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
Wednesday, March 5, Chapter 4.3
Problems 4,10,50
Page 173
Main Idea: Just look at the coefficients.
Key Words: Representation of a vector with
respect to a basis.
VB
PCB
PCB TBB PBC
Goal: Set up the constructions symbolically
before you try to do them with the actual
matrices.
Previous Assignment
Monday, March 3 Chapter 4.2
Problems 6,14,16
Page 164
Page 164 Problem 6
Find out which of the transformations in Exercises
1 through 25 are linear. For those that are
linear, determine whether they are
isomorphisms.
T(A) = BA where B = | 1 2 |
|36|
The fact that T is linear is a direct result of the fact
that matrix multiplication is linear. We can write
out a few details as follows.
T(X+Y) = B(X+Y) = BX + BY = T(X) + T(Y)
T(cX) = B(cX) = c BX = c T(X)
So T is linear.
To show that T is not invertible, we show that the
kernel is not zero.
Since T( |-2 0| ) = | 1 2 | |-2 0 | = | 0 0 |
| 1 0|
|36||10| |00|
T has a non trivial kernel so T will not be
invertible.
Page 164 Problem 14
T(a+bt+ct2) = a-bt+ct2.
What T does is to change the sign of the linear
term.
The additive part of the proof that T is linear:
T((a+bt+ct2) + (a'+b't+c't2))
= T((a+a') + (b+b')t + (c+c')t2)
= (a+a') -(b+b')t + (c+c')t2
=
a-bt+ct2 +
a'-b't+c't2
= T(a+bt+ct2) + T(a'+b't+c't2).
The scalar multiplication part of the proof that T is
linear.
T( d(a+bt+ct2) )
= T( da + dbt + dct2)
=
da -dbt +dct2
=
d(a-bt+ct2)
= d T(a+bt+ct2)
To show that T is invertible, we will actually find
the inverse of T.
Since T(a-bt+ct2) = a+bt+ct2
the inverse of T exists and is T itself.
Page 164 Problem 16
T(f(t)) = t f '(t).
We first show the additive part of linearity.
T( f(t)+g(t) )
= t ( f(t)+g(t) )'
= t ( f '(t) + g '(t) )
= t f '(t) + t g '(t)
= T( f(t)) + T(g(t)).
We next show the scalar multiplication part of
linearity.
T( c f(t) )
= t (c f(t))'
= c t f '(t)
= c T( f(t) ).
We know that T is not invertible because it has
non-zero elements in its kernel.
T kills off all of the constant functions.
New Material.
The numbers in a vector have to be interpreted through
the appropriate basis. We keep track of the particular
basis that is being used by giving a subscript on the
vector.
(1) If we have a Basis B = [B1 B2 ... Bn] we can express
something in terms of the basis by giving the
coefficients of the B's.
| c1 |
| c2 |
| c3 |
V = [B1 B2 ... Bn] | . | = c1 B1 + c2 B2 + ... + cn Bn.
| . |
| cn |
The coefficient vector VB is called the
representation of V with respect to the basis
[B1 B2 ... Bn].
VB =
| c1 |
| c2 |
| c3 |
| . |
| . |
| cn |
For example:
(a) Represent the polynomial x4 + 3 x + 2 with
respect to the basis [ x4 x3 x2 x 1 ].
|1|
|0|
answer: | 0 |
|3|
|2|
(b) Represent x Sin[x] + 3 Cos[x] with respect to
the basis [ x Sin[x], x Cos[x], Sin[x], Cos[x] ].
|1|
|0|
answer: | 0 | .
|3|
(2) If you choose a different basis for whatever
reason, you have to change the representation
of the vector. This change is easily done by
using a matrix called the change of basis
matrix.
| a11 a12 ... a1n |
| a21 a22 ... a2n |
[B1 B2 ... Bn] | . .
. | = [C1 C2 ... Cn]
| .
.
. |
| an1 an2 ... ann |
You create this matrix column by column.
Column i will be the coefficients needed to make
Ci a linear combination of the B’s.
The matrix is called PBC
To get this straight, derive what you are doing.
| a11 a12 ... a1n | | x1 |
| x1 |
| a21 a22 ... a2n | | x2 |
| x2 |
[B1 B2 ... Bn] | . .
. | | . | = [C1 C2 ... Cn] | . |
| . .
. || . |
| . |
| an1 an2 ... ann | | xn |
| xn |
We multiply the expression by the vector X. This gives
us B A X = C X. On the right hand side, X is a
representation of a vector in the C basis. On the left
hand side AX is the representation of the vector in the
B basis. So X B = A XC . Thus A is P BC
For example.
