# Definition of a basis $v_1,\dots,v_n$ of $\textsf{V}$ that makes $\textsf{T}$ lower triangular

Let $$\textsf V$$ be a finite-dimensional vector space, and let $$\textsf{T}∈\mathcal{L}(\textsf{V})$$.

I'm trying to define what it means for a basis $$\{v_1,\dots,v_n\}$$ of $$\textsf{V}$$ to make $$\textsf{T}$$ lower triangular.

My thought is let $$\dim(\textsf{V})=n$$. Then the basis $$\{v_1,\dots,v_n\}$$ is lower triangular if $$\textsf{T}(v_i)∈\operatorname{span}\{v_i,v_{i+1},\dots,v_n\}$$ for all $$i=1,\dots,n.$$

Also, if $$\{v_1,\dots,v_n\}$$ is a basis for $$\textsf{V}$$ that makes $$\textsf{T}$$ upper triangular, then how can I prove that the basis $$\{v_n,v_{n−1},\dots,v_1\}$$ makes $$\textsf{T}$$ lower triangular?

My first thought was "If the first $$k$$ components of $$v$$ are $$0$$, then the first $$k$$ components of $$Tv$$ are zero." That's basically the same thing, though.