

C =

([

22 70 2

]

,

[

12 40 1

]

,

[

33 120 2

])

of U2 (set of 2 × 2 upper triangular matrices).



B =

([

7 30 0

]

,

[

1 20 −1

]

,

[

1 −10 1

])

and

C =

([

22 70 2

]

,

[

12 40 1

]

,

[

33 120 2

])

of U2 (set of 2 × 2 upper triangular matrices). Find

[v]B and [v]C, where v =

[

25 240 −9

]

.


Solution: Clearly,[

25 240 −9

]

= 4

[

7 30 0

]

+ 3

[

1 20 −1

]

− 6

[

1 −10 1

]


Solution: Clearly,[

25 240 −9

]

= 4

[

7 30 0

]

+ 3

[

1 20 −1

]

− 6

[

1 −10 1

]

Hence, [v]B = [4, 3,−6]T .


Solution: Clearly,[

25 240 −9

]

= 4

[

7 30 0

]

+ 3

[

1 20 −1

]

− 6

[

1 −10 1

]

Hence, [v]B = [4, 3,−6]T . Now, since [v]C = P [v]Band

P =

1 −2 1−4 1 11 1 −1

(see Example 23)

implies


Solution: Clearly,[

25 240 −9

]

= 4

[

7 30 0

]

+ 3

[

1 20 −1

]

− 6

[

1 −10 1

]


P =

1 −2 1−4 1 11 1 −1

(see Example 23)

implies [v]C = [−8,−19, 13]T .


Solution: Clearly,[

25 240 −9

]

= 4

[

7 30 0

]

+ 3

[

1 20 −1

]

− 6

[

1 −10 1

]


P =

1 −2 1−4 1 11 1 −1

(see Example 23)

implies [v]C = [−8,−19, 13]T . Clearly,[

25 240 −9

]

= −8

[

22 70 2

]

− 19

[

12 40 1

]

+ 13

[

33 120 2

]


Theorem: Let B and C be be ordered bases for anontrivial finite dimensional vector space V, and letP be the transition matrix from B to C. Then P isnonsingular, and P−1 is the transition matrix from Cto B.


Example 26: For an ordered basisB = ([1,−4, 1, 2, 1], [6,−24, 5, 8, 3], [3,−12, 3, 6, 2]) of asubspace V of R

5.



5.Use the Simplified Span Method to find asecond ordered basis C.




Solution: Consider

B =

1 −4 1 2 16 −24 5 8 33 −12 3 6 2




Solution: Consider

B =

1 −4 1 2 16 −24 5 8 33 −12 3 6 2

Note that

RREF(B) =

1 −4 0 −2 00 0 1 4 00 0 0 0 1


C = ([1,−4, 0,−2, 0], [0, 0, 1, 4, 0], [0, 0, 0, 0, 1]) .


C = ([1,−4, 0,−2, 0], [0, 0, 1, 4, 0], [0, 0, 0, 0, 1]) .

Find the transition matrix P from B to C.


C = ([1,−4, 0,−2, 0], [0, 0, 1, 4, 0], [0, 0, 0, 0, 1]) .

Find the transition matrix P from B to C.

Answer:

P =

1 6 31 5 31 3 2


Find the transition matrix Q from C to B.


Find the transition matrix Q from C to B.

Answer:

Q = P−1 =

1 −3 31 −1 0−2 3 −1


For the given vector v = [2,−8,−2,−12, 3] ∈ V,calculate [v]B and [v]C.



