MAT3701/201/1/2012
Tutorial Letter 201/1/2012
Linear Algebra
MAT3701
Semester 1
Department of Mathematical Sciences
Solutions to Assignment 01
Bar code
, 2
ONLY FOR SEMESTER 1 STUDENTS
ASSIGNMENT 01
Based on Study Units 1 - 9
FIXED CLOSING DATE: 7 MARCH 2012
UNIQUE NUMBER: 771265
Please note that we will only mark a selection of the questions. It is therefore in your own interest to do all the
questions. The fact that a question is not marked does not mean that it is less important than one that is marked.
Worked solutions to all the questions will be sent to all students shortly after the due date.
For this assignment Questions 2, 4, 6, and 8 will be marked.
QUESTION 1
(a) If {v, w} is a basis for R 2 over R, show that {v, w, iv, iw} is a basis for C 2 over R.
(b) Without any further computations, use (a) to explain why ; = {(1, 0) , (i, 0) , (0, 1) , (0, i)} is a basis for C 2
over R.
SOLUTION
(a)
av + bw + civ + diw = 0, a, b, c, d + R
% av + bw + i(cv + dw) = 0
% av + bw = 0 and cv + dw = 0, since av + bw and cv + dw lie in R 2
% a = b = 0 and c = d = 0 since {v, w} is a basis over R
Therefore, {v, w, iv, iw} is linearly independent, and hence also a basis for C 2 over R, since it contains four vectors
and dim(C 2 ) = 4 over R.
(b) Let v = (1, 0) and w = (0, 1) and apply the result in (a).
QUESTION 2
Let V be the vector space C 2 with scalar multiplication over the real numbers R, and let T : V V be the
mapping de¿ned by
T (z 1 , z 2 ) = (z 1 + z 1 , z 2 z 2 ) .
(a) Show that T is a linear operator. (9)
(b) Find a basis for N (T ) . (7)
(c) Find a basis for R (T ) . (7)
(d) Determine whether V = N (T ) c R (T ) (2)
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SOLUTION
Let (z 1 , z 2 ), (z 3 , z 4 ) + V and a + R.
T ((z 1 , z 2 ) + (z 3 , z 4 )) = T (z 1 + z 3 , z 2 + z 4 )
= (z 1 + z 3 + z 1 + z 3 , z 2 + z 4 z 2 + z 4 )
= (z 1 + z 3 + z 1 + z 3 , z 2 + z 4 z 2 z 4 )
= (z 1 + z 1 , z 2 z 2 ) + (z 3 + z 3 , z 4 z 4 )
= T (z 1 , z 2 ) + T (z 3 , z 4 )
and
T (a(z 1 , z 2 )) = T (az 1 , az 2 )
= (az 1 + az 1 , az 2 az 2 )
= (az 1 + az 1 , az 2 az 2 ) since a is real
= a(z 1 + z 1 , z 2 z 2 )
= aT (z 1 , z 2 )
Since both these conditions are satis¿ed, T is a linear operator (9)
(b) By de¿nition,
N (T ) = {(z 1 , z 2 ) + V : T (z 1 , z 2 ) = (0, 0)}.
When is T (z 1 , z 2 ) = (0, 0)?
T (z 1 , z 2 ) = (0, 0)
% (z 1 + z 1 , z 2 z 2 ) = (0, 0)
% z 1 + z 1 = 0 and z 2 z 2 = 0
% z 1 = z 1 and z 2 = z 2 .
Note that the only complex numbers that equal their conjugates are the real numbers: If a + ib = a ib, then
b = b, hence b = 0. (See also Theorem D2(e) in the Appendix of Friedberg).
Similarly, the only complex numbers that are equal to minus their conjugates are the imaginary numbers: If
a + ib = (a ib), then a = a, hence a = 0.
It follows that
N (T ) = {(i x, y) : x, y + R}.
We write an element of N (T ) as a linear combination with real coef¿cients (the scalar ¿eld is still R):
(i x, y) = x(i, 0) + y(0, 1).
It follows that N (T ) = span{(i, 0), (0, 1)}. Since (i, 0) and (0, 1) are clearly linearly independent, a basis for
N (T ) is
; 2 = {(i, 0), (0, 1)}.
