Characteristic Polynomials and Minimal polynomials

xfunctionx · Apr 22, 2009

Hi, there are a few questions and concepts I am struggling with. The first question comes in 3 parts. The second question is a proof.

Question 1: Please Click on the link below

Three-part linear algebra problem: (a) matrices A and B with the same minimal polynomial q but characteristic polynomials p1 = x^3(x-1)^3 and p2 = x^2(x-1)^4; (b) a linear map T on R^3 with eigenvalues 1, 2, 3, find real a1, a2, a3 with I + a1T + a2T^2 + a3T^3 = 0; (c) prove T^2 is diagonalisable if T is.

Question 2: Please Click on the link below

$Handwritten proof sketch: Let T: V→V be linear. Let T(v_i) = λ_i v_i for i = 1,…,k (v_i ≠ 0). If λ_i (i = 1,…,k) are distinct then {v_1,…,v_k} are linearly independent.$

For Q2, could you please show me how to prove this. If possible, could you also link me to a web page where the full proof has already been provided?

I would appreciate the help.

dx · Apr 22, 2009

Hi xfunctionx,

Show us what you've done so far.

xfunctionx · Apr 22, 2009

sure ... let me just write it up

xfunctionx · Apr 22, 2009

Question 1 attempt: I got stuck early.

Handwritten working on ruled paper: p1(x) = x^3(x-1)^3, p2(x) = x^2(x-1)^4, q(x) = x^2(x-1)^2, then p1(x)/q(x) = x(x-1) = x^2-x and p2(x)/q(x) = x^2-2x+1, followed by A^2-2A+I=0 and A^2-A=0.

Question 2 attempt/proof from lecture notes:

I don't understand what my lecturer did right at the end. Or how his conclusion proved anything. Please could you help me understand, or show me a better proof?

Handwritten proof by contradiction that eigenvectors v1 through vk are linearly independent, starting from a dependence relation with fewest nonzero coefficients and applying the linear map T.

Handwritten derivation of a linear combination 0 = sum from i=1 to k of alpha_i v_i, rewritten with eigenvalue weights lambda_i and then manipulated by multiplying by lambda_j and subtracting to give 0 = sum of alpha_i (lambda_i - lambda_j) v_i = 0 when i = j.

Handwritten red text on lined paper reading: So this expression has atleast one more 'a' wsf. than * hence contradicts our assumption.

dx · Apr 22, 2009

For question two:

You can prove this by induction. Clearly, it's true for k = 0 since eigenvectors are always non-zero. Now, to prove it for k = n, we assume it's true for k = n - 1.

Suppose that,

(a_1)(v_1) + (a_2)(v_2) + ... + (a_n)(v_n) = 0

Apply T to the LHS, and you get

(λ_1)(a_1)(v_1) + (λ_2)(a_2)(v_2) + ... + (λ_n)(a_n)(v_n) = 0

Now, multiply the first equation by λ_n, and subtract it from the second. The last term will cancel out. You will be left with

(λ_1 - λ_n)(a_1)(v_1) + (λ_2 - λ_n)(a_2)(v_2) + ... + (λ_(n-1) - λ_n)(a_(n-1))(v_(n-1)) = 0.

By the inductive hypothesis, each of the coefficients here must be zero. Can you show that this implies the the a_i from i=1 to (n-1) must be 0? (Hint: Use the fact that the eigenvalues are distinct). Then the original first equation becomes (a_n)(v_n) = 0, so a_n too must be zero.

dx · Apr 23, 2009

dx said:

Clearly, it's true for k = 0 ...

Sorry, I meant k = 1.

xfunctionx · Apr 23, 2009

dx said:

Sorry, I meant k = 1.

Thank you for your help dx, I will attempt the proof and try to understand it using induction.

xfunctionx · Apr 23, 2009

Can anyone help me with question 1?

VKint · Apr 27, 2009

Have you gone over the rational canonical form of a matrix? Or the Cayley-Hamilton theorem? If you have, question 1 should be straightforward. The answer to part (a) is yes (use the rational canonical form). For (b), use the Cayley-Hamilton theorem. For (c), write D = S^-1 T S, where S is an invertible matrix and D is diagonal. What's D²?

mfarooq52003 · Apr 30, 2009

Hello xf...nx

Your proof of 2nd question which according to you is from Lecture notes is quite straight forward using definition of Linear independence and eigen vectors and the defined LT and given hypothesis of the theorem. It is the best and simplest proof you have. Just read a bit about Linear independence, eigen vectors and linear transformation and you will find that the proof is quite straight forward and simple.

Characteristic Polynomials and Minimal polynomials

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