Diagonalization with nilpotent matrices

[wwm]

So my professor gave me an extra problem for Linear Algebra and I can't find anything about it in his lecture notes or textbooks or online. I think I've made it through some of the more difficult stuff, but I am running into a catch at the end.

Homework Statement

Find [;T(p(x))^{500};] when [;T(p(x))= p(x)-2p'(x)+3p''(x);] and [;p(x)=\alpha+\beta x+\gamma x^{2};] where [;p'(x);] is the first derivative, etc.

The Attempt at a Solution

A is a matrix representing the transformation. [;T(y)^{500}=A^{500}y;]

Normally we would simply find A by plugging in the std. basis and then find a diagonal matrix, D, similar to A s.t. [;A^{500}=SD^{500}S^{-1};] with D constructed out of the eigenvalues and the transition matrix, S, constructed out of the corresponding eigenvectors.

The trick is that this matrix (and all upper or lower triangle matrices) are not properly diagonizable. We find this out in the last step mentioned because the dimension of the eigenvectors (in this case, 2) do not equal the multiplicity the eigenvalues (in this case 3).

Our professor says, then, that we must add a nilpotent matrix to our equation to get [;A^{500}=S(D+N)^{500}S^{-1};] where N is the nilpotent matrix made of the non-diagonal numbers from our original matrix A. Because the nilpotent matrix goes to zero after several powers, the [;(D+N)^{500};] can be reduced to only the first several terms, which is still a bit hairy, but not too bad. so we finally get our matrix [;(D+N)^{500};], great, hard part over.

2. Relevant questions

(1) But now that we are ready to do our calculations I realize that I don't have enough eigenvectors to create S and thus its inverse as well. Where do I get the last eigenvector?

(1a.) A friend suggested that I don't need the transition matrices since in this case the diagonals in A are the eigenvalues. Is that correct? More importantly, if that wasn't the case where would I get them (I expect a similar problem on my final)?

(1b.)My professor said I need to solve the non-homogenous system using the previous eigenvectors, but I am not sure I fully understand it. What I copied down from him was [;(A-\lambda I)u1=0;] and [;(A-\lambda I)u2=u1;] where u1 is one of the original eigenvectors and u2 is our new one. However, when I go through this process I get the same two eigenvectors, no new third one. What am I doing wrong?

(2) Does the method I described above to tackle the problem make sense? I am going about it correctly?

 

[wwm]

also, even if you don't know how to solve the problem, if you know of anywhere I might be able to find more information on diagonalization using nilpotent matrices, that would be a huge help too.

 

[Dick]

You can't diagonalize a nilpotent matrix, unless it's the zero matrix. And you don't really even need the eigenvectors. Write out the matrix in the standard basis {1,x,x^2}. What is it? You can write that as A=(I+N) where I is the identity matrix and N is nilpotent. In fact, N^3=0. Now you can pretty easily work out (I+N)^500 by using the binomial theorem and ignoring powers of N greater than two.

 

[wwm]

Thanks so much for your help. It is sincerely appreciated.

You can't diagonalize a nilpotent matrix, unless it's the zero matrix.

right, we aren't trying to diagonalize a nilpotent. Rather we are trying to diagonalize A using a nilpotent to assist us.

And you don't really even need the eigenvectors. Write out the matrix in the standard basis {1,x,x^2}. What is it? You can write that as A=(I+N) where I is the identity matrix and N is nilpotent. In fact, N^3=0. Now you can pretty easily work out (I+N)^500 by using the binomial theorem and ignoring powers of N greater than two.

This is what's confusing me. I get the binomial theorem aspect for (I+N)500, canceling the terms after N^2. But when we normally are trying to solve a problem like this we have A^500=S(D^500)S^-1. Here we use the nilpotent since it is not properly diagonizable, so we get A^500=S(D+N)^500S^-1. I understand how to deal with the (D+N)^500 part, but we still need S (and S^-1). Those are normally created with the eigenvectors, but we don't have enough eigenvectors. As I mention in (1a) we might not need S and S inverse in this particular case, although even that is not clear to me, but I want to understand what I would do if D=/=I and I did need to use S and S^-1.

 

[Dick]

Thanks so much for your help. It is sincerely appreciated.
right, we aren't trying to diagonalize a nilpotent. Rather we are trying to diagonalize A using a nilpotent to assist us.
This is what's confusing me. I get the binomial theorem aspect for (I+N)500, canceling the terms after N^2. But when we normally are trying to solve a problem like this we have A^500=S(D^500)S^-1. Here we use the nilpotent since it is not properly diagonizable, so we get A^500=S(D+N)^500S^-1. I understand how to deal with the (D+N)^500 part, but we still need S (and S^-1). Those are normally created with the eigenvectors, but we don't have enough eigenvectors. As I mention in (1a) we might not need S and S inverse in this particular case, although even that is not clear to me, but I want to understand what I would do if D=/=I and I did need to use S and S^-1.

