I am trying to understand the proof of this theorem which states

Suppose

has an upper-triangular matrix with respect to some basis in V. Then T is invertible if and only if all the entries on the diagonal of that upper-triangular matrix are nonzero.

I understand the proof in the first direction. However I am a bit lost on proving that the converse is true.

The proof follows:

Suppose that T is not invertible. Thus T is not injective, and hence there exists a nonzero vector v ∈ V such that Tv = 0. Because

is a basis of V, we can write

where

and

(represent v as a linear combination of

and then choose k to be the largest index with a nonzero coefficient). Thus

I understand everything so far until they say

The last term in the parentheses is in

.