Math 307
Spring, 2003
Hentzel
Time: 1:10-2:00 MWF
Room: 1324 Howe Hall
Instructor: Irvin Roy Hentzel
Office 432 Carver
Phone 515-294-8141
E-mail: hentzel@iastate.edu
http://www.math.iastate.edu/hentzel/class.307.ICN
Text: Linear Algebra With Applications, Second Edition
Otto Bretscher
Friday,
Feb 28 Chapter 4.1
Page 157 Problems 4,10,20 and the Leontief
Problem
Main Idea: You can work with all kinds of objects,
not just n-tuples.
Key Words: Vector Space, Derivative, Leontief,
Goal: Use the ideas you already know to study
derivatives and other systems.
Example: Write the matrix for differentiation on
the space spanned by x2 ex, x ex, ex }
x2 ex
x ex
ex
x2 ex
x ex
ex
1
0
0
2
1
0
0
1
1
|1 0 0|
D= |2 1 0|
|0 1 1|
What good is D? Find the derivative of
5 x2 ex + 3 x ex + ex
|1 0 0||5| | 5|
D = | 2 1 0 | | 3 | = |13 |
|0 1 1||1| | 4|
or 5 x2 ex + 13 x ex + 4 ex.
Find the 4th derivative of x2 ex.
D4 = (D2)2
|100|2 |1 0 0|
= |410| = |8 1 0|
|221|
|12 4 1 |
| 1 0 0 || 1 |
| 1|
| 8 1 0 || 0 | = | 8 |
|12 4 1 || 0 |
|12 |
or x2 ex + 8 x ex + 12 ex
Basis:
Find a basis of the 2x2 matrices:
| a b | = a| 1 0 | + b| 0 1 | + c| 0 0 | + d| 0 0 |
|c d|
|00|
|00|
|1 0|
|0 1|
What is the dimension of the 2x2 matrices?
What is the dimension of the 2x3 matrices?
What is the matrix of the linear transformation
X --------> A X - X A
for A = | 0 1 |
2x2
|1 0|
|1 0| |0 1| |0 0| |0 0|
|0 0| |0 0| |1 0| |0 1|
| 1 0 | | 0 -1 | 0
-1
1
0
|0 0| |1 0|
| 0 1 | |-1 0 |
|0 0| |0 1|
-1
0
0
1
| 0 0 | | 1 0 |
| 1 0 | | 0 -1 |
1
0
0
-1
| 0 0 | | 0 1 |
| 0 1 | |-1 0 |
0
1
-1
0
The matrix for the linear transformation is
| 0 -1 1 0 |
|-1 0 0 1 |
| 1 0 0 -1 |
| 0 1 -1 0 |
What are the matrices that commute with A? We
simply need to find the null space of the matrix.
| 0 -1 1 0 |
|-1 0 0 1 |
| 1 0 0 -1 |
| 0 1 -1 0 |
| 0 -1 1 0 |
|-1 0 0 1 |
| 1 0 0 -1 |
| 0 1 -1 0 |
| 1 0 0 -1 |
| 0 1 -1 0 |
| 0 -1 1 0 |
|-1 0 0 1 |
|1
|0
|0
|0
a b
0 0 -1 |
1 -1 0 |
0 0 0|
0 0 0|
|x|
|0|
|1|
|y|=a|1|+b|0|
|z|
|1|
|0|
|w|
|0|
|1|
Or written out as a 2x2 matrix
|b a|
|a b|
Check | 0 1 | | b a | = | a b |
|1 0||a b| |b a|
|b a||0 1| =|a b|
|a b||1 0| |b a|
Since both products are the same, the
matrices commute with A.
Example. The teacher evaluation form for the math
department used to have 12 questions asking to rate
the teacher on a scale of 1 to 5 with 1 being good and
5 being bad. The questions were on
1. Preparation
2. Presentation
3. Speaking and Writing
4. Interest
5. Questions.
6. Classtime
7. Availability
8. Homework
9. Testing
10. Grading
11. Policies
12. Overall
What the math department was interested in was
the answer to question 12.
That determined whether the professor was a
good teacher or not.
The department kept averages for each course
so that a person could be compared with others
in the department.
If the teacher was below average, then the
department advised the professor to look at the
first 11 questions to find what he was doing
wrong.
Since the first 11 questions represent what the
department thinks goes into good teaching, one
ought to be able to predict the answer to the
twelfth question using the information from the
first 11.
