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Consider the following condition from this other post

  1. Define $S_k = \operatorname{span} (e_1, \ldots, e_k)$, where $e_i$ the standard basis vectors. Clearly, the linear map $T$ is upper triangular if and only if $T S_k \subset S_k$.

From this condition, wouldnt any linear map of dimension $\leq k$ be upper triangular? If not, what am I not understanding?

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CodeKingPlusPlus
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    ... if and only if $TS_k \subset S_k$ for all $k$. – Daniel Fischer Jul 18 '13 at 20:11
  • @DanielFischer What I am thinking is this: The transformation corresponding to matrix $A=\begin{pmatrix}0&0\1&0\end{pmatrix}$ is definitely not upper triangular. But it is a subset of $\mathbb{R}^2$ – CodeKingPlusPlus Jul 18 '13 at 20:16
  • But for that, you don't have $AS_1 \subset S_1$. – Daniel Fischer Jul 18 '13 at 20:17
  • I think I am confused by $AS_1$. What is the meaning? – CodeKingPlusPlus Jul 18 '13 at 20:19
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    Identifying the matrix with the linear map. The matrix for $T$ is upper triangular if and only if 1. $T e_1 = \lambda e_1$, 2. $T e_2 \in \operatorname{span}(e_1,, e_2)$, 3. $T e_3 \in \operatorname{span}(e_1,,e_2,,e_3)$ 4. etc. – Daniel Fischer Jul 18 '13 at 20:33
  • @CodeKingPlusPlus Daniel is right. Let $V$ be your space or dimension $n$. Then, for every square matrix $T$ of order $n$ and for every $x \in V$ of order $n$, $TX \in V$, which would make any $T$ triangular (just replace $x$ by the vectors $e_1,\dots,e_n$). You need the "for all $k$" part. – Vedran Šego Jul 18 '13 at 21:11

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The correct criterion is the following:

An $n\times n$ matrix $T$ is upper triangular if and only if $TS_k \subseteq S_k$ for all $k\in\{1,\ldots,n\}$.

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