[B|v] =

1 6 3 2−4 −24 −12 −8

1 5 3 −22 8 6 −121 3 2 3

=

1 0 0 170 1 0 40 0 1 −130 0 0 00 0 0 0



[B|v] =

1 6 3 2−4 −24 −12 −8

1 5 3 −22 8 6 −121 3 2 3

=

1 0 0 170 1 0 40 0 1 −130 0 0 00 0 0 0

Thus,[v]B = [17, 4,−13]



[B|v] =

1 6 3 2−4 −24 −12 −8

1 5 3 −22 8 6 −121 3 2 3

=

1 0 0 170 1 0 40 0 1 −130 0 0 00 0 0 0

Thus,[v]B = [17, 4,−13]

Since P [v]B = [v]C implies



[B|v] =

1 6 3 2−4 −24 −12 −8

1 5 3 −22 8 6 −121 3 2 3

=

1 0 0 170 1 0 40 0 1 −130 0 0 00 0 0 0

Thus,[v]B = [17, 4,−13]

Since P [v]B = [v]C implies

[v]C = [2,−2, 3].


Exercise: For the ordered bases

B =(

2x2 + 3x − 1, 8x2 + x + 1, x2 + 6)

and

C =(

x2 + 3x + 1, 3x2 + 4x + 1, 10x2 + 17x + 5)

of P2, find the transition matrix P from B to C.


Exercise: For the ordered bases

B =(

2x2 + 3x − 1, 8x2 + x + 1, x2 + 6)

and

C =(

x2 + 3x + 1, 3x2 + 4x + 1, 10x2 + 17x + 5)

of P2, find the transition matrix P from B to C.

Answer: P =

20 −30 −6924 −24 −80−9 11 31


Exercise: Let P =

2 2 11 −1 21 1 1

be the transition

matrix from B to C. If C =

201

,

120

,

111

, find

the basis B.


Exercise: Let P =

2 2 11 −1 21 1 1

be the transition

matrix from B to C. If C =

201

,

120

,

111

, find

the basis B.

Answer: B =

633

,

4−13

,

552


Exercise: For an ordered basis

B = ([3,−1, 4, 6], [6, 7,−3,−2], [−4,−3, 3, 4], [−2, 0, 1, 2])

of a subspace W of R4, perform the following steps:

1 Use the Simplified Span Method to find asecond ordered basis C.

2 Find the transition matrix P from B to C.3 Find the transition matrix Q from C to B.4 For the given vector v = [10, 14, 3, 12], calculate

[v]B and [v]C.


Section 5.2

The Matrix of a linear transformation:


Section 5.2

The Matrix of a linear transformation: Let V andW be two finite dimensional real vector spaces suchthat dim(V) = n and dim(W) = m. LetB = {v1, v2, . . . , vn} and C = {w1, w2, . . . , wm} be anordered basis of V and W, respectively.


Section 5.2


Let L : V → W be any linear transformation.


Section 5.2


Let L : V → W be any linear transformation. Forvj ∈ V, L(vj) ∈ W. For each j, 1 ≤ j ≤ n. Since C isa basis of W, for aij ∈ R, we can write


Section 5.2


Let L : V → W be any linear transformation. Forvj ∈ V, L(vj) ∈ W. For each j, 1 ≤ j ≤ n. Since C isa basis of W, for aij ∈ R, we can write

L(vj) = a1jw1 + a2jw2 + · · · + amjwm


Thus, we have

L(v1) = a11w1 + a21w2 + · · +am1wm

L(v2) = a12w1 + a22w2 + · · +am2wm

· · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · ·

L(vn) = a1nw1 + a2nw2 + · · +amnwm


Define

ABC =

a11 a12 · · · a1n

a21 a22 · · · a2n... ... ... ...

am1 am2 · · · amn

m×n

.

The matrix ABC is called the matrix of lineartransformation L w.r.t. the bases B and C.


Define

ABC =

a11 a12 · · · a1n

a21 a22 · · · a2n... ... ... ...

am1 am2 · · · amn

m×n

.

The matrix ABC is called the matrix of lineartransformation L w.r.t. the bases B and C.

Remark: ith column of the matrix ABC is [L(vi)]C.


Theorem: Let V and W be non-trivial vector spaces,with dim(V) = n and dim(W) = m. LetB = (v1, . . . , vn) and C = (w1, . . . , wm) be orderedbases for V and W, respectively. Let L : V → W be aLT.


