Other

MAT3701 - Linear Algebra III

0 purchase

Module
MAT3701 Linear Algebra (MAT3701)

Institution
University Of South Africa (Unisa)

MAT3701 is a third year Linear Algebra module with a pre-requisite of MAT2611. The purpose of this module is to acquire a basic knowledge concerning inner product spaces, invariant subspaces, cyclic subspaces, operators and their canonical forms

[Show more]

Preview 4 out of 460 pages

View example

Uploaded on November 8, 2022
Number of pages 460
Written in 2021/2022
Type Other
Person Unknown

mat3701
previous assignments
exam preparation
revision

Institution
University of South Africa (Unisa)
Module
MAT3701 Linear Algebra (MAT3701)

£3.55

Also available in package deal from £7.99

Add to cart

Add to wishlist

100% satisfaction guarantee
Immediately available after payment
Both online and in PDF
No strings attached

Also available in package deal (1)

APM3711 AND MAT3701 PREVIOUS ASSIGNMENTS AND EXAMS

£ 10.21 £ 7.99 3 items

1. Other - Apm3711 numerical methods ii
2. Other - Mat3701 - linear algebra iii
3. Exam (elaborations) - Mat3701 linear algebra exam revision material
Show more

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)

, 3 MAT3701/201/1

[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 } .

The benefits of buying summaries with Stuvia:

Guaranteed quality through customer reviews

Stuvia customers have reviewed more than 700,000 summaries. This how you know that you are buying the best documents.

Quick and easy check-out

You can quickly pay through credit card for the summaries. There is no membership needed.

Focus on what matters

Your fellow students write the study notes themselves, which is why the documents are always reliable and up-to-date. This ensures you quickly get to the core!

Frequently asked questions

What do I get when I buy this document?

You get a PDF, available immediately after your purchase. The purchased document is accessible anytime, anywhere and indefinitely through your profile.

Satisfaction guarantee: how does it work?

Our satisfaction guarantee ensures that you always find a study document that suits you well. You fill out a form, and our customer service team takes care of the rest.

Who am I buying these notes from?

Stuvia is a marketplace, so you are not buying this document from us, but from seller znyele. Stuvia facilitates payment to the seller.

Will I be stuck with a subscription?

No, you only buy these notes for £3.55. You're not tied to anything after your purchase.

Can Stuvia be trusted?

4.6 stars on Google & Trustpilot (+1000 reviews)

69411 documents were sold in the last 30 days

Founded in 2010, the go-to place to buy revision notes and other study material for 15 years now

Start selling