Vtyjkyuk

I am currently reading Jim Hefferon's Linear Algebra.

In chapter 5, nilpotence, strings, he goes through the process of finding a string basis of a map, and proves that there exists a string basis for any nilpotent transformation. He then suddenly states that from this string basis, you can create a matrix consisting of only 0's, except subdiagonal 1's in blocks. Also this block-diagonal form, is the canonical form of similar nilpotent matrices. He doesn't go into many details, and latter on the exercises, he says that the canonical form can be "immediately" found just by knowing the amount of strings and their length (not even knowing the actual basis vectors).

So my two questions:
1) Is it obvious that the canonical form is a block-diagonal? Or does it need a rigorous proof that the text simply ignores?
2) How do we find the canonical form, if we know the strings and their lengths?

My answer is based on Wikipedia articles "Nilpotent matrix" and "Jordan normal form."

1. It is obvious in the sense that the Jordan canonical form is block diagonal by definition. (Any square matrix, nilpotent or not, over an algebraically-closed field has a Jordan canonical form. Because zero is the only eigenvalue of a nilpotent matrix, the diagonal entries of each Jordan block are zero; hence, the Jordan-canonical-form matrix has all zeros except for possibly some ones as subdiagonal entries.)


2. The number of strings equals the number of Jordan blocks. The length of a string gives the size of the corresponding Jordan block. For example, in Exercise 2.20(b), there are one three-by-three and three one-by-one Jordan blocks in the canonical form:
   $$left( beginarrayc
   0 & 0 & 0 & 0 & 0 & 0\
   1 & 0 & 0 & 0 & 0 & 0\
   0 & 1 & 0 & 0 & 0 & 0\ hline
   0 & 0 & 0 & 0 & 0 & 0\ hline
   0 & 0 & 0 & 0 & 0 & 0\ hline
   0 & 0 & 0 & 0 & 0 & 0\
   endarray right).$$