Theorem: Let V and W be non-trivial vector spaces,with dim(V) = n and dim(W) = m. LetB = (v1, . . . , vn) and C = (w1, . . . , wm) be orderedbases for V and W, respectively. Let L : V → W be aLT. Then there is a unique m × n matrix ABC suchthat ABC[v]B = [L(v)]C, for all v ∈ V.


Theorem: Let V and W be non-trivial vector spaces,with dim(V) = n and dim(W) = m. LetB = (v1, . . . , vn) and C = (w1, . . . , wm) be orderedbases for V and W, respectively. Let L : V → W be aLT. Then there is a unique m × n matrix ABC suchthat ABC[v]B = [L(v)]C, for all v ∈ V. Furthermore,for 1 ≤ i ≤ n, the ith column of ABC = [L(vi)]C.


Example: Consider the LT L : P1 → P2, given by

L(p(x)) = xp(x)

with ordered bases B = (x, 1) andC = (x2, x − 1, x + 1) of P1 and P2, respectively.



L(p(x)) = xp(x)

with ordered bases B = (x, 1) andC = (x2, x − 1, x + 1) of P1 and P2, respectively.Compute ABC .



L(p(x)) = xp(x)


Solution: SinceL(x) = x2 = 1(x2) + 0(x − 1) + 0(x + 1) so that



L(p(x)) = xp(x)


Solution: SinceL(x) = x2 = 1(x2) + 0(x − 1) + 0(x + 1) so that

[L(x)]C =

100

.


Similarly, L(1) = x = 0(x2) + 1

2(x − 1) + 1

2(x + 1)

implies


Similarly, L(1) = x = 0(x2) + 1

2(x − 1) + 1

2(x + 1)

implies

[L(1)]C =

01/21/2

.


Similarly, L(1) = x = 0(x2) + 1

2(x − 1) + 1

2(x + 1)

implies

[L(1)]C =

01/21/2

.

Hence,

ABC =

1 00 1/20 1/2

.


Method for computing ABC: Let B = (v1, . . . , vn)and C = (w1, . . . , wm) be ordered bases for R

n andR

m, respectively. Also, let L : Rn → R

m be a LT.



n andR


m be a LT.Compute L(vi) for all i = 1, 2, . . . , n.



n andR


m be a LT.Compute L(vi) for all i = 1, 2, . . . , n.Form the augmented matrix

[w1 w2 . . . wm | L(v1)| L(v2)| . . . |L(vn)]



n andR



[w1 w2 . . . wm | L(v1)| L(v2)| . . . |L(vn)]

Apply row reduction on

[w1 w2 . . . wm | L(v1)| L(v2)| . . . |L(vn)].



n andR



[w1 w2 . . . wm | L(v1)| L(v2)| . . . |L(vn)]

Apply row reduction on

[w1 w2 . . . wm | L(v1)| L(v2)| . . . |L(vn)].

to produce [Im | ABC].Jitender Kumar (BITS PILANI) Mathematics-II (MATH F112) 103 / 122

Example: Consider the LT L : R2 → P2, given by

L([a, b]) = (−a + 5b)x2 + (3a − b)x + 2b

with ordered bases B = ([5, 3], [3, 2]) and

C = (3x2 − 2x,−2x2 + 2x − 1, x2 − x + 1)

of R2 and P2, respectively.



L([a, b]) = (−a + 5b)x2 + (3a − b)x + 2b


C = (3x2 − 2x,−2x2 + 2x − 1, x2 − x + 1)

of R2 and P2, respectively. Compute ABC.



