Question on generalized eigenspaces

Hi Sir,

I’ve completed the computation of the first part of question 3. However I don’t understand how to find the basis for each generalized eigenspace. Couldyou direct me to which area of the notes that I sould read up on or give me an example please.

Thanks

Reply:

Finding a basis for generalized eigenspaces is simply the old problem of finding a basis for the solution space of a system of equations. That is, given a linear map

L:V-> V

the t-th generalized eigenspace V_t(a) for some eigenvalue a is simply the set of solutions v to

[(L-aI)^t]v=0

For any specific L in matrix form and specific eigenvalue a, (L-aI)^t will again be a matrix, and you are asked to find a basis for the solution space to the linear system above. I’m sure you know how to do it. Look also at example 3.4.61 in the notes.

If you still can’t do it, try asking again.

MK

Question on derivative map

Dear Mr Kim,

I am writing to you as I am stuck on this week’s homework sheet and was wondering if you could help me. I am in your 2201 course and I struggle with number 4 on the Sheet 4.

I basically came as far as showing that D^11 = 0 as I only apply my mapping to

f, f’, f”,…

etc and eventually get 0 for the 11th derivative which does make sense. But for the second part it seems unnatural to write down an eigenvalue equation in a matrix form so am I right that in our case we have dg/dx = lambda*g with g being f,f’,f”,f”’,…..so that this equation is only satisfied if we have lambda=0. Am I on the right track here? But then again I dont know how to find the generalized eigenspaces as I havent got a matrix (or do I?)

It would be great if you could give me some hints on this question as it does bother me not to get anywhere with it 🙂

Have a good reading week, all the best,

Reply:

In some sense, you could work directly with the polynomials, but this is an excellent opportunity to review `the matrix associated to a linear map’. How does that go? If

L:V-> V

is a linear map, whenever you choose a basis B={b_1,b_2, …, b_n} for V, it determines a matrix representative for L that can be computed as follows: Compute L(b_1) and write it in terms of the basis B. That is, we will find some coefficients a_{ij} so that

L(b_1)=a_{11}b_1+a_{21}b_2+…_a_{n-1,1}b_{n-1}+a_{n1}b_n

Then

a_{11}
a_{21}
.
.
.
a_{n1}

is the first column of our matrix. Now compute L(b_2) to get the second column, and so on. After the matrix is completely computed, you proceed to compute all eigenvalues and generalized eigenspaces using the matrix. For the question at hand, you of course have to choose a basis first for the polynomial space. Don’t forget to express your final answer *in terms of polynomials*. If that last comment is confusing, think about it a bit. If you’re still stuck, ask again.

MK

Question on minimal polynomial

Dear Prof Kim,

There is an example about minimal polynomials on the online notes for the course 2201 page 20 which I am not sure about it.

I know how to find the characteristic polynomial, (x-2)^3 but I am not clear how to find m(T)=0. Would you please help me ?
Many thanks.

Reply:

Once you have the characteristic polynomial ch(x) , you use the fact that the minimal polynomial m(x) is monic and divides ch(x). In this case, because

ch()=(x-2)^3

this means

m(x)=(x-2)^3, (x-2)^2, or (x-2).

Note that the matrix (T-2) is

0 1 0
0 0 0
0 0 0

(Recall that (T-2) means `T minus 2 times the identity matrix’.)

Of course we know that (T-2)^3=ch(T) must be zero by the Cayley-Hamilton theorem. So we need only check for the possibility that a smaller power might be zero. In fact, it turns out by direct computation that (T-2)^2=0. Of course (T-2) itself is not zero. This tells us that the minimal polynomial is (x-2)^2.

MK