Thread: Simple eigenvalue / core-nilpotent decomposition

1. Simple eigenvalue / core-nilpotent decomposition

I am reading a book on linear algebra, and I found the following statement:

" Assume z is a simple eigenvalue of the matrix A. Because z is a simple eigenvalue, the core-nilpotent decomposition insures that A-zI is similar to a matrix of the form:
C 0
( ). "
0 01x1

I have trouble seeing why this is the case. This statement is used in the book to prove that the intersection between Range of (A-zI) and Nullspaceof (A-zI) is empty (so index(A-zI) = 1 and y*x <> 0 (where y is a left-hand eigenvector and x is an eigenvector of A). So I can't use these facts as assumptions to construct my argument!

Damien

2. Hi everyone,

On this question, someone was kind enough to suggest the following outline of a proof.

1. Since z is a simple eigenvalue A, dim Nullspace(A-zI) = 1, i.e. there is a single linearly independent eigenvector associated with A-zI.

Let's call this vector x and let's complete it with n-1 vector x1, x2, ..., xn-1, xn = x so that the n vectors form a basis for my vector space (Rn or Cn). (Notice that the eigenvector is placed in the last column of P).

2. Place the n vectors as column in the matrix P: P = (x1|x2|...|xn). Now rank(P) = n so P is non-singular.

3. Show the form of P^(-1)AP. I use block matrix multiplication, so:

P^(-1)AP = transpose((y1|y2|...|yn)) A (x1|x2|...|xn) =

[y1Ax1 y1Ax2 ... y1Axn
y2Ax1 y2Ax2 ... y2Axn
...
ynAx1 ynAx2 ... ynAxn]

I can see straight away that the last column is 0nx1 since Axn = 0, but I fail to see why the last row should be 0nx1 too.

I would really appreciate any help here.