L([a, b]) = (−a + 5b)x2 + (3a − b)x + 2b


C = (3x2 − 2x,−2x2 + 2x − 1, x2 − x + 1)


Solution: Since L[5, 3] = 10x2 + 12x + 6



L([a, b]) = (−a + 5b)x2 + (3a − b)x + 2b


C = (3x2 − 2x,−2x2 + 2x − 1, x2 − x + 1)


Solution: Since L[5, 3] = 10x2 + 12x + 6 andL[3, 2] = 7x2 + 7x + 4. Consider



L([a, b]) = (−a + 5b)x2 + (3a − b)x + 2b


C = (3x2 − 2x,−2x2 + 2x − 1, x2 − x + 1)


Solution: Since L[5, 3] = 10x2 + 12x + 6 andL[3, 2] = 7x2 + 7x + 4. Consider

3 −2 1 10 7−2 2 −1 12 7

0 −1 1 6 4


RREF of the above matrix is

1 0 0 22 140 1 0 62 390 0 1 68 43


RREF of the above matrix is

1 0 0 22 140 1 0 62 390 0 1 68 43

so that

ABC =

22 1462 3968 43


Example: Consider the LT L : P3 → P2, given byL(p) = p′.


Example: Consider the LT L : P3 → P2, given byL(p) = p′. Compute ABC with respect to standardbases B = {x3, x2, x, 1} of P3 and C = {x2, x, 1} ofP2.


Example: Consider the LT L : P3 → P2, given byL(p) = p′. Compute ABC with respect to standardbases B = {x3, x2, x, 1} of P3 and C = {x2, x, 1} ofP2. Using ABC, find L(4x3 − 5x2 + 6x − 7).



Solution: Standard basis of P3 is



Solution: Standard basis of P3 is {x3, x2, x, 1}.



Solution: Standard basis of P3 is {x3, x2, x, 1}. Since

L(x3) = 3x2, L(x2) = 2x, L(x) = 1, L(1) = 0, we have



Solution: Standard basis of P3 is {x3, x2, x, 1}. Since

L(x3) = 3x2, L(x2) = 2x, L(x) = 1, L(1) = 0, we have

ABC =

3 0 0 00 2 0 00 0 1 0


Since

[L(4x3 − 5x2 + 6x − 7)]C = ABC[(4x3 − 5x2 + 6x − 7)]B

Since

[L(4x3 − 5x2 + 6x − 7)]C = ABC[(4x3 − 5x2 + 6x − 7)]B

Since

[L(4x3 − 5x2 + 6x − 7)]C = ABC[(4x3 − 5x2 + 6x − 7)]B

= ABC

4−56−7

=

12−106

Thus, L(4x3 − 5x2 + 6x − 7) = 12x2 − 10x + 6


Since

[L(4x3 − 5x2 + 6x − 7)]C = ABC[(4x3 − 5x2 + 6x − 7)]B

= ABC

4−56−7

=

12−106

Thus, L(4x3 − 5x2 + 6x − 7) = 12x2 − 10x + 6

Also, note that

L(4x3−5x2+6x−7) = (4x3−5x2+6x−7)′ = 12x2−10x+6.


Example: Let the matrix of LT L : P1 → P1 with

respect to basis B = (x + 1, x − 1) be[

2 3−1 −2

]

.




2 3−1 −2

]

.

Find the matrix of L with respect to basis C = (x, 1).




2 3−1 −2

]

.

Find the matrix of L with respect to basis C = (x, 1).

Solution: Since ABB =

[

2 3−1 −2

]

, we have

L(x + 1) = 2(x + 1) − 1(x − 1) = x + 3

L(x − 1) = 3(x + 1) − 2(x − 1) = x + 5


L(ax + b) = L

(

a + b

2(x + 1) +

a − b

2(x − 1)

)


L(ax + b) = L

(

a + b

2(x + 1) +

a − b

2(x − 1)

)

L(ax + b) =

(

a + b

2(x + 3) +

a − b

2(x + 5)

)


L(ax + b) = L

(

a + b

2(x + 1) +

a − b

2(x − 1)

)

L(ax + b) =

(

a + b

2(x + 3) +

a − b

2(x + 5)

)

so that L(x) = x + 4 and L(1) = −1.


