# 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!

Many thanks for your help!!!

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.