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Characteristic/Minimal polynomials - Linear Algebra question

  • Thread starter Locoism
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  • #1
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Homework Statement


Given the matrix:

0 1 0
0 0 1
12 8 -1

(sorry I don't know how to put proper matrix format)

a) find polynomials a(λ)(λ+2)2+b(λ)(λ-3) = 1 (where a(λ) and b(λ) are the polynomials)


The Attempt at a Solution



Well the characteristic polynomial is already given, but can easily be found, as
(λ+2)2(λ-3)
but how do I find such polynomials? Is this what is meant by the minimal polynomial?
 

Answers and Replies

  • #2
Ray Vickson
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Homework Statement


Given the matrix:

0 1 0
0 0 1
12 8 -1

(sorry I don't know how to put proper matrix format)

a) find polynomials a(λ)(λ+2)2+b(λ)(λ-3) = 1 (where a(λ) and b(λ) are the polynomials)


The Attempt at a Solution



Well the characteristic polynomial is already given, but can easily be found, as
(λ+2)2(λ-3)
but how do I find such polynomials? Is this what is meant by the minimal polynomial?
I don't understand. First you say that the characteristic polynomial "can easily be found, as (x+2)^2 *(x-3)" [using 'x' instead of 'lambda'], but then you say "but how do I find such polynomials?". So, do you, or do you not know how to find the characteristic polynomial?

As to minimal polynomial: well, it is a theorem that a matrix satisfies its own characteristic polynomial. For you matrix A the characteristic polynomial is C(x) = x^3 + x^2 - 8x - 12, so A satisfies A^3 + A^2 - 8A - 12I (where I = 3x3 identity matrix). In other words, the polynomial C "annihilates" A: C(A) = 0. The minimal polynomial of A is the polynomial of least degree that annihilates A (and has leading coefficient = 1). Sometimes the characteristic polynomial is the minimal polynomial, and sometimes not. For example, the 3x3 identity matrix I = [[1 0 0],[0 1 0], [0 0 1]] has characteristic polynomial C(x) = x^3 - 1, but has minimal polynomial x-1 (because A - 1*I = 0 when A = I).

One algorithm to find the minimal polynomial is that used by Maple: regard I, A, A^2 and A^3 as 9-dimensional vectors, obtained by writing the rows side-by side (so A <--> [0 1 0 0 0 1 12 8 -1], etc.). The fact that A satisfies the characteristic polynomial means that A^3 is a linear combination of I, A and A^2 with known coefficients. By doing linear algebra row operations, we can find the smallest number of rows among I, A, A^2 and A^3 such that A^k is a linear combination of I, A,..., A^(k-1), and that linear combination gives the minimal polynomial.

RGV
 
  • #3
360
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I don't understand. First you say that the characteristic polynomial "can easily be found, as (x+2)^2 *(x-3)" [using 'x' instead of 'lambda'], but then you say "but how do I find such polynomials?". So, do you, or do you not know how to find the characteristic polynomial?
I think Locoism was referring to part a):
a) find polynomials a(λ)(λ+2)2+b(λ)(λ-3) = 1 (where a(λ) and b(λ) are the polynomials)
These can be found using the Euclidean algorithm: http://en.wikipedia.org/wiki/Euclidean_algorithm#Polynomials.
 
  • #4
81
0
Yes, sorry that was a little ambiguous.
Thank you very much spamiam, if I understand correctly, I just need to apply the algorithm to find the gcd of (x+2)^2 and (x-3)?
 
  • #5
360
0
Well, you're actually using the Euclidean algorithm followed by reverse substitution. I probably should have given you http://en.wikipedia.org/wiki/Extended_Euclidean_algorithm" [Broken] link instead. For simplicity, I'll give you an example using integers instead of polynomials.

Consider 7 and 19. They are relatively prime, so gcd(7,19)=1. By http://en.wikipedia.org/wiki/B%C3%A9zout%27s_identity" [Broken] there are integers a and b such that 7a + 19b = 1. Using the Euclidean algorithm, we get

19=2(7) +5
7=1(5) +2
5=2(2)+1

Solving these equations for the remainders, we have

1=5-2(2)
2=7-5
5=19-2(7)

giving us
[tex]
1=5-2 \cdot 2 = 5-2(7-5)=3 \cdot 5 - 2 \cdot 7 = 3(19-2 \cdot 7) -2 \cdot 7 = 3\cdot 19 -8 \cdot 7 \, .
[/tex]
The process is analogous for polynomials, I believe.
 
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  • #6
81
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Thank you so much!
Ok so I end up with a(λ) = 1/25 and b(λ) = -1/25(λ-7) (it works out)
So if I use these with the matrix A as in
E1 = 1/25(A+2)2
E2 = -1/25(A-7)(A-3)
and I want to calculate the Kernel, would that just be (A-3) and (A+2)2 respectively seeing as E1*E2 = 0? And then ImageE1 would just be KernelE2 and vice-versa?
 
  • #7
360
0
I'm not quite sure I follow... Typically you just need to set up [itex] E_1 x = 0[/itex] and solve for x to compute the kernel. What is the question you're trying to answer?
 

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