I looked for the coefficients {c1, c2, ..., c11} so that
c1 a1 + c2 a2 + ... + c11 a11 = a12.
where a1, a2, ..., a12 are the class averages for
the evaluation.
The Correlation of this approximation was
0.973641. That means that 97% of the concept
of good teaching can be explained by the skills
measured in the first eleven questions.
The department was happy because it says that
the score a12 actually represents qualities which
are believed to be good teaching.
The next step is to look at the coefficients. The
coefficients tell how much each of the 11
qualities contribute to good teaching.
Coefficient Quality
0.32
Presentation
0.18
Interest
0.15
Grading
0.15
Questions
0.13
Preparation
0.08
Classtime
0.05
Testing
0.03
Homework
0.02
Speak/Writing
-0.06
Policies
-0.10
Availability
The largest contributor to the overall grade was
Presentation:
"The instructor presented the course material in a
clear and appropriate manner."
The next highest was Interest
"The instructor's enthusiasm and presentation
made the course interesting."
It seems unintuitive that Testing, Homework, and
Speaking/Writing should count for relatively
little.
It seems strange that Policies should have a
negative influence.
"The instructor was timely and clear in
announcing policies on attendance, homework,
tests, and grading."
The worse the instructor was in this area, the
better the students liked him.
Availability counts more than Classtime, Testing,
Homework, Speaking/Writing but in the
opposite direction.
"The instructor was available for help outside of
class."
The less the instructor was available, the better a
teacher he was perceived to be.
Maybe the students felt that if a professor had
time to talk with them, he must not be a very
important person. If he was unavailable, he
must be very busy and important.
On the other hand, if he presented the material
flawlessly, the students had no need to find him
for questions and clarifications.
Since they did not know if he was available or
not, they simply gave him an average rating.
The importance of this example is to show how
one can use ideas of linear algebra to dissect
data to look for hidden messages.
The study of the correlation coefficient will be
done in the study of the "Line of Regression"
which is taught in many courses as an exercise
in formula manipulation.
Hopefully, we will approach it with some actual
understanding when we cover it.
Along with the correlation coefficient we also
learn how to solve inconsistent systems for the
best possible answer.
It often happens that the system AX = B has no
solution. So you want the closest thing to a
solution.
That is, the X such that the error is minimized.
The closest thing to a solution will be the X0
where |B-AXo| is the smallest possible.
Now for the Leontief Open model. We are studying
the internal workings of a country. We study the
manufacturing of goods, the production of food,
and the production of energy like dams and hydro
electric plants.
The interaction between the sectors is very hard to
understand. For example, during the petroleum
shortage back in the 1970 the airlines and the
trucking industry was badly hurt and suffered
many layoffs. What happened to the truck repair
industry? Strangely, they had a boom time.
Since the trucks were idle anyway, this was a good
time to get the needed work done.
The prime example of this tinkering with the
economy came with Russia and the five year plans
where they tried to plan the economy to meet
projected goals.
The same was true in the USA during the war.
We had to make sure that what was produced
could actually reach the front.
It was pointless to concentrate on producing
tanks if there were no ships to deliver them, and
it was pointless to make ships if there was
nothing to transport. The idea was to divide the
resources so that there were no bottle necks
and everything flowed smoothly.
Now days, one can use the idea to work the stock
market. When you read in the paper that there
is a new oil embargo, or that there is a ban on
DDT, or what ever, the big boys have already
had that information and the stocks market
prices have already been adjusted.
What is left is for one to get in on the second tier.
Those are the industries which are not so
obviously affected.
You can buy the data on the US market and use
that to predict what will happen if the current
supplies are adjusted.
The Leontief method works with an intact
economy. We simply do the same things we
always did, but do more or less of them.
We do not introduce new radical ways of doing
things.
For example, before world war two, air planes
were made a dozen at a time. If a part did not
fit, the mechanic would get out his file and
adjust it. Then he would do the next part and
the next part after that.
In the auto industry, if a part did not fit, they went
back and changed the die for making that part.
During the war, the airplanes were made in the
car plants. This was not just a scaling up of the
way things were, but was a new type of
operation.
You have an economy that is working. The
following table tells how the goods produced by
sectors R,S,T are interrelated. We only use
three sectors. The real economy uses about 40
sectors.