, 4
(7)
(c) We’ll use Theorem 2.2 in Friedberg. In order to apply this theorem, we ¿rst determine a basis for V . Write
an element (z 1 , z 2 ) + V as a linear combination with coef¿cients in R (since V is a vector space over R): If
we write z 1 = a + ib and z 2 = c + id, then
(z 1 , z 2 ) = (a + ib, c + id)
= (a, 0) + (ib, 0) + (0, c) + (0, id)
= a(1, 0) + b(i, 0) + c(0, 1) + d(0, i).
Therefore, V = span{(1, 0), (i, 0), (0, 1), (0, i)}. To check that it is linearly independent, suppose that
a(1, 0) + b(i, 0) + c(0, 1) + d(0, i) = (0, 0)
for some a, b, c, d + R. Adding up, we obtain (a +ib, c+id) = (0, 0). Therefore, a +ib = 0 and c+id = 0,
which implies a = b = c = d = 0. It follows that
; = {(1, 0), (i, 0), (0, 1), (0, i)}
is a basis for V . By Theorem 2.2, T (;) will span R(T ):
T (;) = {T (1, 0), T (i, 0), T (0, 1), T (0, i)}
= {(2, 0), (0, 0), (0, 0), (0, 2i)}.
To ¿nd a basis contained in this set, we can clearly ignore (0, 0). The remaining two vectors (2, 0) and (0, 2i)
are clearly linearly independent. Therefore, a basis of R(T ) is {(2, 0), (0, 2i)}. It will be easier later if we get
rid of the 2’s now. Multiply each element of the basis by 1/2 to obtain the basis
; 1 = {(1, 0), (0, i)}.
(7)
(d) Note that ; 1 C ; 2 = ;, the basis of V mentioned in (c). It therefore follows that V is the direct sum of the
two subspaces N (T ) and R(T ). (2)
[25]
QUESTION 3
Let f a , f b , f c be the Lagrange polinomials associated with the distinct real numbers a, b, c respectively.
De¿ne T : P2 (R) P2 (R) by T (g) = g (a) f a + g(b) f b
(a) Show that T is a linear operator.
(b) Explain whether or not T is a projection.
(c) Find [T ]; , where ; = { f a , f b , f c } .
Tutorial Letter 201/1/2012
Linear Algebra
MAT3701
Semester 1
Department of Mathematical Sciences
Solutions to Assignment 01
Bar code
, 2
ONLY FOR SEMESTER 1 STUDENTS
ASSIGNMENT 01
Based on Study Units 1 - 9
FIXED CLOSING DATE: 7 MARCH 2012
UNIQUE NUMBER: 771265
Please note that we will only mark a selection of the questions. It is therefore in your own interest to do all the
questions. The fact that a question is not marked does not mean that it is less important than one that is marked.
Worked solutions to all the questions will be sent to all students shortly after the due date.
For this assignment Questions 2, 4, 6, and 8 will be marked.
QUESTION 1
(a) If {v, w} is a basis for R 2 over R, show that {v, w, iv, iw} is a basis for C 2 over R.
(b) Without any further computations, use (a) to explain why ; = {(1, 0) , (i, 0) , (0, 1) , (0, i)} is a basis for C 2
over R.
SOLUTION
(a)
av + bw + civ + diw = 0, a, b, c, d + R
% av + bw + i(cv + dw) = 0
% av + bw = 0 and cv + dw = 0, since av + bw and cv + dw lie in R 2
% a = b = 0 and c = d = 0 since {v, w} is a basis over R
Therefore, {v, w, iv, iw} is linearly independent, and hence also a basis for C 2 over R, since it contains four vectors
and dim(C 2 ) = 4 over R.
(b) Let v = (1, 0) and w = (0, 1) and apply the result in (a).
QUESTION 2
Let V be the vector space C 2 with scalar multiplication over the real numbers R, and let T : V V be the
mapping de¿ned by
T (z 1 , z 2 ) = (z 1 + z 1 , z 2 z 2 ) .
(a) Show that T is a linear operator. (9)
(b) Find a basis for N (T ) . (7)
(c) Find a basis for R (T ) . (7)
(d) Determine whether V = N (T ) c R (T ) (2)
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[25]
SOLUTION
Let (z 1 , z 2 ), (z 3 , z 4 ) + V and a + R.
