SOLUTION: We know that if p(lambda) is the characteristic polynomial of an n x x matrix A, then p(A)=0. Define the minimum polynomial m(lambda) of A by requiring: 1)The degree of m(lambda)

Algebra ->  College  -> Linear Algebra -> SOLUTION: We know that if p(lambda) is the characteristic polynomial of an n x x matrix A, then p(A)=0. Define the minimum polynomial m(lambda) of A by requiring: 1)The degree of m(lambda)      Log On


   

Question 22608: We know that if p(lambda) is the characteristic polynomial of an n x x matrix A, then p(A)=0. Define the minimum polynomial m(lambda) of A by requiring:
1)The degree of m(lambda) is > 0, and m(A)=0
2)m(lambda) is monic(leading coefficient is 1) and it divides any polynomial f(lambda) such that f(A)=0.
(a)Prove that every eigenvalue of A is a root of m(lamda).
(b)Find the minimum polynomial for matrix A = [1201]
[2110]
[0012]
[0021]
Answer by khwang(438) About Me  (Show Source):
You can put this solution on YOUR website!
Define the minimum polynomial m(lambda) of A by requiring:
1)The degree of m(lambda) is > 0, and m(A)=0
2)m(lambda) is monic(leading coefficient is 1) and it divides any polynomial f(lambda) such that f(A)=0.
(a)Prove that every eigenvalue of A is a root of m(lamda).
(b)Find the minimum polynomial for matrix A =
[1201]
[2110]
[0012]
[0021]
Here,I use x to replace lambda.
Proof: a) If c is an eigenvalue of A, then Av = cv for some non-zero
(eigen)vector v.
Hence p(A)v = p(c) v for all polynomial p(x).
In particular, by 1), m(A)v = m(c) v = 0.
Since, v is non-zero, we have m(c) =0.
This proves that c is a root of m(x).
[Also, m(x) divides the char. Poly of A.
b)Compute the char. Poly. of A = de(x I –A) =
|x-1 –2 0 -1|
|-2 x-1 –1 0|
|0 0 x-1 –2|
|0 0 -2 x-1| (by expaning alongthe 1st column,we have)
=
=
= %28x%2B1%29%5E2%28x-3%29%5E2+ ... char. poly. of A
By a), we have m(x) = %28x%2B1%29%5Ei+%2A%28x-3%29%5Ej+, 1<=i, j <=2
Compute eigenvector wrt to -1
A + I = [2 2 0 1]
[2 2 1 0]
[0 0 2 2]
[0 0 2 2]
[2 2 0 1] [x y z w]T : 2x + 2y +w = 0
[2 2 1 0] 2x + 2y + z = 0
[0 0 2 2] w +z = 0 --> w=z=0,(x,-x,0,0)
[0 0 2 2]
dim eigenspace of A wrt eigenvalue -1 = 1 (generated by (1,-1,0,0)^T )
note, the geometric mult. the eigenvalue -1 = 1 < 2 = alg. mult. in char. poly.

Then compute eigenvector wrt to 3.
A-3I =
[1 1 0 0]
[-2 2 0 1] [x y z w]T : -2x + 2y +w = 0
[2 –2 1 0] 2x - 2y + z = 0
[0 0 –2 2] -z +w = 0 -->w=z =0,x=y
dim eigenspace of A wrt eigenvalue 3 = 1 (generated by (1,1,0,0)^T )
Also, the geometric mult. the eigenvalue 3 =1 < 2 = alg. mult. in char. poly.
Hence, the min. poly. is %28x%2B1%29%5E2%28x-3%29%5E2+.
Or (A+I)(A-3I) =
[0 0 2 -2]
[0 0 0 4]
[0 0 4 0]
[0 0 0 0]
[0 0 0 0]
%28A%2BI%29%28A-3I%29%5E2 =
[0 0 8 – 8]
[0 0 -8 8]
[0 0 0 0]
[0 0 0 0]
Also, (A+I)^2(A-3I)^2 not 0
Hence, m(x) = %28x%2B1%29%5E2%28x-3%29%5E2+.[same as the char. poly.]

Kenny