Linear Algebra and Partial Differential Equations T Veerarajan Download PDF Chapter
Linear Algebra and Partial Differential Equations T Veerarajan Download PDF Chapter
Linear Algebra and Partial Differential Equations T Veerarajan Download PDF Chapter
Equations T Veerarajan
Visit to download the full and correct content document:
https://ebookmass.com/product/linear-algebra-and-partial-differential-equations-t-veer
arajan/
Linear Algebra and
Partial Differential Equations
About the Author
T Veerarajan
Dean (Retd)
Department of Mathematics
Velammal College of Engineering and Technology
Viraganoor, Madurai
Tamil Nadu
Information contained in this work has been obtained by McGraw Hill Education (India), from sources
believed to be reliable. However, neither McGraw Hill Education (India) nor its authors guarantee the accuracy
or completeness of any information published herein, and neither McGraw Hill Education (India) nor its
authors shall be responsible for any errors, omissions, or damages arising out of use of this information. This
work is published with the understanding that McGraw Hill Education (India) and its authors are supplying
information but are not attempting to render engineering or other professional services. If such services are
required, the assistance of an appropriate professional should be sought.
Typeset at Text-o-Graphics, B-1/56, Aravali Apartment, Sector-34, Noida 201 301, and printed at
Linear Algebra and Partial Differential Equations, has been designed specifically
to cater to the needs of third semester B Tech students. The current edition aims
at preparing the students for examination alongside strengthening the fundamental
concepts related to Partial Differential Equations. Lucidity of the text, ample worked
examples and notes highlighted within the text help students navigate through
complex topics seamlessly. Stepwise explanation, use of multiple methods of
problem solving, and additional information presented by the means of appendices
are few other notable features of the content.
Salient Features
∑ Strict adherence to the syllabus
∑ Stepwise solutions of solved problems which will enable students to score
marks
Chapter Organization
The book is organised into 5 units. Unit 1 deals with Vector Spaces. Unit 2 explains
in detail about Linear Transformation. Unit 3 discusses the Inner Product Spaces.
Unit 4 focuses on Partial Differential Equations while Chapter 5 elaborates on Fourier
Series Solutions of Partial Differential Equations.
Acknowledgements
I have great pleasure in dedicating this book to the students and teachers. I hope that
both the faculty and the students will receive the present edition as willingly as the
earlier editions and my other books.
A number of reviewers took pains to provide valuable feedback for the book. We
are grateful to all of them.
T Veerarajan
Contents
Preface v
Roadmap to the Syllabus xi
Unit 1
Vector Spaces
Note
The Vector space is also referred to as the vector space over the field F or
linear space.
Note
The zero vector of Fn is 0 = (0, 0, ..., 0)
2. The set of all (m × n) matrices with entries from any field F is a vector space
over F w.r.t. the operations of matrix addition and scalar multiplication is
denoted by F m × n
Note
F1 × n = Fn
3. The set of all polynomials c0 + c1x + c2x2 + L + cnxn, with the coefficients ci
from any field F with respect to additions of polynomials and multiplication
by a constant.
4. The set V of all function from a non-empty set X into any arbitrary field F
for which addition and scalar multiplication are defined as follows is a vector
space.
The sum of any two functions f and g Œ V is the function (f + g) (x) = f(x) + g(x)
The product of a scalar c Œ F and a function f Œ V in the function cf ŒV,
defined by (cf) (x) = cf(x).
Subspaces
If W is a subset of a vector space V over a field F, such that W is itself a vector space
over F w.r.t. vector addition and scalar multiplication [viz., (1) W is non-empty,
(2) v, w Œ W implies v + w Œ W and (3) v Œ W implies c v Œ W for every c Œ F], then
W is called a sub-space of V.
Examples of Subspaces
1. If V is R3, then the set W consisting of those vectors whose first component
is zero. i.e., W = {(0, a, b): a, b Œ R} is a sub-space of V.
2. If V is the space of all n × n matrices, then the set of all symmetric matrices
of order n is a sub-space of V.
3. If V is any space, then the set {0} consisting of the zero vector alone and the
entire space V are sub-spaces V.
