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MATH 2270 Practice Final Solutions: QR Factorization, Linear Independence, Inner Product, Exams of Algebra

Solutions to practice problems from a university-level mathematics course, covering topics such as qr factorization, linear independence, and the definition of inner product. The solutions include detailed explanations and computations.

Typology: Exams

Pre 2010

Uploaded on 07/23/2009

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MATH 2270 PRACTICE FINAL SOLUTIONS
Problem 1
Let A=๎˜”1 3
2โˆ’5๎˜•.Find a QR-factorization of A.
Solution 1.We first find a suitable orthonormal basis for the span of
the columns of A. Let x1=๎˜”1
2๎˜•. Let x2=๎˜”3
โˆ’5๎˜•. Next, let
v1=x1, and
(1) v2=x2โˆ’x2ยทv1
v1ยทv1
v1=๎˜”3
โˆ’5๎˜•โˆ’โˆ’7
5๎˜”1
2๎˜•=11
5๎˜”2
โˆ’1๎˜•.
We now normalize v1and v2. Let u1=v1
||v1|| =1
โˆš5๎˜”1
2๎˜•. Now, ||v2|| =
11โˆš22+12
5=11
โˆš5. Let u2=v2
||v2||. Thus,
(2) u2=1
โˆš5๎˜”2
โˆ’1๎˜•.
Let Q=๎˜‚u1u2๎˜ƒ.
We now want to find an upper triangular matrix Rwith positive
entries on the diagonal such that A=QR.
Let R=๎˜”a b
0c๎˜•.We compute:
A=QR
๎˜”1 3
2โˆ’5๎˜•=1
โˆš5๎˜”1 2
2โˆ’1๎˜•๎˜” a b
0c๎˜•
๎˜”1 3
2โˆ’5๎˜•=1
โˆš5๎˜”a b + 2c
2a2bโˆ’c๎˜•
(3)
Hence a=โˆš5. Equating the terms in the second columns we obtain
the system:
3 = b+ 2c
โˆš5
โˆ’5 = 2bโˆ’c
โˆš5.
(4)
1
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Problem 1

Let A =

[

]

. Find a QR-factorization of A.

Solution 1. We first find a suitable orthonormal basis for the span of

the columns of A. Let x 1 =

[

]

. Let x 2 =

[

]

. Next, let

v 1 = x 1 , and

(1) v 2 = x 2 โˆ’

x 2 ยท v 1 v 1 ยท v 1

v 1 =

[

]

[

]

[

]

We now normalize v 1 and v 2. Let u 1 = (^) ||vv^11 || = โˆš^15

[

]

. Now, ||v 2 || = 11 โˆš 22 +1^2 5 =^ โˆš^11

  1. Let^ u^2 =^

v 2 ||v 2 ||. Thus,

(2) u 2 =

[

]

Let Q =

[

u 1 u 2

]

We now want to find an upper triangular matrix R with positive entries on the diagonal such that A = QR.

Let R =

[

a b 0 c

]

. We compute:

A = QR

[

]

[

] [

a b 0 c

]

[

]

[

a b + 2c 2 a 2 b โˆ’ c

]

Hence a =

  1. Equating the terms in the second columns we obtain the system:

3 =

b + 2c โˆš 5

โˆ’5 =

2 b โˆ’ c โˆš 5

1

Multiplying the first equation by 2 gives the following system:

2 b + 4c โˆš 5

โˆ’5 =

2 b โˆ’ c โˆš 5

Subtracting the first equation from the second equation, we get:

11 =

5 c โˆš 5 11 โˆš 5

= c.

Plugging this back into the first equation in (5), we get;

b + โˆš^225 โˆš 5

3

= b

= b.

A quick inspection shows that R has the desired properties, and that A = QR.

Remark 2. This problem is computationally a bit challenging without a calculator. (I did not use one.) Iโ€™ll make sure that the computations are a bit nicer on the actual exam.

Problem 2 Let A be an m ร— n matrix such that AT^ A is invertible. Show that the columns of A are linearly independent.

Solution 3. Consider the equation Ax = 0. Weโ€™d like to show that this equation has only the trivial solution, since that is a necessary and sufficient condition for A to have linearly independent columns. We apply AT^ to both sides to get AT^ Ax = AT^0 which is equivalent to AT^ Ax = 0. Since AT^ A is invertible, this last equation has only the trivial solution. Thus, x = 0 , as desired.

Remark 4. This was a homework problem. It is part of the proof of Theorem 14 in section 6.5.