math 307 spring, 2003 hentzel time: 1:10-2:00 mwf room: 1324 howe hall
DESCRIPTION
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: [email protected] http://www.math.iastate.edu/hentzel/class.307.ICN Text: Linear Algebra With Applications, - PowerPoint PPT PresentationTRANSCRIPT
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Math 307Spring, 2003
Hentzel
Time: 1:10-2:00 MWFRoom: 1324 Howe Hall
Instructor: Irvin Roy HentzelOffice 432 Carver
Phone 515-294-8141E-mail: [email protected]
http://www.math.iastate.edu/hentzel/class.307.ICN
Text: Linear Algebra With Applications, Second Edition Otto Bretscher
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Monday, Mar 3 Chapter 4.2 Page 164 Problems 6,14,16
Main Idea: Dejavu! We are doing everything all over again.
Key Words: Kernel, Image, Linear Transformation, rank, nullity
Goal: We want to expand the ideas of Matrices into a broader context.
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Previous Assignment
The Leontief Problem
The production of the plants R, S, and T for some period oftime is given below.
R S T Consumer Total R 10 10 30 30 80 S 10 20 10 20 60 T 60 20 10 10 100
The Leontief input-output model for the open model is: X = AX + D.
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(1) What is the matrix A?(2) Solve the equation X = AX+D for X | 145 | when D = | 290 | | 145 |
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A | 10/80 10/60 30/100 | | 10/80 20/60 10/100 | | 60/80 20/60 10/100 |
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Check with original system:
A X + D = X |10/80 10/60 30/100|| 80| |30| | 80| |10/80 20/60 10/100|| 60| + |20| = | 60| |60/80 20/60 10/100||100| |10| |100|It checks.
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Calculating X the new system where consumer | 145 |demands is | 290 | | 145 |
AX+D=X
D = (I-A)X
(I-A) -1 D = X
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-1 || 1 0 0 | |10/80 10/60 30/100| | |145| |616||| 0 1 0 | - |10/80 20/60 10/100| | |290| =|690||| 0 0 1 | |60/80 20/60 10/100| | |145| |930|
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-1 | 7 1 3 | | - -(--) -(---) | | 8 6 10 | | | | 1 2 1 | | 145 | X = | -(-) -- -(---) | | 290 | | 8 3 10 | | 145 | | | | 3 1 9 | | -(-) -(--) --- | | 4 3 10 |
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| 272 24 104 | | ----- ---- ---- | | 145 29 145 | | | | 18 54 12 | | 145 | X = | ---- ---- ---- | | 290 | | 29 29 29 | | 145 | | | | 52 40 54 | | ---- ---- ---- | | 29 29 29 |
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| 616 | X = | 690 | | 930 |
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Previous Assignment
Page 157 Problem 4
Which of the subsets of P2 given in Exercises 1 through 5 are subspaces of P2? Find a basis for those that are subspaces.
t=1{ p(t) | INT p(t) dt = 0} t=0
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t=1It is closed under addition since if INT p(t) dt = 0 t=1 t=0and INT q(t) dt = 0, t=0 t=1 t=1 t=1then INT(p(t) + q(t)) dt = INT p(t) dt + INT q(t) dt t=0 t=0 t=0 = 0 + 0 = 0So p(t) + q(t) is also in the subspace.
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It is closed under scalar multiplication since if c is a number and
t=1 t=1 t=1INT p(t) dt = 0, then INT c p(t) dt = c INT p(t) dt = t=0 t=0 t=0 c 0 = 0 so c p(t) is also in the subspace.
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Now we want to find a basis of the subspace.
{1,x,x2} of P2. If p(x) = a + bx + cx2 is any element of p2,
| | x=1then p(x) is in the space if |ax+bx2 /2 +c x3/3|= 0 | | x=0
That is, if and only if a + b/2 + c/3 = 0
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This is a linear system [ 1 1/2 1/3 | 0 ] in Row Canonical Form.
|a| |-1/2| |-1/3| |b| = u | 1 | + v | 0 | |c| | 0 | | 1 |
So the basis is -1/2 + x, and -1/3 + x2.
