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Let $\{f_1,f_2,f_3\}$ be the dual basis of $\{e_1,e_2,e_3\}$, where $e_1 = (1,1,1),$ $e_2 = (1,1,-1),$ and $e_3 = (1,-1,-1).$ Find $f_1(x),$ $f_2(x)$, $f_3(x),$ where $x = (1,0,0).$

Just to check I'm understanding this correctly - by definition a dual basis is biorthogonal with the set of basis vectors. So let $F = \{f_1,f_2,f_3\}$ and $B = \{e_1,e_2,e_3\}$, then $$F^TB = I_3,$$

which I found to be $$ \left[\begin{matrix} \dfrac{1}{2} & 0 & \dfrac{1}{2} \\ 0 & \dfrac{1}{2} & -\dfrac{1}{2} \\ \dfrac{1}{2} & -\dfrac{1}{2} & 0\end{matrix} \right]$$

Hence $f_1(x) = \frac{1}{2}, f_2(x) = 0$ and $f_3(x) = \frac{1}{2}$.

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Your work looks good to me. Here's another way to do it (just for fun).

First recognize that $(1,0,0) = \frac 12e_1 + \frac 12 e_3$

Thus $$f_1(x) = f_1\left(\frac 12e_1 + \frac 12 e_3\right) = \frac 12f_1(e_1) + \frac 12f_1(e_3) = \frac 12 (1) + \frac 12 (0) = \frac 12 \\ f_2(x) = 0f_2(e_2) = 0 \\ f_3(x) = \frac 12f_3(e_3) = \frac 12$$

(For simplicity I just ignored the irrelevant terms in the last two lines.)

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  • $\begingroup$ Thanks, nice way to do it! $\endgroup$ Commented Nov 19, 2015 at 16:33

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