(a) Find the change of basis matrix for
[ 1 x x2 x3 ]
[ 1 x x2
and [ x3 x2 x 1]
|0001|
x3 ] | 0 0 1 0 | = [ x3 x2 x 1]
|0100|
|1000|
(b) Find the change of basis matrix for
[ ex, e -x]
and
[ Sinh(x), Cosh(x) ].
ex – e -x
Sinh(x) = -------------2
ex + e -x
Cosh(x) = --------------2
[ e x e -x] | ½ ½ | = [ Sinh(x) Cosh(x) ].
|-½ ½ |
(c) Find the change of basis matrix which rotates
the axis sending (1,0) into (5/13, 12/13) and
(0,1) into (-12/13, 5/13)
| 1 0 | | 5/13 -12/13 | = | 5/13 -12/13 |
| 0 1 | | 12/13 5/13 | |12/13 5/13 |
P BB'
What does (3,5) in the B' system correspond to in
the B system?
| 5/13 -12/13 | | 3 | = | -45/13 |
|12/13 5/13 | | 5 |
| 61/13 |
P BB’
VB’
VB
What does the equation 2 x y = 4 become in the B'
basis.
[xy]|0 1 ||x| =4
|1 0 ||y|
[ x' y‘ ] | 5/13 12/13 | | 0 1 | | 5/13 -12/13 | | x' | = 4
|-12/13 5/13 | | 1 0 | |12/13 5/13 | | y' |
[ x’ y’ ] | 120 /169
| -119 /169
- 119/169 | | x’ | = 4
-120/169 | | y’ |
(120/169) (x')2 -238/169 x' y' -120/169 (y')2 = 4
(3) The matrix of a linear transformation is written
with respect to a particular basis. If you change
basis, you have to change the matrix of the
linear transformation.
Suppose T(V) = W and you know how to
compute this linear transformation in the B
basis. Then using matrix multiplication we
know that:
T BB VB = WB
Now we want to represent T in the C Basis.
We first change VC to VB and then use TBB to
get WB and then change WB to WC
VOILA:
PCB TBB PBC VC
|____________|
TCC
= WC
For Example:
Write the matrix for differentiation using the basis
[ ex, e -x ]
and [ Sinh(x), Cosh(x) ]
ex
ex
e-x
1
0
e-x
0
-1
Sinh(x)
Sinh(x) | 0
Cosh(x) | 1
Cosh(x)
1
|
0
|
Using the change of basis matrix.
x
[ e e x ] | 1/2 1/2 | = [ Sinh[ x ] Cosh[ x ] ]
 B  |-1/2 1/2 | 
C

PBC
PCB TBB PBC VC
= WC
| 1 -1 | | 1 0 | | 1/2 1/2 |
| 1 1 | | 0 -1 | |-1/2 1/2 |
| 1 1 | | 1/2 1/2 | = | 0 1 |
| 1 -1 | |-1/2 1/2 | | 1 0 |
Page 175 Problem 61. Let V be the linear
space of all functions of the form
c1Cos[t] + c2Sin[t] + c3t Cos[t] + c t Sin[t].
Solve this differential equation:
f’’ + f = Cos[t]
We are looking to solve (D2+I) f(t) = Cos[t].
|
D=|
|
|
0
-1
0
0
1 1
0 0
0 0
0 -1
0|
1|
1|
0|
We write out the matrix of differentiation with
respect to the basis:
Cos[t]
Sin[t]
t Cos[t]
t Sin[t]
Cos[t]
0
1
1
0
Sin[t]
-1
0
0
1
t Cos[t]
0
0
0
1
t Sin[t]
0
0
-1
0
|
D=|
|
|
0
-1
0
0
|
D2 = |
|
|
1 1
0 0
0 0
0 -1
0|
1|
1|
0|
-1 0 0
0 -1 -2
0 0 -1
0 0 0
2
0
0
-1
|
|
|
|
|
D2 + I = |
|
|
0
0
0
0
0 0
0 -2
0 0
0 0
2
0
0
0
|
|
|
|
We wish to solve (D2+ I) f(t) = Cos[t]
This is just a typical linear equation once
the basis has been chosen.
|
|
|
|
0
0
0
0
0 0
0 -2
0 0
0 0
2
0
0
0
| | c1 |
| | c2 |
| | c3 |
| | c4 |
=
|
|
|
|
1
0
0
0
|
|
|
|
The Row Canonical Form is
c1=a c2=b c3 c4 RHS
| 0
0 1 0 0 |
| 0
0 0 1 ½ |
| 0
0 0 0 0 |
| 0
0 0 0 0 |
| c1 |
| 0|
| 1 |
| 0 |
| c2 | = | 0 |+ a | 0 | + b | 1 |
| c3 |
| 0|
| 0 |
| 0 |
| c4 |
|½|
| 0 |
| 0 |
f(t) = a Cos[t] + b Sin[t] + ½ t Sin[t]
check:
f’(t) = -a Sin[t] + b Cos[t] + ½ Sin[t] + ½ t Cos[t]
f”(t) = -a Cos[t] –b Sin[t] + ½ Cos[t] + ½ Cos[t]
-1/2 t Sin[t]
f”(t) + f(t) = Cos[t]. It checks.
Graph your solution(s).
(The differential equation f” + f = Cos[t] describes
a forced undamped oscillator. In this example,
we observe the phenomenon of resonance.