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

Suppose $\displaystyle T\in\mathcal{L}(V)$ 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 $\displaystyle (v_1,\dots,v_n)$ is a basis of V, we can write $\displaystyle v=a_1v_1+\cdots+a_kv_k$

where $\displaystyle a_1,\dots,a_k\in\mathbb{F}$ and $\displaystyle a_k\ne0$ (represent v as a linear combination of $\displaystyle (v_1,\dots,v_n)$ and then choose k to be the largest index with a nonzero coefficient). Thus

$\displaystyle 0=Tv$

$\displaystyle 0=T(a_1v_1+\cdots+a_kv_k)$

$\displaystyle 0=T(a_1v_1+\cdots+a_{k-1}v_{k-1})+a_kTv_k$

I understand everything so far until they say

The last term in the parentheses is in $\displaystyle span(v_1,\dots,v_{k-1})$.