# Express the new basis in terms of the old basis

I am looking at a homework problem I turned in and the posted solution and I still am having trouble making sense of the question.

Given two vectors:

$v_1$ = $\begin{bmatrix}1 \\ 0 \\ 1\end{bmatrix}$ $\>$$\> v_2 = \begin{bmatrix}0 \\ -1 \\ 1\end{bmatrix} a) use the Gram-Schmidt procedure to perform orthonormalization on the basis {v_1,v_2}. I was able to do this with the result of: q_1 = \frac{1}{\sqrt{2}} \begin{bmatrix}1 \\ 0 \\ 1\end{bmatrix} \>$$\>$ $q_2$ = $\frac{1}{\sqrt{1.5}}$$\begin{bmatrix}\frac{-1}{2} \\ -1 \\ \frac{1}{2}\end{bmatrix} b) Express the new basis in terms of the old basis However, here I get turned around. How do I generate the change of basis matrix Q that will transform from v_i to q_i and back? • The given expression for q_2 is wrong. q_2 is not orthogonal to q_1 and does not have unity norm. – Math Lover Sep 17 '17 at 19:34 • You are right, I had a typo for the q2 matrix which I just changed. – varlotbarnacle Sep 17 '17 at 19:38 ## 1 Answer The second part of the question asks you to find scalars a_{ij} such that q_1=a_{11}v_1+a_{12}v_2 and q_2=a_{21}v_1+a_{22}v_2. The Gram-Schmidt process that you just performed gives you a way to find these coefficients. You just have to save some information from the steps you took. You no doubt began by setting q_1={1\over\|v_1\|}v_1, so obviously a_{11}={1\over\|v_1\|} and a_{12}=0. When computing q_2, you would first have computed$$w_2=v_2-(v_2\cdot q_1)q_1=v_2-{v_2\cdot v_1\over v_1\cdot v_1}v_1$$and then normalized the result. It should be clear, then, that the remaining two coefficients$a_{21}$and$a_{22}$that you’re looking for are the coefficients of$v_1$and$v_2$in the right-hand side above, divided by$\|w_2\|$. • With this I get$\begin{bmatrix} q_1 & q_2 \end{bmatrix}$=$\begin{bmatrix} v_1 & v_2 \end{bmatrix}\begin{bmatrix} \frac{1}{\sqrt{2}} & 0 \\ 0 & \frac{1}{\sqrt{1.5}} \end{bmatrix}$But this is not the correct answer. I am missing the$Q_{12}$component – varlotbarnacle Sep 18 '17 at 12:42 • @varlotbarnacle Yes, you’re obviously missing a coefficient. Take another careful look at the equation for$w_2$above. It involves both$v_1$and$v_2$, so you should be getting two coefficients from that, not one. – amd Sep 18 '17 at 18:31 • Okay, if I take the coefficient of$v_1$and the coefficient from$v_2$in the equation$w_2$and then divide by the norm of$w_2$for each vector$v_i$I get the resulting Q matrix:$\begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{-1}{\sqrt{6}} \\ 0 & \frac{1}{\sqrt{1.5}} \end{bmatrix}\$ This matches the posted solution. Thanks for all the help! – varlotbarnacle Sep 18 '17 at 19:03
• @varlotbarnacle Looks good. Note that the change-of-basis matrix for a basis that you’ve orthonormalized via G-S will always be triangular. – amd Sep 18 '17 at 21:30