L(ax + b) = L

(

a + b

2(x + 1) +

a − b

2(x − 1)

)

L(ax + b) =

(

a + b

2(x + 3) +

a − b

2(x + 5)

)

so that L(x) = x + 4 and L(1) = −1.Hence,

ACC =

[

1 04 −1

]

.


Exercise: Consider the LT L : R3 → R

2, given byL([x, y, z]) = [x + y, y − z].



2, given byL([x, y, z]) = [x + y, y − z]. Compute ABC withrespect to bases B = ([1, 0, 1], [0, 1, 1], [1, 1, 1]) andC = ([1, 2], [−1, 1]).



2, given byL([x, y, z]) = [x + y, y − z]. Compute ABC withrespect to bases B = ([1, 0, 1], [0, 1, 1], [1, 1, 1]) andC = ([1, 2], [−1, 1]).

Answer: ABC =

[

0 1/3 2/3−1 −2/3 −4/3

]


Exercise: Consider the LT L : P3 → M22, given by

L(ax3 + bx2 + cx + d) =

[

−3a − 2c −b + 4d4b − c + 3d −6a − b + 2d

]

.

Compute ABC with respect to standard bases for P3

and M22.


Exercise: Consider the LT L : P3 → M22, given by

L(ax3 + bx2 + cx + d) =

[

−3a − 2c −b + 4d4b − c + 3d −6a − b + 2d

]

.

Compute ABC with respect to standard bases for P3

and M22.

Answer:

ABC =

−3 0 −2 00 −1 0 40 4 −1 3−6 −1 0 2


Exercise: Consider the LT L : R2 → P2, given by

L([a, b]) = (−a + 5b)x2 + (3a − b)x + 2b.



L([a, b]) = (−a + 5b)x2 + (3a − b)x + 2b.

Compute ABC with respect to bases B = ([5, 3], [3, 2])and C = (3x2 − 2x,−2x2 + 2x − 1, x2 − x + 1).



L([a, b]) = (−a + 5b)x2 + (3a − b)x + 2b.

Compute ABC with respect to bases B = ([5, 3], [3, 2])and C = (3x2 − 2x,−2x2 + 2x − 1, x2 − x + 1).

Answer:

ABC =

22 1462 3968 43


Exercise: Let B = ([1, 2], [2,−1]) andC = ([1, 0], [0, 1]) be ordered bases for R

2. If

L : R2 → R

2 be a LT such that ABC =

[

4 32 −4

]

. Find

L([5, 5]), Also, find L([x, y]) for all [x, y] ∈ R2.


Exercise: Let B = ([1, 2], [2,−1]) andC = ([1, 0], [0, 1]) be ordered bases for R

2. If

L : R2 → R


[

4 32 −4

]

. Find

L([5, 5]), Also, find L([x, y]) for all [x, y] ∈ R2.

Answer: L([5, 5]) = [15, 2].


Exercise: Let

B = ([1, 1, 0, 0], [0, 1, 1, 0], [0, 0, 1, 1], [0, 0, 0, 1]) and

C = ([1, 1, 1], [1, 2, 3], [1, 0, 0])

be ordered bases for R4 and R

3, respectively. If

L : R4 → R


1 1 0 00 1 1 00 1 0 1

.

Find L?


Exercise: Let

B = ([1, 1, 0, 0], [0, 1, 1, 0], [0, 0, 1, 1], [0, 0, 0, 1]) and

C = ([1, 1, 1], [1, 2, 3], [1, 0, 0])

be ordered bases for R4 and R

3, respectively. If

L : R4 → R


1 1 0 00 1 1 00 1 0 1

.

Find L?

Answer:

L([x1, x2, x3, x4]) = [−2x1+3x2 +x4, x2 +2x3, x2 +3x3].


Matrix for the composition of LinearTransformations:



Theorem: Let V1,V2 and V3 be nontrivial finitedimensional vector spaces with ordered bases B, Cand D, respectively.



