Upper Triangular Matrix and Gram-Schmidt

Question

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?

Answer 1

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. $\mathcal B_\textrm{GS}.$