Exercise 6.4.24

Question

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.

Jephian C.-H. Lin · Accepted Answer

\begin{enumerate}
\item Let 
$$\beta =\left\{v_1,v_2,\ldots ,v_n\right\}$$
and 
$$\gamma=\left\{w_1,w_2,\ldots ,w_n\right\}$$
be the two described basis. Denote $A$ to be $[T]_{\beta}$. We have 
$$T(w_1)=T(\frac{v_1}{\|v_1\|})=\frac{A_{11}}{\|v_1\|}T(v_1)=A_{11}T(v_1).$$
Let $t$ be the maximum integer such that $T(w_{t})$ is an element in 
$$\operatorname{span}\left\{w_1,w_2,\ldots ,w_{t}\right\}.$$
If $t=\dim (V)$, then we've done. If not, we have that 
$$w_{t+1}=\frac{1}{L}(v_t-\sum_{j=1}^{t}{\left\langle v_{t+1},w_j\right\rangle w_j}),$$
where 
$$L=\|v_t-\sum_{j=1}^{t}{\left\langle v_{t+1},w_j\right\rangle w_j}\|.$$
By the definition of $w_i$'s we may define 
$$W_i=\operatorname{span}\left\{v_1,v_2,\ldots ,v_n\right\}=\operatorname{span}\left\{w_1,w_2,\ldots,w_2\right\}.$$
Now we have $T(w_t)\in W_t$ since 
$$T(w_t)=\frac{1}{L}(T(v_t)-\sum_{j=1}^{t-1}{\left\langle v_t,w_j\right\rangle T(w_j)})$$
and $T(v_t)\in W_t$ and $T(w_j)\in W_j\subset W_t$ for all $j<t$. This is a contradiction to our choice of $i$. So $[T]_{\gamma}$ is an upper triangular matrix.
\item If the characteristic polynomial of $T$ splits, we have an ordered basis $\beta $ such that $[T]_{\beta }$ is upper triangular. Applying the previous argument, we get an orthonormal basis $\gamma $ such that $[T]_{\gamma}$ is upper triangular.
\end{enumerate}

Exercise 6.4.24

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Exercise 6.4.24

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