Span
If S is a non-empty sub set of a vector space V, the set of all linear combinations of
vectors in S is a subspace of V and is called the span of S and denoted by L(S). The
subspace L(S) is said to be generated by S.
If L(S) = V, then V is said to be finitely generated by S.
Examples
1. The vector e1 = (1, 0, 0), e2 = (0, 1, 0) and e3 = (0, 0, 1) and span the vector
space R, for, any vector (a, b, c) in R3 can be expressed as a linear combination
of e1, e2, e3 as (a, b, c) = ae1 + be2 + ce3
2. The polynomial 1, t, t2, ... generate the vector space of all polynomials in t,
as any polynomial can be expressed as a linear combination of 1, t, t2, ... .
Vector Spaces
1-3
Examples
1. The vectors e1, (1, 0, 0, ..., 0), e2 = (0, 1, 0, ..., 0), e3 = (0, 0, 1, ..., 0), en = (0,
0, 0, ..., 1) form a basis of Rn, called the standard basis and dim (Rn) = n.
2. If V is the vector space of all ( m × n) matrices over F, then dim V = mn.
In particular, if V is the vector space of all (2 × 2) matrices over R, then dim
V = 4.
Ê 1 0ˆ Ê 0, 1ˆ Ê 0 0ˆ Ê 0 0ˆ
The matrices Á ˜ ,Á ˜ ,Á ˜ and Á from the basis of V.
Ë 0 0¯ Ë 0, 0¯ Ë 1 0¯ Ë 0 1˜¯
3. If V is the vector space of polynomials in t of degree n, then dim (V) = n + 1,
for the linearly independent set {1, t, t2, ..., tn} is a basis of V.
Example 1
Determine whether the vector v = (3, 9, –4, –2) belongs to the space spanned by
u1 = (1, –2, 0, 3), u2 = (2, 3, 0, –1) and u3 = (2, –1, 2, 1)
If v belongs to the space spanned by u1, u2 and u3, then constants k1, k2, k3 should
exist such that v = k1u1 + k2u2 + k3u3.
viz., (3, 9, –4, –2) = k1(1, –2, 0, 3) + k2(2, 3, 0, –1) + k2(2, –1, 2, 1)
viz., k1 + 2k2 + 2k3 = 3 (1)
–2k1 + 3k2 – k3 = 9 (2)
2k3 = –4 (3)
Linear Algebra and Partial Differential Equations
1-4
and 3k1 – k2 + k3 = –2 (4)
Equations (1), (2), (3) and (4) are satisfied by k1 = 1, k2 = 3 and k3 = –2
\ The vector v belongs to the space spanned by the vectors u1, u2, u3.
Example 2
Find whether the vector (2, 4, 6, 7, 8) is in the subspace of R5 spanned by (1, 2, 0, 3,
0), (0, 0, 1, 4, 0) and (0, 0, 0, 0, 1)
If possible, let (2, 4, 6, 7, 8) = k1(1, 2, 0, 3, 0) + k2(0, 0, 1, 4, 0) + k3(0, 0, 0, 0, 1).
Then k1 = 2, 2k, = 4, k2 = 6, 3k1 + 4k2 = 7, k3 = 8
There equations are not satisfied by the same set of values of k1, k2 and k3,
\ The given vector does not belong to the subspace of R5.
Example 3
Examine the linear dependence or independence of the following vectors:
u1 = (1, –2, 3, 4), u2 = (–2, 4, –1, –3) and u3 = (–1, 2, 7, 6)
Writing the vectors as row vectors, one below the other, we have
Ê 1, -2, 3, 4 ˆ Ê 1, -2, 3, 4 ˆ
Á -2, 4, 1, -3˜ Á 0, 0, 5, 5 ˜ (u , u + 2u , u + u )
Á ˜ Á ˜ 1 2 1 3 1
Ë -1, 2, 7, 6 ¯ Ë 0, 0, 10, 10¯
Ê 1, -2, 3, 4ˆ
Á 0, 0, 5, 5˜ [u1 , u2 + 2u1 , u3 + u1 - 2 (u2 + 2u1 )
Á ˜
Ë 0, 0, 0, 0¯
Example 4
Find the maximum number of linearly independent vectors among the following and
express each of the remaining vectors as a linear combination of these.
u1 = (1, 2, 1); u2 = (4, 1, 2); u3 = (6, 5, 4) and u4 = (–3, 8, 1).
