Exercise 1.10. Verify that with sigma=(12),tau=(123), the standard representation las a basis alpha=(omega,1,omega^(2)),beta=(1,omega,omega^(2)), with tau alpha=omega alpha,quad tau beta=omega^(2)beta,quad sigma alpha=beta,quad sigma beta=alpha.

Question

Exercise 1.10. Verify that with  sigma=(12),tau=(123), the standard representation las a basis  alpha=(omega,1,omega^(2)),beta=(1,omega,omega^(2)), with  tau alpha=omega alpha,quad tau beta=omega^(2)beta,quad sigma alpha=beta,quad sigma beta=alpha.

Bytelearn · Accepted Answer

Define permutations sigma and tau: First, let's define the permutations sigma and tau. Sigma is the permutation (12), which swaps the first and second elements, and tau is the permutation (123), 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 omega ≠ 1.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 * alpha = (omega^2, omega, 1), which is the same as the result of applying tau to alpha. Therefore, tau alpha = omega alpha.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.

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 $\tau \alpha=\omega \alpha$ , $\tau \beta=\omega^{2}\beta$ , $\sigma \alpha=\beta$ , $\sigma \beta=\alpha$ .

Full solution

More problems from Write equations of cosine functions using properties

Related topics