# Thread: Upper-Triangular Matrix - Invertible Operator Theorem

1. ## Upper-Triangular Matrix - Invertible Operator Theorem

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 $\displaystyle span(v_1,\dots,v_{k-1})$. Thus the last equation shows that $\displaystyle a_kTv_k\inspan(v_1,\dots,v_{k-1})$

I don't quite understand why this is true.

Thank you.

2. Originally Posted by 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 $\displaystyle span(v_1,\dots,v_{k-1})$.

Thus the last equation shows that $\displaystyle a_kTv_k\inspan(v_1,\dots,v_{k-1})$
Do you mean that $\displaystyle a_kTv_k$ is in the span of $\displaystyle (v_1, \cdots, v_{k-1})$?
That follows from the fact that $\displaystyle a_kTv_k= T(a_1v_1+ \cdots+ a_{k-1}v_{k-1})$ and the right side is in the span.

I don't quite understand why this is true.

Thank you.
A simpler proof is that a matrix is invertible if and only if its determinant is not 0. And the determinant of a triangular matrix is the product of the numbers on the diagonal.