| Title: | Mathematics at DEC |
| Moderator: | RUSURE::EDP |
| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
A problem in Linear Algebra which demonstrates use of CaYley
Hamiltion Theorem (Every matrix satisfies it own characteristic
equation)
GIVEN a 2 x 2 matrix A
2
a polynomial P(x)= x + 2x + 1
If matrix A where used in this polynomial,ie P(A)= A +2A+I
the resultant MATRIX (call it B) has eigenvalues 4 and 1 .
Now we are given a new polynomial---f(x)= x
------ = x
2 ----
x +2x+1 P(x)
THEN, THE FOLLOWING RELATIONSHIP is GIVEN
3 2
f(A)= k3A + k2A + k1 A + k0I
QUESTION: find scalars k3,k2,k1,k0 ???
note A is original 2 x 2 matrix..
Here is how I approached the problem but ran into difficulty--
I thought it useful to know the characteristic equation of A..If
I had the characteristic equation of A,I could determine relationships
between A,A*A,A inverse etc...using the Cayley Hamilton
So given that P(A) had eigenvalues 4 and 1, I know the eigenvalues
of A by solving this equation:
2
since P(A) was derived from x +2X+1 its eigenvalues were
also dervived from this relationship!!
Let us call the Eigenvalues of A L1 and L2
therefore the Eigenvalues of P(A) were derived as follows:
2 2
L1 + 2L1 +1 = 4 and L2 +2L2 +1 = 1
the above is simply a substitution
into the original polynomial P(x)
solving for L1 and L2 we get:
L1= -3 L2= -2
+1 0
Well at this point I'm at a little bit of a quandry...what to do
with 4 eigenvalues when I only need 2 to get characteristic
equation of A . Here I make a decsion to use -3 and -2
because they generate eigenvalues of P(A)
Now I can get characteristic equation of A
Call new lamda L3
(L3+3)(L3+2)=0
2
yields L3 +5L3 +6=0
now using CAYLEY HAMILTON
2
A+5A+6I= 0
2
or A = -5A-6I
2 2
A = AA = -5A -6A= 19A+30I
3 2
MY PLAN AFTER THIS WAS TO SUBSTITUE THE VALUES FOR A AND A INTO
3 2
f(A) = k3A +k2A+k1A +k0I
but f(A)= A
--------
2
A +2A+I
now I have equation
A
------ = k3(19A-30I)-k2(5A+6I)+k1A +k0I
-3A-5I
Can't solve this equation ...don't know how to clear the denominator??
Is my approach correct???
Thanks
JOHN
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 1203.2 | problem stated verbatim | DUGGAN::J_FERRARA | Mon Mar 05 1990 12:47 | 28 | |
Okay ..here is the problem verbatim
Given A a 2 x 2 matrix
2
Given P(x) = x +2x +1
Given P(A) has eigenvalues 4 and 1
Given f(x)= x
---
P(x)
3 2
then f(A)= k3A + k2A +k1A+k0I where k3,k2,k1 and k0 are scalars
Find k3,k2,k1,k0?
2. Assume P(A) = D a diagonal matrix (|d11|>|d22|) Find an A (maybe
more than one A exists)
3 Given k's found above does A 3 2
----- =k3A +k2A +k1A +k0I for all A??
that's it..
j
P(A)
| |||||
| 1203.3 | HPSTEK::XIA | In my beginning is my end. | Mon Mar 05 1990 14:51 | 28 | |
Well, the way, I am going to do it is using the brute force
undeterminded coefficient method. As usual, I am quite lazy about the
details, so I am just going to find k0 for ya and let ya do the rest.
Well, we first let P(A) = [4 0]
[0 1]
and one choice for A then is [1 0]
[0 0]
Ok, so A/P(a) = [1/4 0]
[0 0]
So we have:
k3 + k2 + k1 + k0 = 1/4 and
k0 = 0
So we have found k0 = 0 and the relation: k3 + k2 + k1 = 1/4.
Now all you have to do is to come up with other A's that will
give similar equations. Make sure the A's you come up has some off
diagnol elements; otherwise, it is no good.
Eugene
| |||||
| 1203.4 | is this therefore true? | DUGGAN::J_FERRARA | Mon Mar 05 1990 15:44 | 18 | |
Since your selection of A= 1 0
0 0
3 2
then A = 1 0 and A = 1 0
0 0 0 0
therefore
1/4 0 = k3 [1 0] + k2[1 0] +k1 [ 1 0]
0 0 [0 0] [0 0] [ 0 0] doesn't this require
k3 = 1/12 k2= 1/12 k1 = 1/12 ???
john
| |||||
| 1203.5 | AITG::DERAMO | Dan D'Eramo, nice person | Mon Mar 05 1990 19:00 | 4 | |
I don't believe his choice determines k3, k2, and k1. That's why you have to keep trying with other A's. Dan | |||||
| 1203.6 | AITG::DERAMO | Dan D'Eramo, nice person | Mon Mar 05 1990 19:51 | 29 | |
1) If P(A) has eigenvalues 4 and 1, then P(A) satisfies the characteristic equation with roots 4 and 1, i.e., (x-4)(x-1) = x^2 - 5x + 4 yields the zero matrix when you substitute P(A) for x. So (P(A))^2 - 5P(A) + 4I = 0, where I is the identity matrix. This gives P(A) (-P(A) + 5I)/4 = I, or 1/P(A) = (-P(A) + 5I)/4, and f(A) = A/P(A) = A(-P(A) + 5I)/4 = A(-A^2 - 2A - I + 5I)/4 = (-1/4)A^3 + (-1/2)A^2 + A, from which you read off k3 = -1/4, k2 = -1/2, k1 = 1, k0 = 0. 2) Try diagonal A = (a 0) (0 b) Then P(A) = (a^2 + 2a + 1 0) = (4 0) (0 b^2 + 2b + 1) (0 1) (as P(A) is diagonal with eigenvalues 4 and 1 and the upper left entry greater in absolute value than the lower right entry). Solving yields |a+1| = 2 and |b+1| = 1, so four examples are ( 1 0) ( 1 0) (-3 0) (-3 0) ( 0 -2) ( 0 0) ( 0 -2) ( 0 0) 3) I doubt that A/P(A) = (-1/4)A^3 + (-1/2)A^2 + A for all A, because those numbers were derived assuming that P(A) has eigenvalues 4 and 1. Choose an A with P(A) having different eigenvalues to find trial counterexamples. Let A = 2I, then P(A) = 7I and f(A) = A/P(A) = (2/7)I but (-1/4)A^3 + (-1/2)A^2 + A = -2I. So A = 2I is a counterexample. Dan | |||||
| 1203.7 | A's with non zero eigenvectors?? | DUGGAN::J_FERRARA | Tue Mar 06 1990 10:25 | 8 | |
Responding to Dan's solution...It part 2.. the A's you suggested
ie 1 0 -3 0
0-2 0-2 etc satisfy P(A)= Diagonal matrix 4 0
0 1
Would it be possible to find an A that has at least 2 eigenvectors??
the above A's all have zero eigenvectors..
| |||||
| 1203.8 | AITG::DERAMO | Dan D'Eramo, nice person | Tue Mar 06 1990 11:03 | 5 | |
They don't have zero eigenvectors. I thought their unit eigenvectors would be (1) and (0). (0) (1) Dan | |||||
| 1203.9 | You're right.. | DUGGAN::J_FERRARA | Tue Mar 06 1990 11:28 | 8 | |
re .8
that's right 4 0 1 0 4 0 1 0
0 1 = 0 1 0 1 0 1
S D S inverse
Thanks Dan
J
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