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Page 157 Problem 10
Which of the subsets of R3x3 given in Exercises 6 through 11 are subspaces of R3x3.
| 1 |The 3x3 matrices A such that vector | 2 | is in the
kernel of A. | 3 |
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The set is closed under addition since if AV = 0, and BV = 0, then
(A+B)V = AV + BV = 0 so A+B is also in the set.
The set is closed under scalar multiplication since if c is a number and AV = 0, then
(cA)V = c(AV) = c0 = 0 so cA is also in the subset.
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A basis of the space is
| a b c | | 1 | | 0 || d e f | | 2 | = | 0 || g h i | | 3 | | 0 |
a + 2 b + 3 c = 0d + 2 e + 3 f = 0g + 2 h + 3 i = 0
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x1 x2 x3 x4 x5 x6 a b c d e f g h i RHS |1 2 3 0 0 0 0 0 0 0 | |0 0 0 1 2 3 0 0 0 0 | |0 0 0 0 0 0 1 2 3 0 |
|a| |-2| |-3| | 0| | 0| | 0| | 0| |b| | 1| | 0| | 0| | 0| | 0| | 0| |c| | 0| | 1| | 0| | 0| | 0| | 0| |d| | 0| | 0| |-2| |-3| | 0| | 0| |e| = x1 | 0| + x2 | 0| + x3 | 1| +x4 | 0| +x5 | 0|+x6 | 0| |f| | 0| | 0| | 0| | 1| | 0| | 0| |g| | 0| | 0| | 0| | 0| |-2| |-3| |h| | 0| | 0| | 0| | 0| | 1| | 0| | i| | 0| | 0| | 0| | 0| | 0| | 0|
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| -2 1 0 | | 0 0 0 | | 0 0 0 || 0 0 0 | | -2 1 0 | | 0 0 0 || 0 0 0 | | 0 0 0 | | -2 1 0 |
| -3 0 1 | | 0 0 0 | | 0 0 0 || 0 0 0 | | -3 0 1 | | 0 0 0 || 0 0 0 | | 0 0 0 | | -3 0 1 |
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Page 157 Problem 20 Find a basis for each of the spaces in Exercises
16 through 31 and determine its dimension.
The space of all matrices A = | a b | in R2x2
such that a+d = 0. | c d |
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The basis written as 2x2 matrices is: | 0 1 | | 0 0 | |-1 0 | | 0 0 | | 1 0 | | 0 1 |
The dimension is 3.
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New MaterialA linear transformation T:V W requires
(1) V and W to be vector spaces.
(2) T(V1+V2) = T(V1)+T(V2).
(3) T(c V) = c T(V).
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_________ ____________ | V | | W | | | | iiiii | | kk | T |iiiiiiii | | kkkk |------------>|0iiiiii | | kkk | |iiii | | | | ii | |_________| |____________|
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Points of interest.
(4) Ker(T) = { V | T(V) = 0}.
(5) Im(T) = {T(V) | V is in V}
(6) T is invertible is equivalent to
(a) T(V) = 0 ==> V = 0. Ker(T) = 0
(b) T(X) = W is solvable for every W. Im(T)=W
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Main Theorem: Dim(V) = Dim( IMAGE ) + Dim(KERNEL).
Pretty much that is what you would expect. The quantity you started with is how much you have left plus how much was lost.
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We now discuss the kernel and the range for various functions.
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Example 1.
For differentiation, what is the kernel?
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The kernel is the set of all functions which map to zero.
That is, those functions whose derivative is 0. Those are the constant functions like:
y = constant.
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The range is the set of all functions which are derivatives of something.
This includes all continuous functions.
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Is differentiation invertible?
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No because you cannot recover the constant.
Both y = x2 and y = x2+1 have the same derivative.
Thus having been given only y = 2x, you cannotdetermine the original function.
Thus differentiation does not have an inverse.
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Example 2.
Is integration invertible?
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Let INT f be the indefinite integral of f.
INT f+g = INT f + INT g INT cf = c INT f
So integration satisfies the requirements to be a linear transformation.
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There is one small problem. INT is not a function since as it stands, there is not one, but many possible integrals.
We can patch up this problem by saying INT f is the function F such that F' = f and F[0] = 0.
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Notice that d/dx INT f = f so that INT has a left inverse.
However INT does not have a right inverse since INT d/dx f will only equal f(x) - f(0).
