Thread: Proving a Matrix Nilpotent

A matrix is called nilpotent if there is some positive integer d such that N^d =O (the zero matrix).
The smalles value of d which fullfills this is called degree of nilpotency.

Consider the following matrix:

N = 0 1 0 0
0 0 1 0
0 0 0 1
0 0 0 0

Consider the following principle:

Let e1, e2...en be the collumns of the n x n identitiy matrix I. Then for any n x n matrix A, the product Aej is exactly the jth collumn of A.

Now for the proof.
(1) Explain why Ne1 = 0 (a collumn matrix of 0s)
Ne2 = e1
Ne3 = e2
Ne4 = e4
(2) Use the four equations of part (1) to prove that
N^2e1 = 0 ( collumn matrix of 0s).
N^2e2 = 0 " "
N^2e3 = e1
N^2e4 = e2
(3) By this similar reasoning, find a set of four equations that characterizes N^3 and likewise N^4.

(4) Interpret the results of part (2) and (3) with the principle mentioned above to show that

N^2 =
0 0 1 0
0 0 0 1
0 0 0 0
0 0 0 0

N^3 =
0 0 0 1
0 0 0 0
0 0 0 0
0 0 0 0

N^4 =
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0



(5) Finally prove that if Ni nilpotent of degree 4, then
N^d= N^(d+1) = N^(d+2)= ... = O (the zero matrix).

Thanks if anyone can help me with this

For part 1 merely apply the principle given to the matrix. is, by the principle, the first column of your matrix. This column is all zeros. Similarly, the first column is , etc.

For part 2, apply part 1. , and we know what is. Part 3 immediately follows.

For part 4, note that is the column of . For example, we wish to show that all the columns of are zero but (just adding a formula to part 2 - ) and we are done as and is the zero vector.