I'm trying to understand this one proof about the matrix of a nilpotent operator and I'm stuck on this one part. Here is the theorem I'm trying to prove and how the proof goes:

Suppose N is a nilpotent operator on a vector space V. Then there is a basis of V with respect to which the matrix of N has the form:

where all entries on and under the diagonal are 0 and the asterisk indicates the other entries.

The proof starts by creating a basis for V. First a basis for

is chosen and then extended to a basis of

. Then our basis is extended to a basis for

and so on until

. Now we have a basis of V.

The text then says:

"Now let’s think about the matrix of N with respect to this basis. The

first column, and perhaps additional columns at the beginning, consists

of all 0’s because the corresponding basis vectors are in

. The

next set of columns comes from basis vectors in

. Applying N

to any such vector, we get a vector in

; in other words, we get a

vector that is a linear combination of the previous basis vectors. Thus

all nonzero entries in these columns must lie above the diagonal. The

next set of columns come from basis vectors in

. Applying N

to any such vector, we get a vector in

; in other words, we get a

vector that is a linear combination of the previous basis vectors. Thus,

once again, all nonzero entries in these columns must lie above the

diagonal. Continue in this fashion to complete the proof."

I understand how the entries in the first set of columns are all 0. However, I do not understand how the second set of columns are determined.