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Eigenvalues of nilpotent matrices are always 0
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Tue Apr 09 2002 at 2:01:55
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
eigenvalue
s 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
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