Verify that with $\sigma=(1\,2)$ , $\tau=(1\,2\,3)$ , the standard representation l as a basis $\alpha=(\omega,1,\omega^{2})$ , $\beta=(1,\omega,\omega^{2})$ , with $\tau \alpha=\omega \alpha$ , $\tau \beta=\omega^{2}\beta$ , $\sigma \alpha=\beta$ , $\sigma \beta=\alpha$ .

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Algebra 2

Write equations of cosine functions using properties

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Q. Verify that with

\sigma=(1\,2)

\tau=(1\,2\,3)

, the standard representation l as a basis

\alpha=(\omega,1,\omega^{2})

\beta=(1,\omega,\omega^{2})

, with

\tau \alpha=\omega \alpha

\tau \beta=\omega^{2}\beta

\sigma \alpha=\beta

\sigma \beta=\alpha

Define permutations $\sigma$ and $\tau$ : First, let's define the permutations $\sigma$ and $\tau$ . $\sigma$ is the permutation $(1 2)$ , which swaps the first and second elements, and $\tau$ is the permutation $(1 2 3)$ , which cycles the first element to the second position, the second to the third, and the third to the first.
Define basis vectors $\alpha$ and $\beta$ : Next, we define the basis vectors $\alpha$ and $\beta$ . $\alpha$ is given by $(\omega, 1, \omega^2)$ , and $\beta$ is given by $(1, \omega, \omega^2)$ , where $\omega$ is a complex cube root of unity, meaning $\omega^3 = 1$ and $\beta$ $0$ .
Verify action of tau on alpha: We need to verify the action of $\tau$ on $\alpha$ . Since $\tau$ is the cycle $(123)$ , applying $\tau$ to $\alpha$ gives us $(1, \omega^2, \omega)$ . We know that $\omega \cdot \alpha = (\omega^2, \omega, 1)$ , which is the same as the result of applying $\tau$ to $\alpha$ . Therefore, $\alpha$ $0$ .
Verify action of tau on beta: Now, we verify the action of tau on beta. Applying tau to beta gives us $(\omega, \omega^2, 1)$ . Multiplying beta by $\omega^2$ gives us $(\omega^2, 1, \omega)$ , which is the same as the result of applying tau to beta. Therefore, $\tau \beta = \omega^2 \beta$ .
Verify action of sigma on alpha: Next, we verify the action of sigma on alpha. Sigma swaps the first and second elements, so applying sigma to alpha gives us $(1, \omega, \omega^2)$ , which is exactly beta. Therefore, $\sigma \alpha = \beta$ .
Verify action of sigma on beta: Finally, we verify the action of sigma on beta. Applying sigma to beta swaps the first and second elements, resulting in $(\omega, 1, \omega^2)$ , which is exactly alpha. Therefore, $\sigma \beta = \alpha$ .