An nxn

matrix A is called

nilpotent if A

^{k} = 0 for some k. k is called the index of nilpotency.
Consider the matrix A =

`
|0 1| `

|0 0|

|0 1|^2 =

|0 0|

|0 1|*|0 1|=|0 0|

|0 0| |0 0| |0 0|
The index of nilpotency in this case is 2.

**Claim:** All

eigenvalues of a nilpotent matrix are 0.

**Proof:** Let A be an nxn nilpotent matrix with index of nilpotency k, and let λ be an eigenvalue of A, with corresponding

eigenvector v. Then by definition of

eigenvalue and

eigenvector, Av= λ v.

Consider the

polynomial p(x)=x

^{k} . Then p(A)=A

^{k} = 0.

p(A)v = (A

^{k} )*v = A

^{k-1}*A*v = A

^{k-1}*(A*v) = A

^{k-1}*( λ *v) = A

^{k-1}* λ *v

Note at this point that λ is a

scalar, and so

commutes with the matrix A

^{k-1}.

A

^{k-1}* λ *v = λ *A

^{k-1}*v = .... = λ

^{k}*v = p( λ )v

In other words, p(A)v = p( λ )v.

Since A is nilpotent, A

^{k}=0 , and λ v = Av = 0v = 0. It is important to note that eigenvectors may NOT be the zero

vector, and if λ v = 0, with v non-zero, λ must be equal to zero. Since we chose an

arbitrary eigenvalue, all eigenvalues are equal to zero.

Graded homework from a class called Matrix Theory at the University of Iowa, 2002