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 . Thus the last equation shows that
I don't quite understand why this is true.