Prove that $[T]_\gamma$ is an upper triangular matrix (Question 6.4.24 (a) in Linear algebra by Friedberg, Insel, and Spence).

Question

Question 6.4.24 (a) (in Linear algebra by Friedberg, Insel, and Spence): Suppose that $\beta$ is an ordered basis for $V$ such that $[T]_{\beta}$ is an upper triangular matrix. Let $\gamma$ be the orthonormal basis for $V$ obtained by applying the Gram-Schmidt orthogonalization process to $\beta$ and then normalizing the resulting vectors. Prove that $[T]_\gamma$ is an upper triangular matrix.

Let $\beta = \{v_1, ... , v_n\}$ and $\gamma = \{w_1, ..., w_n\}$, and $[T]_{\beta}=A$ and $[T]_\gamma=B$. I know that $w_1 = \frac{v_1}{||v_1||}$. Thus, $T(w_1) = \frac{1}{||v_1||} T(v_1) = \frac{1}{||v_1||} A_{11}v_1$, and this implies that $B_{11} = A_{11}$. I guess I need to proceed from here, using induction on $\dim(V)$, but I am struggling with it. I appreciate if you give some help.

Answer 1

Hint: It suffices to apply (and prove if necessary) the following two facts.

• For a basis $\alpha = \{u_1,\dots,u_n\}$, $[T]_\alpha$ is upper triangular if and only if for every $j$ from $1$ to $n$, $\operatorname{span}(\{u_1,u_2,\dots,u_j\})$ is an invariant subspace of $T$.
• By the Gram-Schmidt process, $\{v_1,\dots,v_j\}$ and $\{w_1,\dots,w_j\}$ have the same span for all $j$.