Theorem: Let V1,V2 and V3 be nontrivial finitedimensional vector spaces with ordered bases B, Cand D, respectively. Let L1 : V1 → V2 be a lineartransformation with matrix ABC and let L2 : V2 → V3

be a linear transformation with matrix ACD. Thenmatrix

ABD = ACDABC

is the matrix of linear transformationL2 ◦ L1 : V1 → V3 with respect to the bases B and D.


Example: Let L1 : R2 → R

2 and L2 : R2 → R

3

defined byL1([x, y]) = [y, x]

L2([x, y]) = [x + y, x − y, y]

Find the matrix of L1 and L2 with respect to thestandard basis in each case.


Example: Let L1 : R2 → R

2 and L2 : R2 → R

3

defined byL1([x, y]) = [y, x]

L2([x, y]) = [x + y, x − y, y]

Find the matrix of L1 and L2 with respect to thestandard basis in each case.Find the matrix of L2 ◦ L1 with respect tostandard basis of R

2 and R3.


Answer: The matrix of L1 w.r. to B = {[1, 0], [0, 1]} is

ABB =

[

0 11 0

]

.


Answer: The matrix of L1 w.r. to B = {[1, 0], [0, 1]} is

ABB =

[

0 11 0

]

.

The matrix of L2 w. r. to the bases C = [1, 0], [0, 1]and D = {[1, 0, 0], [0, 1, 0], [0, 0, 1]} is

ACD =

1 11 −10 1


Thus, the matrix ABD of the linear transformationL2 ◦ L1 : R

2 → R3 w.r. to the bases B and D is

ABD = ACDABB =

1 11 −10 1

[

0 11 0

]

=

1 1−1 11 0


Theorem: Let L : V → W be a linear transformationbetween n-dimensional vector spaces V and W andlet B and C are ordered bases for V and W,respectively. Then L is an isomorphism (orinvertible) if and only if the matrix representationABC for L with respect to B and C is nonsingular.


Theorem: Let L : V → W be a linear transformationbetween n-dimensional vector spaces V and W andlet B and C are ordered bases for V and W,respectively. Then L is an isomorphism (orinvertible) if and only if the matrix representationABC for L with respect to B and C is nonsingular.

In this case If DCB is the matrix for L−1 with respectto C and B then A−1

BC = DCB.


Example: Let L1 and L2 be linear operators on R3.

Let

A =

0 −2 10 −1 01 0 0

and B =

1 0 0−2 0 10 −3 0

be matrices for L1 and L2 respectively, with respectto standard basis.



Let

A =

0 −2 10 −1 01 0 0

and B =

1 0 0−2 0 10 −3 0


Show that L1 and L2 are isomorphisms.



Let

A =

0 −2 10 −1 01 0 0

and B =

1 0 0−2 0 10 −3 0



Answer: Since rank(A) = 3 and rank(B) = 3, thematrices A and B are nonsingular.



Let

A =

0 −2 10 −1 01 0 0

and B =

1 0 0−2 0 10 −3 0



Answer: Since rank(A) = 3 and rank(B) = 3, thematrices A and B are nonsingular. Hence, L1 and L2

are isomorphisms.


Find matrices for L−1

1and L−1

2.



1and L−1

2.

Answer: Since L−1

1(v) = A−1(v).



1and L−1

2.

Answer: Since L−1

1(v) = A−1(v). Using row

reduction (see Chapter 3), we have

A−1 =

0 0 10 −1 01 −2 0



1and L−1

2.

Answer: Since L−1

1(v) = A−1(v). Using row

reduction (see Chapter 3), we have

A−1 =

0 0 10 −1 01 −2 0

Similarly, L−1

2(v) = B−1(v), where

B−1 =

1 0 00 0 −1/32 1 0

.


Thank You


Mathematics-II (MATH F112)universe.bits-pilani.ac.in/uploads/L1_Chapter 5.pdfMathematics-II (MATH F112) Linear Algebra Jitender Kumar Department of Mathematics Birla Institute of Technology

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