#algebra/linear-algebra # Definition Suppose $T \in$ [[vector space of operators on a vector space]] and $v_1, \dots, v_n$ is a [[basis]] of $V$. Then the following are **equivalent**: 1. The [[matrix]] of $T$ with respect to $v_1, \dots, v_n$ is [[upper-triangular matrix|upper-triangular]]; 2. $Tv_j \in \span(v_1, \dots, v_j)$ for each $1 \leq j \leq n$; 3. $\span(v_1, \dots, v_j)$ is [[invariant subspace|invariant]] under $T$ for each $1 \leq j \leq n$. # Proof That (1) $\iff$ (2) is quickly deduced from the definitions: we have $\MM(T)=\begin{pmatrix} |& | & | \\ \MM(Tv_1) & \dots & \MM(Tv_n) \\ | & | & | \end{pmatrix},$ where $\MM(Tv_j)$ by definition consists of the [[scalar]]s $A_{1,j}, \dots, A_{j,j}$ needed to write $Tv_j$ as a [[linear combination]] of $v_1, \dots, v_j$. Hence because $Tv_j=A_{1,j}v_1 + \dots + A_{j,j}v_j+ 0v_{j+1} + \dots + 0v_n$, we have $\MM(Tv_j)=[A_{1,j}, \dots, A_{j,j}, 0, \dots, 0]^\top$ as desired. Obviously (3) $\implies$ (2). So we're done if we can show (2) $\implies$ (3). Suppose that (2) holds. Fix $j \in \{1, \dots, n\}$. From (b) we know that $\begin{align} Tv_1 &\in \span(v_1) \subset \span(v_1, \dots, v_j)\\ Tv_2 &\in \span(v_1, v_2) \subset \span(v_1, \dots, v_j)\\ &\vdots \\ Tv_j &\in \span(v_1, \dots, v_j) \subset \span(v_1, \dots, v_j). \end{align}$ Thus if $v \in \span(v_1, \dots, v_j)$ is a [[linear combination]] of $v_1, \dots, v_j$, then $Tv = T(a_1v_1 + \dots + a_jv_j) = a_1Tv_1 + \dots + a_jTv_j$ is a [[linear combination]] of elements $Tv_1, \dots, Tv_j \in \span(v_1, \dots, v_j)$, hence $Tv \in \span(v_1, \dots, v_j)$. In other words, $\span(v_1, \dots, v_j)$ is [[invariant subspace|invariant]] under $T$. $\qedin$ #notFormatted