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