# Upper Triangular Matrix and Gram-Schmidt

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?

• 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. Commented Nov 13, 2018 at 20:58
• @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? Commented Nov 13, 2018 at 21:00
• 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". Commented Nov 13, 2018 at 21:02

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