You don't need S and S^(-1) since you aren't trying to diagonalize, because you know you can't. A^(500)=(I+N)^500. Just work out (I+N)^500 DIRECTLY. If D isn't equal to I, then you can just use the binomial theorem, like before (assuming they commute). Diagonalization (or a change of basis) is no help here. Call S=I if you really want to have an S in the problem.

 

[Dick]

right, we aren't trying to diagonalize a nilpotent. Rather we are trying to diagonalize A using a nilpotent to assist us.

That's not what you are trying to do. You are trying work out a large power of A using that it's the sum of a simple matrix and a nilpotent matrix. You can't diagonalize A any more than you can diagonalize the nilpotent. Put diagonalization out of your mind for this problem.

 

[wwm]

That's not what you are trying to do. You are trying work out a large power of A using that it's the sum of a simple matrix and a nilpotent matrix. You can't diagonalize A any more than you can diagonalize the nilpotent. Put diagonalization out of your mind for this problem.

That's interesting, because in class when we did similar problems but D=/=I we had to use the transitions matrices (S and S^-1) in order to get the correct answer, but its possible that I am misunderstanding.

For example, when we have [; A = \left[ \begin{array}{ccc}3 & 1 & 0 \\
0 & 3 & 0 \\
0 & 0 & 4 \end{array}\right];] and we are looking for e^A.

In class our professor used [;e^{A}=Se^{D+N}S^{-1};] where D is a matrix with the eigenvalues in the diagonal, N is a matrix with the non-diagonal values in A and S and S^-1 are formed using the eigenvectors. If I do that problem without using S and S^-1 i get a significantly different answer right? In that case I need to be able to find additional eigenvectors beyond [;(A - \lambda I);] offers for each [;\lambda;].

Perhaps at this point we are not doing diagonalization, strictly speaking (but we are find D the same way, correct?), but the process is similar except that I don't know how to form S without additional eigenvectors.

 

[Dick]

That's interesting, because in class when we did similar problems but D=/=I we had to use the transitions matrices (S and S^-1) in order to get the correct answer, but its possible that I am misunderstanding.

For example, when we have [; A = \left[ \begin{array}{ccc}3 & 1 & 0 \\
0 & 3 & 0 \\
0 & 0 & 4 \end{array}\right];] and we are looking for e^A.

In class our professor used [;e^{A}=Se^{D+N}S^{-1};] where D is a matrix with the eigenvalues in the diagonal, N is a matrix with the non-diagonal values in A and S and S^-1 are formed using the eigenvectors. If I do that problem without using S and S^-1 i get a significantly different answer right? In that case I need to be able to find additional eigenvectors beyond [;(A - \lambda I);] offers for each [;\lambda;].

Perhaps at this point we are not doing diagonalization, strictly speaking (but we are find D the same way, correct?), but the process is similar except that I don't know how to form S without additional eigenvectors.

That matrix has only two linearly independent eigenvectors, so I don't know what you are using for S. If you are trying to find exp(A) you should get the answer no matter what you do as long as it's correct. But that's a whole different problem. Why aren't you just trying to solve the given problem? (I+N)^500. Just do it. It's EASIER than a diagonalization problem. Do it DIRECTLY.

 

[wwm]

That matrix has only two linearly independent eigenvectors, so I don't know what you are using for S. If you are trying to find exp(A) you should get the answer no matter what you do as long as it's correct. But that's a whole different problem. Why aren't you just trying to solve the given problem? (I+N)^500. Just do it. It's EASIER than a diagonalization problem. DIRECTLY.

Thanks so much for all your help.

As I mentioned in the original post, I've already done (I+N)^500. What I am trying to understand is why we don't use the transition matrices here and if I need them for similar problems how I would go about getting them if there aren't enough eigenvectors.

 

[Dick]

Thanks so much for all your help.

As I mentioned in the original post, I've already done (I+N)^500. What I am trying to understand is why we don't use the transition matrices here and if I need them for similar problems how I would go about getting them if there aren't enough eigenvectors.

If by 'transition matrices' you mean the S matrices, sometimes it's useful to choose a different basis. Which is what the S's are doing. Sometimes it's not. If there aren't enough eigenvectors then you can't diagonalize, period. You can't fix the lack of eigenvectors. They won't magically appear. You have to work around it, like we are doing here.

 

[wwm]

I see. Thanks a bunch.