If $T$ is an operator on the finite dimensional vector space $V$ which has a basis $\mathcal{B}$ for which the matrix representation of $T$ with this basis is upper-triangular, does it follow that $T$ represented with respect to the basis obtained from $\mathcal{B}$ by Gram-Schmidt is also upper-triangular?
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$\begingroup$ Let $\mathcal{C}$ be the basis obtained from $\mathcal{B}$ by Gram-Schmidt. Is the change-of-basis matrix between $\mathcal{B}$ and $\mathcal{C}$ upper-triangular? (The answer depends somewhat on how exactly you define Gram-Schmidt.) If the answer is "yes", then the answer to your question is "yes" as well, because products and inverses of upper-triangular matrices are upper-triangular. $\endgroup$– darij grinbergCommented Nov 13, 2018 at 20:58
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$\begingroup$ @darijgrinberg By Gram-Schmidt, I mean by procedure whereby one transforms a basis into an orthonormal basis. Is that what you had in mind, too? $\endgroup$– user193319Commented Nov 13, 2018 at 21:00
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$\begingroup$ Yes, but there are several of these procedures. In your version, is the $j$-th vector of $\mathcal{C}$ a linear combination of the first $j$ vectors of $\mathcal{B}$ ? If so, the answer is "yes". $\endgroup$– darij grinbergCommented Nov 13, 2018 at 21:02
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This has been discussed in Axler's LADR. Specifically, if $T$ is upper-triangular w.r.t. $\mathcal B:=\{v_1,v_2,\ldots,v_n\},$ then by the characterization of upper-triangular operators, $\forall~j\in\{1,2,\ldots,n\},~\operatorname{span}(v_1,v_2,\ldots,v_j)$ is invariant under $T.$
As is noted, one then applies Gram-Schmidt on $\mathcal B$ to obtain an orthonormal basis $\mathcal B_\textrm{GS}:=\{e_1,e_2,\ldots,e_n\}.$ Now observe that $\forall~j\in\{1,2,\ldots,n\}$ $$\operatorname{span}(e_1,e_2,\ldots,e_j)=\operatorname{span}(v_1,v_2,\ldots,v_j), $$ which implies $\operatorname{span}(e_1,e_2,\ldots,e_j)$ is invariant under $T$ for all $j$ ranging over $\{1,2,\ldots,n\}.$ Hence, $T$ is again upper-triangular w.r.t. $\mathcal B_\textrm{GS}.$