Writing the vectors as row vectors one below the other, we have
Ê 1, 2, 1ˆ Ê 1, 2, 1 ˆ
Á 4, 1, 2˜ Á 0, 7, -2˜ [u1 , u2 , - 4u1 , u3 - 6u1 , u4 + 3u1 ]
Á ˜ Á ˜
Á 6, 5, 4˜ Á 0, -7, -2˜ [u1¢, u2¢ , u3¢ , u2¢ say]
Á - 3, 1˜¯ ÁË 0, 14, 4 ˜¯
Ë 8,
Ê 1, 2, 1 ˆ
Á 0, -7, -2˜
Á ˜ (u1¢, u2¢ , u3¢ - u2¢ , u4¢ + 2 n2¢ )
Á 0, 0, 0 ˜
Á 0, 0, 0 ˜
Ë ¯
Vector Spaces
1-5
Example 5
Determine whether the set of vectors (4, 1, 2, 0), (1, 2, –1, 0), (1, 3, 1, 2) and
(6, 1, 0, 1) is linearly independent.
Let k1(4, 1, 2, 0) + k2(1, 2, –1, 0) + k3(1, 3, 1, 2) + k4(6, 1, 0, 1) = 0 (A)
Then 4k1 + k2 + k3 + 6k4 = 0 (1)
k1 + 2k2 + 3k3 + k4 =0 (2)
2k1 – k2 + k3 =0 (3)
and 2k3 + k4 =0 (4)
using (4) in (1); 4k1 + k2 – 11k3 =0 (5)
using (4) in (2); k1 + 2k2 + k3 =0 (6)
Eliminating k3 from (3) and (5); 26k1 – 10k2 = 0 or 13k1 – 5k2 =0 (7)
Eliminating k3 from (3) and (6); k1 – 3k2 =0 (8)
Solving (7) and (8), we get k1 = 0 and k2 =0
From (3), k3 = 0 and from (4), k4 =0
viz., the only values satisfying (A) are k1 = k2 = k3 = k4 = 0.
\ The given system in linearly independent.
Example 6
Show that the vectors u = (1, 2, 3), v = (0, 1, 2) and w = (0, 0, 1) generate R3.
If u, v, w generate R3, a general vectors (a, b, c) in R3 should expressed as a linear
combination of u, v, w.
Let (a, b, c) = k1(1, 2, 3) + k2(0, 1, 2) + k3(0, 0, 1)
\ k1 = a; 2k1 + k2 = b \ k2 = b – 2a
and 3k1 + 2k2 + k3 = c \ k3 = c – 3a – 2(b – 2a) = c – 2b + a
Hence the three given vectors generate R3.
Example 7
Find the condition on a, b, c so that (a, b, c) ŒR3 belong to the space generated by
u = (2, 1, 0), v = (1, –1, 2) and w = (0, 3, –4)
Let (a, b, c) = k1(2, 1, 0) + k2(1, –1, 2) + k3(0, 3, –4)
Linear Algebra and Partial Differential Equations
1-6
Then 2k1 + k2 = a (1)
k1 – k2 + 3k3 = b (2)
2k2 – 4k3 = c (3)
Eliminating k3 from (2) and (3), 4k1 + 2k2 = 4b + 3c
viz., 2a = 4b + 3c from (1)
Example 8
Show that the vectors u = (1, 0, –1), v = (1, 2, 1) and w = (0, –3, 2) form a basis for
R3. Express each of the standard basis vectors as a linear combination of u, v, w.
Writing u, v, w as row vectors one below the other and row reducing, we get
Ê 1, 0, -1ˆ Ê 1, 0, -1ˆ
Á 1, 2, 1 ˜ Á 0, 2, 2 ˜ ( R , R - R , R )
Á ˜ Á ˜ 1 2 1 3
Ë 0, -3, 2 ¯ Ë 0, -3, 2 ¯
Ê 1, 0, -1ˆ
3
Á 0, 2, 2 ˜ ( R1¢, R2¢ , R3¢ + R2¢ )
Á ˜ 2
Ë 0, 0, 5 ¯
Example 9
Find a basis and the dimension of the sub space W of R4, generated by the vectors
(1, –2, 5, –3), (2, 3, 1, –4) and (3, 8, –3, –5).