R S
R 50 20
S 20 30
T 30 20
T
40
20
20
Consumer
70
90
50
Total
180
160
120
The total produced by sector R was 180 units. Of
this, the consumer got 70, the rest was used up
by the other industries. In this data the columns
are what is consumed. Think of the Consumer
column. This tells how much of R,S,T are
eaten by the consumers.
The R column tells you how much Plant R eats.
The COLUMN says that in the process of
making 180 units, the plant R used:
50 units of its own output,
20 units of the S output, and
30 units of the T output
To make 1 unit of R, Plant R uses
50/180 of R,
20/180 of S, and
30/180 of T.
Divide the COLUMN under R by the total R output.
Divide the COLUMN under S by the total S output.
Divide the COLUMN under T by the total T output.
| 50/180 20/160 40/120|
A = | 20/180 30/160 20/120|
| 30/180 20/160 20/120|
To make x units of R, and y units of S and z units
of T requires internal consumption of
| 50/180 20/160 40/120| |x|
| 20/180 30/160 20/120| |y|
| 30/180 20/160 20/120| |z|
| 60 |
Now, suppose that we wish to provide | 110 |
| 60 |
for the customers.
|x|
What values of production | y | should be ordered so
|z|
| 60 |
that after internal consumption, | 110 | remains for the
| 60 |
customers.
|x|
Of the total output of | y | some has to be used as
|z|
internal consumption and the rest goes to demand.
We write AX+D = X.
The left hand side is the internal consumption
plus what the consumers use.
The right hand side is the total produced.
In matrix form:
| 50/180 20/160 40/120| |x|
| 60 |
|x|
| 20/180 30/160 20/120| |y| + |110 | = | y |
| 30/180 20/160 20/120| |z|
| 60 |
|z|
the new
demand
To solve the system AX+D=X for the new
demand vector one does these steps.
AX+D=X
D = X-AX
D = IX-AX
D = (I-A)X
(I-A) -1 D = X.
Hand-in-homework
The production of the plants R, S, and T for some period of time
is given below.
R
R 10
S 10
T 60
S
10
20
20
T Consumer Total
30
30
80
10
20
60
10
10
100
The Leontief input-output model for the open model is:
X = AX + D.
(1) What is the matrix A?
(2) Solve the equation X = AX+D for X
| 145 |
when D = | 290 |.
|145 |
If you do not have a calculator to do the
computations, simply leave the answer in some
expression involving A,I,D.
42. If two nxn matrices A and B have the
same rank, then they must be similar.
False.
| 1 0 | and | 0 1 | both have rank
|0 0|
|00|
one, but are not similar.
43. If A is similar to B, and A is invertible,
then B must be invertible as well.
True.
If P-1 A P = B then
P-1 A-1 P = B-1
44. If A2 = 0 for a 10x10 matrix A, then the
inequality rank(A) <= 5 must hold.
True. 10 = rank(A) + nullity(A)
Since A is contained in the null space of A,
10 >= 2 rank(A). So rank(A) <= 5.
45. For every subspace V of R3 there is a 3x3
matrix A such that V = im(A).
True. Just pick 3 vectors which span V. Use
these as the columns of the matrix.
46. There is a nonzero 2x2 matrix A that is
similar to 2A.
True.
| 2 0| |01| |½ 0| =|0 2|
| 0 1| |00| |0 1| | 0 0|
47. If the 2x2 matrix R represents the
reflection across a line in R2, then R must be
similar to the matrix | 0 1 |.
| 1 0 |
True. Use the basis
| /
_____|/_______
|\
| \
48. If A is similar to B, then there is one and
only one invertible matrix S such that S-1
A S = B.
False.
(A-1S)-1 A (A-1 S) will also work.
49. If the kernel of a 5x4 matrix A consists
of the zero vector alone, and if AB = AC for
two 4x5 matrices B and C, then the
matrices B and C must be equal.
True. A(B-C) = 0 so B-C = 0 and so B=C.
50. If A is any nxn matrix such that A2 = A, then
the image of A and the kernel of A have only the
zero vector in common.
True. If A(AV) = 0, that is, if AV is in the image
of A and also in the kernel of A, then
0 = A2V = AV.
51. There is a 2x2 matrix A such that A2 =/= 0
and A3 = 0.
False. If A2 V =/= 0, and A3 V = 0, then
V, AV, A2V must be linearly independent.
This is impossible in R2.