T ((z 1 , z 2 ) + (z 3 , z 4 )) = T (z 1 + z 3 , z 2 + z 4 )
= (z 1 + z 3 + z 1 + z 3 , z 2 + z 4 z 2 + z 4 )
= (z 1 + z 3 + z 1 + z 3 , z 2 + z 4 z 2 z 4 )
= (z 1 + z 1 , z 2 z 2 ) + (z 3 + z 3 , z 4 z 4 )
= T (z 1 , z 2 ) + T (z 3 , z 4 )
and
T (a(z 1 , z 2 )) = T (az 1 , az 2 )
= (az 1 + az 1 , az 2 az 2 )
= (az 1 + az 1 , az 2 az 2 ) since a is real
= a(z 1 + z 1 , z 2 z 2 )
= aT (z 1 , z 2 )
Since both these conditions are satis¿ed, T is a linear operator (9)
(b) By de¿nition,
N (T ) = {(z 1 , z 2 ) + V : T (z 1 , z 2 ) = (0, 0)}.
When is T (z 1 , z 2 ) = (0, 0)?
T (z 1 , z 2 ) = (0, 0)
% (z 1 + z 1 , z 2 z 2 ) = (0, 0)
% z 1 + z 1 = 0 and z 2 z 2 = 0
% z 1 = z 1 and z 2 = z 2 .
Note that the only complex numbers that equal their conjugates are the real numbers: If a + ib = a ib, then
b = b, hence b = 0. (See also Theorem D2(e) in the Appendix of Friedberg).
Similarly, the only complex numbers that are equal to minus their conjugates are the imaginary numbers: If
a + ib = (a ib), then a = a, hence a = 0.
It follows that
N (T ) = {(i x, y) : x, y + R}.
We write an element of N (T ) as a linear combination with real coef¿cients (the scalar ¿eld is still R):
(i x, y) = x(i, 0) + y(0, 1).
It follows that N (T ) = span{(i, 0), (0, 1)}. Since (i, 0) and (0, 1) are clearly linearly independent, a basis for
N (T ) is
; 2 = {(i, 0), (0, 1)}.
, 4
(7)
(c) We’ll use Theorem 2.2 in Friedberg. In order to apply this theorem, we ¿rst determine a basis for V . Write
an element (z 1 , z 2 ) + V as a linear combination with coef¿cients in R (since V is a vector space over R): If
we write z 1 = a + ib and z 2 = c + id, then
(z 1 , z 2 ) = (a + ib, c + id)
= (a, 0) + (ib, 0) + (0, c) + (0, id)
= a(1, 0) + b(i, 0) + c(0, 1) + d(0, i).
Therefore, V = span{(1, 0), (i, 0), (0, 1), (0, i)}. To check that it is linearly independent, suppose that
a(1, 0) + b(i, 0) + c(0, 1) + d(0, i) = (0, 0)
for some a, b, c, d + R. Adding up, we obtain (a +ib, c+id) = (0, 0). Therefore, a +ib = 0 and c+id = 0,
which implies a = b = c = d = 0. It follows that
; = {(1, 0), (i, 0), (0, 1), (0, i)}
is a basis for V . By Theorem 2.2, T (;) will span R(T ):
T (;) = {T (1, 0), T (i, 0), T (0, 1), T (0, i)}
= {(2, 0), (0, 0), (0, 0), (0, 2i)}.
To ¿nd a basis contained in this set, we can clearly ignore (0, 0). The remaining two vectors (2, 0) and (0, 2i)
are clearly linearly independent. Therefore, a basis of R(T ) is {(2, 0), (0, 2i)}. It will be easier later if we get
rid of the 2’s now. Multiply each element of the basis by 1/2 to obtain the basis
; 1 = {(1, 0), (0, i)}.
(7)
(d) Note that ; 1 C ; 2 = ;, the basis of V mentioned in (c). It therefore follows that V is the direct sum of the
two subspaces N (T ) and R(T ). (2)
[25]
QUESTION 3
Let f a , f b , f c be the Lagrange polinomials associated with the distinct real numbers a, b, c respectively.
De¿ne T : P2 (R) P2 (R) by T (g) = g (a) f a + g(b) f b
(a) Show that T is a linear operator.
(b) Explain whether or not T is a projection.
(c) Find [T ]; , where ; = { f a , f b , f c } .