Vector Spaces
1-7
We form the matrix with the given vectors as rows and then row-reduce to echelon
form as given below:
Ê 1, -2, 5, -3ˆ Ê 1, -2, 5, -3ˆ
Á 2, 3, 1, -4˜ Á 0, 7, -9, 2 ˜ ( R , R - 2 R , R - R - R )
Á ˜ Á ˜ 1 2 1 3 1 2
Ë 3, 9, -3, -5¯ Ë 0, 7, -9, 2 ¯
Ê 1, -2, 5, 3ˆ
Á 0, 7, -9, 2˜ ( R1 , R2 , R3 - R2 )
Á ˜
Ë 0, 0, 0, 0¯
The non zero row vectors in the echelon form, namely, (1, –2, 5, –3) and (0, 7, –9, 2)
form a basis of W and dim (W) = 2
Example 10
If W is the space spanned by the polynomial v1 = t3 – 2t2 + 4t + 1, v2 = t3 + 6t – 5, v3
= 2t3 – 3t2 + 9t – 1 and v4 = 2t3 – 5t2 + 7t + 5, find a basis and dimension of W.
The coefficient vectors relative to the basis (t3, t2, t, 1) are (1, –2, 4, 1), (1, 0, 6, –5),
(2, –3, 9, –1) and (2, –5, 7, 5)
We form the matrix with these coefficient vectors as rows and row-reduce to the
echelon form as given below:
Ê 1 -2 4 1 ˆ Ê 1 -2 4 1 ˆ
Á 1 0 6 -5˜ Á 0 2 2 -6˜
Á ˜ Á ˜ ( R1 , R2 - R1 , R3 - 2 R1 , R4 - 2 R1 )
Á 2 -3 9 -1˜ Á 0 1 1 -3˜
Á ˜ Á ˜
Ë 2 -5 7 5 ¯ Ë 0 -1 -1 3 ¯
Ê 1 -2 4 1 ˆ
Á 0 1 1 -3˜
Á ˜ ( R1 , R2 ∏ 2, R3 , R4 )
Á 0 1 1 -3˜
Á 0 -1 -1 3 ˜
Ë ¯
Ê 1 -2 4 1ˆ
Á0 1 1 -3˜
Á ˜ ( R , R , R - R2 , R4 + R2 )
Á0 0 0 0˜ 1 2 3
Á0 0 0 0 ˜¯
Ë
\ (1, –2, 4, 1) and (0, 1, 1, –3) form a basis for the space generated by the coefficient
vectors
viz., t3 –2t2 + 4t + 1 and t2 + t –3 form a basis for the space W and dim (W) = 2.
Example 11
Find the dimension and a basis for the solution space W of the system of homogeneous
equation given below.
x1 + 2x2 + 2x3 – x4 + 3x5 = 0
Linear Algebra and Partial Differential Equations
1-8
x1 + 2x2 + 3x3 + x4 + x5 = 0
3x1 + 6x2 + 8x3 + x4 + 5x5 = 0
Row-reducing the given system of equation, we get
x1 + 2x2 + 2x3 – x4 + 3x5 = 0 (1)
x3 + 2x4 – 2x5 = 0 (2)
dim (W) = No. of unknowns – No. of non-zero equations
= 5 – 2 = 3.
The free variables are taken as x2, x4 and x5 (\ x2 is not present in (2))
Taking x2 = 1, x4 = 0, x5 = 0; v1 = (x1, x2, x3, x4, x5) = (–2, 1, 0, 0, 0), using (1) and (2)
Taking x2 = 0, x4 = 1, x5 = 0; v2 = (5, 0, –2, 1, 0), using (1) and (2)
Taking x2 = 0, x4 = 0, x5 = 1; v3 = (–7, 0, 2, 0, 1), using (1) and (2)
v1, v2, v3 form a basis for the solution space W.
