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An nxn matrix A is called nilpotent if Ak = 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)=xk . Then p(A)=Ak = 0.

p(A)v = (Ak )*v = Ak-1*A*v = Ak-1*(A*v) = Ak-1*( λ *v) = Ak-1* λ *v

Note at this point that λ is a scalar, and so commutes with the matrix Ak-1.
Ak-1* λ *v = λ *Ak-1*v = .... = λ k*v = p( λ )v
In other words, p(A)v = p( λ )v.

Since A is nilpotent, Ak=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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