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?

  • $\begingroup$ The given expression for $q_2$ is wrong. $q_2$ is not orthogonal to $q_1$ and does not have unity norm. $\endgroup$ – Math Lover Sep 17 '17 at 19:34
  • $\begingroup$ You are right, I had a typo for the q2 matrix which I just changed. $\endgroup$ – varlotbarnacle Sep 17 '17 at 19:38

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 $\|w_2\|$.

  • $\begingroup$ 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 $\endgroup$ – varlotbarnacle Sep 18 '17 at 12:42
  • $\begingroup$ @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. $\endgroup$ – amd Sep 18 '17 at 18:31
  • $\begingroup$ 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! $\endgroup$ – varlotbarnacle Sep 18 '17 at 19:03
  • $\begingroup$ @varlotbarnacle Looks good. Note that the change-of-basis matrix for a basis that you’ve orthonormalized via G-S will always be triangular. $\endgroup$ – amd Sep 18 '17 at 21:30

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