Example 12
Find a homogenous system of equations whose solution set W is spanned by (1, –2,
0, 3 –1), (2, –3, 2, 5, –3) and (1, –2, 1, 2, –2).
v = (x1, x2, x3, x4, x5) ŒW, if and only if v is a linear combination of the given vectors.
\ (x1, x2, x3, x4, x5) = k1(1, –2, 0, 3, –1) + k2 (2, –3, 2, 5, –3) + k3(1, –2, 1, 2, –2)
viz., k1 + 2k2 + k3 = x1 (1)
–2k1 – 3k2 – 2k3 = x2 (2)
2k2 + k3 = x3 (3)
3k1 + 5k2 + 2k3 = x4 (4)
–k1 –3k2 – 2k3 = x5 (5)
Row-reducing the above equation, we get
k1 + 2k2 + k3 = x1 (1¢)
k2 = 2x1 + x2 (2¢)
2k2 + k3 = x3 (3¢)
–k2 – k3 = x4 – 3x1 (4¢)
–k2 –k3 = x1 + x5 (5¢)
(3¢) + (4¢) gives k2 = –3x1 + x3 + x4
Also k2 = 2x1 + x2
\ 2x1 + x2 = –3x1 + x3 + x4
viz; 5x1 + x2 – x3 – x4 = 0 (6)
Equating (4¢) and (5¢), we also get
x1 + x5 = x4 – 3x1
Vector Spaces
1-9
viz; 4x1 – x4 + x5 = 0
(OR) 4x1 – (5x1 + x2 – x3) + x5 = 0 viz., x1 + x2 – x3 – x5 = 0 (7)
v Œ W, if and only if the above system has a solution.
viz., if 5x1 + x2 – x3 – x4 = 0 (6)
and x1 + x2 – x3 – x5 = 0 (7¢)
Equations (6) and (7¢) form the required homogeneous system of equations.
Exercise 1
Part A (Short-Answer Questions)
1. Define vector space with two examples.
2. Define subspace with two examples.
3. Define span of a vector space.
4. Define standard vectors in R3 and prove that they span the vector space R3
5. Define linear dependence and independence of vectors.
6. Define basis and dimension of a vector space.
7. If V is the vector space of all (2 × 2) matrices over R, give a basis of V and
dimension of V.
8. Determine whether the vectors (1, 1, 1) and (1, –1, 5) form a basis for the
vector space R3.
9. Find whether the vectors (1, 1, 2), (1, 2, 5) and (5, 3, 4) form a basis for the
vector space R3.
10. Find whether the vector (1, 1, 1), (1, 2, 3) and (2, –1, 1) form a basis for the
vector space R3.
Part B
11. Is the vector (3, –1, 0, –1) in the sub-space of R4 spanned by the vectors
(2, –1, 3, 2), (–1, 1, 1, –3) and (1, 1, 9, –5)?
12. Find whether the vector (–3, –6, 1, –5, 2) is in the sub space of R5 spanned
by (1, 2, 0, 3, 0), (0, 0, 1, 4, 0) and (0, 0, 0, 0, 1).
13. Examine the linear dependence or independence of the following vectors:
(i) u1 = (2, –1, 3, 2), u2 = (1, 3, 4, 2) and u3 = (3, –5, 2, 2).
(ii) u1 = (1, –1, 0, 1), u2 = (–1, –1, –1, 2) and u3 = (2, 0, 1, –1)
14. Find the maximum number of linearly independent vectors among the
following and express each of the remaining vectors as a linear combination
of these:
u1 = (3, 1, –4); u2 = (2, 2, –3); u3 = (0, –4, 1) and u4 = (–4, –4, 6)
15. Show that the vectors u1 (2, 3, –1, –1); u2 = (1, –1, –2, –4); u3 = (3, 1, 3, –2)
and u4 = (6, 3, 0, –7) form a linearly dependent system, also express u4 as a
linear combination of other.
16. Determine whether the vector (4, 2, 1, 0) is a linear combination of
the vectors u1 = (6, –1, 2, 1), u2 = (1, 7, –3, –2), u3 = (3, 1, 0, 0) and
u4 = (3, 3, –2, –1).
