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In an orthonormal set with $ q_1= \left[\begin{matrix} \frac 13 \\ \frac 23 \\ \frac {-2}3 \\ \end{matrix}\right] $ $ q_2= \left[\begin{matrix} 0 \\ \frac {1}{\sqrt{2}} \\ \frac {1}{\sqrt{2}} \\ \end{matrix}\right] $ and $ q_3= \left[\begin{matrix} \frac {-4}{\sqrt{18}} \\ \frac {1}{\sqrt{18}} \\ \frac {-1}{\sqrt{18}} \\ \end{matrix}\right] $

Express the vector $ w= \left[\begin{matrix} 1 \\ 2 \\ 3 \\ \end{matrix}\right] $ in terms of q1, q2, and q3. That is, find c1, c2,and c3 such that:

$ w= \left[\begin{matrix} 1 \\ 2 \\ 3 \\ \end{matrix}\right] = c_1q_1 + c_2q_2 + c_3q_3 $

I don't understand the c variables' function in this question. I assume they represent scalars, but I can't say for certain. Also, assuming that they do represent scalars, does a systematic approach exist for finding a solution, or do I have to rely on guessing and checking?

I feel like I might have missed a fundamental relationship that exists in the problem.

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  • $\begingroup$ Did you mean 18 rather than 16 for q3? $\endgroup$ Feb 11 '19 at 0:32
  • $\begingroup$ Oops yes I did, thanks $\endgroup$ Feb 11 '19 at 0:33
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They are scalars. Here is how to formulate this problem

$$ w= \left[\begin{matrix} 1 \\ 2 \\ 3 \\ \end{matrix}\right] = c_1 \left[\begin{matrix} \frac 13 \\ \frac 23 \\ \frac {-2}3 \\ \end{matrix}\right] + c_2 \left[\begin{matrix} 0 \\ \frac {1}{\sqrt{2}} \\ \frac {1}{\sqrt{2}} \\ \end{matrix}\right] + c_3 \left[\begin{matrix} \frac {-4}{\sqrt{18}} \\ \frac {1}{\sqrt{18}} \\ \frac {-1}{\sqrt{18}} \\ \end{matrix}\right] = \left[ \matrix{ \frac{1}{3} & 0 & -\frac{4}{\sqrt{18}} \\ \frac{2}{3} & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{18}} \\ -\frac{2}{3} & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{18}} } \right] \left[\matrix{c_1 \\ c_2 \\ c_3}\right]$$

You can solve the above with matrix inversion (which in case it is equal to the transpose since the matrix is orthonormal).

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    $\begingroup$ Thanks, I appreciate the matrix visualization $\endgroup$ Feb 12 '19 at 17:37
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if the set of vectors really is orthonormal, the coefficient $c_i = w \cdot q_i$ the dot product

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  • $\begingroup$ That’s so intuitive, thank you for dragging me out of the haze of problem set fatigue haha. $\endgroup$ Feb 11 '19 at 0:59
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Since they are orthonormal, so $$\underline{q_i}\cdot\underline{q_j}=0 \ \ \ for \ i \neq j \ and \\ \underline{q_i}\cdot\underline{q_i}=1$$

Consider $$ \underline{w} = c_1 \underline{q_1} + c_2\underline{q_2} + c_3\underline{q_3} \\ \implies \underline{w}\cdot\underline{q_1}= c_1 \underline{q_1}\cdot\underline{q_1} + c_2\underline{q_2}\cdot\underline{q_1} + c_3\underline{q_3}\cdot\underline{q_1} $$

$$\implies \underline{w}\cdot\underline{q_1}= c_1 \underline{q_1}\cdot\underline{q_1}+0+0$$ so $c_1=\frac{\underline{w}\cdot\underline{q_1}}{\underline{q_1}\cdot\underline{q_1}}=\underline{w}\cdot\underline{q_1}$

You can repeat these steps by dotting other vectors from the orthonormal set to find out $c_2 ,c_3$

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  • $\begingroup$ Thank you for the concrete example! $\endgroup$ Feb 12 '19 at 17:36

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