Since you do not get f(x) back exactly, integration does not have an inverse.
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We once remarked that if a matrix has a left inverse, it also has a right inverse and the two inverses are equal.
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This theorem requires that the matrix be square and the space be finite dimensional.
It is not true for non square matrices nor for infinite dimensional spaces.
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Example 3.Consider the function R2x2 ----> R2x2 given by
|a b| | 1 2 | | a b | |c d| | 2 4 | | c d |
Is this function invertible?
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We see if anything is mapped to zero.
That is, is there any matrix | a b | | c d |
such that | 1 2 | | a b | = 0 ? | 2 4 | | c d |
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Thus a+2c = 0 b+2d = 0 2a+4c = 0 2b+4d = 0
a b c d RHS a b c=u d=v RHS | 1 0 2 0 0 | | 1 0 2 0 0 | | 0 1 0 2 0 | | 0 1 0 2 0 | | 2 0 4 0 0 | | 0 0 0 0 0 | | 0 2 0 4 0 | | 0 0 0 0 0 |
| a | |-2 | | 0 | | b | = u | 0 | + v |-2 | | c | | 1 | | 0 | | d | | 0 | | 1 |
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The kernel contains |-2u -2v | | u v |
So since some information is lost, the function is not invertible.
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Define T[X] = AX-XA for a fixed matrix A.
1. Show that T is a linear transformation.
2. Show that T is never invertible.
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We have to first show that T[X+Y] = T[X]+T[Y].
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T[X+Y] = A(X+Y) – (X+ Y)A
= AX+AY – XA – YA
= AX – XA + AY – YA = T[X] + T[Y].
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T[cX] = A(cX) + (cX)A
= c(AX+XA)
= cT[X].
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The kernel of T contains the identity matrix I so T is not invertible.
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What is the matrix for T if A = | 1 2 | | 3 4 |
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|1 0| |0 1| |0 0| |0 0| |0 0| |0 0| |1 0| |0 1||1 0| |1 2| - |1 0| = | 0 2| | 0 2 -3 0 || 0 0| |0 0| |3 0| |-3 0| | | | ||0 1| |3 4| -| 0 1| = | 3 3| | 3 3 0 -3 ||0 0| |0 0| | 0 3| | 0-3| | | | ||0 0 | |0 0| - | 2 0| =|-2 0| | -2 0 -3 2 ||1 0 | |1 2| | 4 0| |-3 2| | | | ||0 0| |0 0| - | 0 2| =| 0 -2| | 0 -2 3 0 ||0 1| |3 4| | 0 4| | 3 0| | |
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The matrix for T is
| 0 3 -2 0| | 2 3 0 -2| |-3 0 -3 3| | 0 -3 2 0|
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What is the kernel of T?
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|0 3 -2 0| |-3 0 -3 3| |1 0 1 -1||2 3 0 -2| | 0 3 -2 0| |0 3 -2 0||-3 0 -3 3| | 2 3 0 -2| |2 3 0 -2||0 -3 2 0| | 0 -3 2 0| |0 -3 2 0|
|1 0 1 -1| |1 0 1 -1| | 1 0 1 -1 ||0 3 -2 0| |0 3 -2 0| | 0 1 -2/3 0 ||0 3 -2 0| |0 0 0 0| | 0 0 0 0 ||0 -3 2 0| |0 0 0 0| | 0 0 0 0 |
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x y z=a w=b | 1 0 1 -1 | | 0 1 -2/3 0 | | 0 0 0 0 | | 0 0 0 0 |The kernel is | x | | -1 | | 1 | | y | = a |2/3| + b | 0 | | z | | 1 | | 0 | | w | | 0 | | 1 |
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So the elements in the kernel are linear combinations of
| -1 2/3 | and | 1 0 | | 1 0 | | 0 1 |
I.E. linear combinations of A and I.
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Example 4. Suppose we have a linear transformation Twhich reflects across a line. What is the matrix of T with respect to the standard basis? | 1 | | 0 | | 0 | | 1 |
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Case 1. /|\ / | / y = x | / | / -----------------------+----------------------- /| / | / |
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Case 2. \ /|\ \ | y = -x \ | -----------------------+----------------------- | \ | \