Hi Sir
Following on from the 2006 paper qs 4c) I did try to turn the vector space into a matrix by using (xI – D), but I could only think of
(a0 0 0)(x^0)
0 a1 0 (x^1)
0 0 a2 (x^2)
The way I would find a Jordan basis for a 3×3 matrix is to find the det ch(x)= det(xI-A) and then find the m(x)=(A-lamdaI)^b where b is the power that gives m(x)=0. From their the Jordan canonical form is simple, but I can’t see how this relates to this problem, and how D is involved.
Thanks for your help
Kind regards
Reply:
The way you phrase things is rather worrisome. First of all, you can’t `turn a vector space into a matrix.’ Also, it’s hard to make sense of the matrix you’ve written down, even taking into account the difficulty of using precise notation here. Perhaps you should review some notions from Algebra 1 and 2, especially the relation between matrices and linear maps.
That is, what you *can* turn into a matrix is a *linear map*. This is done by choosing a basis. Let me just tell you the result in the case of your question and you should try your best to figure out how it’s obtained.
If you choose the basis B={1,x,x^2} for the vector space of complex polynomials, then the matrix representing D becomes
-4 0 2
0 -4 0
0 0 -4
You can find the JCF for this and a Jordan basis (for C^3), and then translate it back into the original problem. For example, if one of the basis elements is
0
1
0
Then for the original vector space, the corresponding basis element is x.
