# How to express a vector as scalar combinations of a separate orthonormal set?

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.

• Did you mean 18 rather than 16 for q3? Feb 11 '19 at 0:32
• Oops yes I did, thanks Feb 11 '19 at 0:33

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).

• Thanks, I appreciate the matrix visualization Feb 12 '19 at 17:37

if the set of vectors really is orthonormal, the coefficient $$c_i = w \cdot q_i$$ the dot product

• That’s so intuitive, thank you for dragging me out of the haze of problem set fatigue haha. Feb 11 '19 at 0:59

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 $$c_2 ,c_3$$

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