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Equivalent?

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Reply 20
Avatar for TheEd
TheEd
OP
Original post by Hopple
For them to be equivalent, yes.


So if λ=Re(b)\lambda = \text{Re}(b) and μ=Im(b)\mu = \text{Im}(b) the matrices would be given by [−2λλ+μiλ−μi−2λ]\begin{bmatrix} -2\lambda & \lambda + \mu i \\ \lambda - \mu i & -2\lambda \end{bmatrix} where λ<0,∣μ∣<3∣λ∣\lambda <0, |\mu| < \sqrt{3}|\lambda|

which are essentially the same matrices as in the solution.
(edited 11 years ago)
Reply 21
Avatar for Hopple
Hopple
18
Original post by TheEd
So if λ=Re(b)\lambda = \text{Re}(b) and μ=Im(b)\mu = \text{Im}(b) the matrices would be given by [−2λλ+μiλ−μi−2λ]\begin{bmatrix} -2\lambda & \lambda + \mu i \\ \lambda - \mu i & -2\lambda \end{bmatrix} where λ<0,∣μ∣<3∣λ∣\lambda <0, |\mu| < \sqrt{3}|\lambda|

which are essentially the same matrices as in the solution.


I'm not actually sure what's going on. Rho is presumably mentioned earlier in the text? What I can say is that your two inequalities in the OP are equivalent if you remove the minus sign in the second.
Reply 22
Avatar for TheEd
TheEd
OP
[INDENT]
Original post by Hopple
I'm not actually sure what's going on. Rho is presumably mentioned earlier in the text? What I can say is that your two inequalities in the OP are equivalent if you remove the minus sign in the second.


Yeah, ρ:Z3→GL(C2)\rho : \mathbb{Z}_3 \to GL(\mathbb{C}^2) is a representation (homomorphism) of the vector space C2\mathbb{C}^2 in the cyclic group Z3={1,z,z2}\mathbb{Z}_3 = \{ 1 , z , z^2 \} and is defined by (where zz is a generator of Z3\mathbb{Z}_3 satisfying z3=1z^3 = 1) :

ρ(z)=[−11−10]\rho (z) = \begin{bmatrix} -1 & 1 \\ -1 & 0 \end{bmatrix}.

The question asks to describe all ρ\rho-invariant inner products (meaning ⟨x,y⟩=⟨ρ(g)x,ρ(g)y⟩    ∀g∈Z3\langle \boldsymbol{x} , \boldsymbol{y} \rangle = \langle \rho(g) \boldsymbol{x} , \rho(g) \boldsymbol{y} \rangle \;\;\forall g\in \mathbb{Z}_3) in terms of the coordinates of any 2 vectors x,y∈C2\boldsymbol{x},\boldsymbol{y} \in \mathbb{C}^2.

From the matrix in terms of λ\lambda and μ\mu we get that all the ρ\rho-invariant inner products of Z3\mathbb{Z}_3 in the vector space C2\mathbb{C}^2 are ⟨x,y⟩=xtAyˉ\langle \boldsymbol{x} , \boldsymbol{y} \rangle = \boldsymbol{x}^t A \boldsymbol{\bar{y}} where AA is the (positive definite, hermitian) matrix found and xt\boldsymbol{x}^t denotes the conjugate transpose of x\boldsymbol{x}.
(edited 11 years ago)
Reply 23
Avatar for Hopple
Hopple
18
Original post by TheEd
[INDENT]

Yeah, ρ:Z3→GL(C2)\rho : \mathbb{Z}_3 \to GL(\mathbb{C}^2) is a representation (homomorphism) of the vector space C2\mathbb{C}^2 in the cyclic group Z3={1,z,z2}\mathbb{Z}_3 = \{ 1 , z , z^2 \} and is defined by (where zz is a generator of Z3\mathbb{Z}_3 satisfying z3=1z^3 = 1) :

ρ(z)=[−11−10]\rho (z) = \begin{bmatrix} -1 & 1 \\ -1 & 0 \end{bmatrix}.

The question asks to describe all ρ\rho-invariant inner products (meaning ⟨x,y⟩=⟨ρ(g)x,ρ(g)y⟩    ∀g∈Z3\langle \boldsymbol{x} , \boldsymbol{y} \rangle = \langle \rho(g) \boldsymbol{x} , \rho(g) \boldsymbol{y} \rangle \;\;\forall g\in \mathbb{Z}_3) in terms of the coordinates of any 2 vectors x,y∈C2\boldsymbol{x},\boldsymbol{y} \in \mathbb{C}^2.

From the matrix in terms of λ\lambda and μ\mu we get that all the ρ\rho-invariant inner products of Z3\mathbb{Z}_3 in the vector space C2\mathbb{C}^2 are ⟨x,y⟩=xtAyˉ\langle \boldsymbol{x} , \boldsymbol{y} \rangle = \boldsymbol{x}^t A \boldsymbol{\bar{y}} where AA is the (positive definite, hermitian) matrix found and xt\boldsymbol{x}^t denotes the conjugate transpose of x\boldsymbol{x}.


Yes, I think that line in the text is a typo, everything else seems consistent.

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