17. Show that the vector (a, b, 0) in R3 is generated by
(i) u1 = (1, 2, 0) and u2 = (0, 1, 0)
Linear Algebra and Partial Differential Equations
1-10
(ii) u1 = (2, –1, 0) and u2 = (1, 3, 0)
18. Show that the vector (a, b, 0) in R3 is generated by
(i) u1 = (0, 1, 1) and u2 = (0, 2, –1)
(ii) u1 = (0, 1, 2) and u2 = (0, 2, 3)
19. Show that the vectors (1, 1, 1), (1, 2, 3) and (2, 3, 8) form a basis for R3.
Express each of the standard basis vectors as a linear combination of these
vectors.
20. Show that the vectors u = (1, 2, 2), v = (2, 1, –2) and w = (2, –2, 1) form a basis
for R3. Express each of the standard basis vectors as a linear combination of
u, v and w.
21. Find a basis and dimension for the subspace of R4 spanned by the four vectors
v1 = (1, 1, 2, 4), v2 = (2, –1, –5, 9), v3 = (1, –1, –4, 0) and v4 = (2, 1, 1, 6)
22. Find a basis and dimension of the subspace of R4 spanned by
(i) (1, 4, –1, 3); (2, 1, –3, –1) and (0, 2, 1, –5)
(ii) (1, –4, –2, 1); (1, –3, –1, 2) and (3, –8, –2, 7)
23. Find a basis and dimension of the solution space W of the homogeneous
system x + 3y + 2z = 0, x + 5y + z = 0 and 3x + y + 8z = 0.
24. Find a basis and dimension of the solution space W of the homogeneous
system: x1 + 2x2 – x3 + x4 = 0 and x1 – 2x2 + x3 + 2x4 = 0
25. Find a homogeneous system of equations whose solution set W is spanned
by (1, –2, 0, 3), (1, –1, –1, 4) and (1, 0, –2, 5)
Answers
Exercise 1
7 5 1 1 3 1
19. e1 = u - u + u ; e = - u1 + u2 - u3
4 1 4 2 4 3 2 2 2 2
1 1 1
and e3 = - u1 - u2 + u3
4 4 4
1 2 2 2 1 2 2 2 1
20. e1 = u + v + w; e2 = u + v - w; e3 = u - v + w
9 9 9 9 9 9 9 9 9
21. (1, 1, 2, 4), (0, –3, –11, 1) and (0, –2, –6, –4) form a basis and dim = 3.
22. (i) dim (W) = 3; Basis ∫ [(1, 4, –1, 3), (0, –7, –1, –7) and (0, 2, 1, 5)
(ii) dim (W) = 2; Basis ∫ [(1, –4, –2, 1) and (0, 1, 1, 1)]
23. Basis ∫ (7, –1, 2) and dim (W) = 1
24. dim (W) = 2; Basis ∫ [(–5, 1, 0, 3) and (3, 0, 1 –2)]
25. 2x1 + x2 + x3 = 0 and 5x1 + x2 – x4 = 0
Linear Transformation
2-1
Unit 2
Linear Transformation
Note
T(0) = T(0v) = 0. T(v) = 0.
Note
T: V Æ W can be uniquely determined by arbitrarily assigning elements of W
to the elements of a basis of V as per the following theorem which is stated
without proof.
Theorem: If V is a finite dimensional vector space over a field F, with {v1, v2, ..., vn}
as a basis and W is another vector space over the same field containing the arbitrary
vector {w1, w2, ..., wn} (which may be linearly dependent or equal to each other), there
exists a unique linear transformation T : V Æ W such that T(vj) = wj ij = 1, 2, ..., n.
For example, let us find T : R2 Æ R3 defined by T((1, 2) = (3, –1, 5) and T(0, 1)
= (2, 1, –1)
Since (1, 2) and (0, 1) are linearly independent, they form a basis of R2. The vectors
(3, –1, 5) and (2, 1, –1) have been arbitrarily chosen in R3
Now (a, b) = c1(1, 2) + c2(0, 1) \ c1 = a and c2 = b – 2a
\ T(a, b) = a(3, –1, 5) + (b – 2a) (2, 1, –1)
= (2b – a, b – 3a, 7a – b), which is the unique linear transformation
required.
Note
Null space of T is a subspace of V and range space of T is a subspace of
W.)
Dimension Theorem
The sum of the dimension of the range space and null space of a linear transformation
is equal to the dimension of its domain viz., if V and W are vector spaces over the
field F and if T : V Æ W is a linear transformation and if V is finite dimensional, then
rank (T) + nullity (T) = dim (V).
Proof: Let {v1, v2, ..., vk} be a basis for NT, so that dim (NT) = k
Linear Transformation
2-3
We can find vectors vk + 1, vk + 2, ... vn such that (v1, v2, ... vn) is a basis of V, so that
dim (V) = n.
The theorem is proved, if we can prove that (Tvk + 1, Tvk + 2, ... Tvn) is basis for RT
Clearly the vectors Tv1, Tv2, ... Tvn span RT
But Tv1 = Tv2 = ... = Tvk = 0
\ Tvk + 1, Tvk + 2, ..., Tvn span RT
These vector Tvk +1, Tvk +2, Tvn will be a basis of RT, provided they are linearly
independent.
n
Let these be scalars ci such that  ci (Tvi ) = 0
i = k +1
n
viz., T Â ci vi = 0 (Q T is linear)
i = k +1
n n
This means that v = Â ci vi is (in N T ) viz., T Â ci vi = 0 [Q T is linear]
i = k +1 i = k +1 (1)
Since {v1, v2, ..., vk} is a basis of NT, there exist scalars b1, b2, ..., bk such that
k
v= Â bi vi (2)
i =1
k n
From (1) and (2), we get  bi vi -  ci vi = 0
i =1 i = K +1
Since v1, v2, .., vn form a basis of V, they are linearly independent.
\ b1 = b2 = ... = bk = ck + 1 = ck + 2 = ... = cn = 0
\ Tvk + 1, Tvk + 2, ..., Tvn form a basis for RT
\ dim (RT) = n – k and dim (NT) = k
\ Rank (T) + nullity (T) = dim(V)
Example 1
Show that the transformation T : R3 Æ R2 defined by T(x, y, z) = (z, x + y) is linear.
Let v = (x, y, z) and w = (x¢, y¢, z¢)
Then c1v + c2w = (c1x + c2x¢, c1y + c2y¢, c1z + c2z¢)
\ T(c1v + c2w) = {c1z + c2z¢, c1(x + y) + c2(x¢ + y¢)}, by definition of T
= c1(z, x + y) + c2(z¢, x¢ + y¢)
Linear Algebra and Partial Differential Equations
2-4
= c1T(x, y, z) + c2T(x¢, y¢, z¢)
= c1T(v) + c2T(w)
\ T is linear.
Example 2
Show that the transformation T : R2 Æ R2 defined by T(x, y) = (sin x, y) is not linear.
Let v = (x, y) and w = (x¢, y¢)
Then c1v + c2w = (c1x + c2x¢, c1y + c2y¢)
\ T(c1v + c2w) = {sin(c1x + c2x¢), c1y + c2y¢} (1)
But c1T(v) + c2T(w) = c1 (sin x, y) + c2 (sin x¢, y¢)
= {c1 sin x + c2 sin x¢, c1y + c2y¢} (2)
From (1) and (2), we see that T(c1v + c2w) π c1T(v) + c2T(w)
\ T is not linear.
Example 3
Find whether the transformation T : R2 Æ R3 defined by
T(x, y) = (x + 1, 2y, x + y) is linear.
Let v (x, y) and w = (x¢, y¢)
Then c1v + c2w = (c1x + c2x¢, c1y + c2y¢)
\ T(c1v + c2w) = {c1x + c2x¢ + 1, 2(c1y + c2y¢), c1(x + y) + c¢2(x¢ + y¢)
π c1T(v) + c2T(w)
\ T is not linear.
Example 4
If V is the vector space of all n × n matrices over F and if B is an arbitrary matrix in
V, show that the transformation T : V Æ V defined by T(A) = AB – BA, where A Œ V
is linear. Show also that T(A) = A + B is not linear, unless B = 0.
T(c1A + c2A¢) = (c1A + c2A¢)B – B(c1A + c2A¢)
*
* *
*
* *
*
* *
*
* *
*
* *
*
* *