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Exercise 7.1 · Q6

Q.Consider the matrix Aα=(cos⁡α−sin⁡αsin⁡αcos⁡α)A_\alpha = \begin{pmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha\end{pmatrix}.

(i) Show that AαAβ=Aα+βA_\alpha A_\beta = A_{\alpha+\beta}.
(ii) Find all possible real values of α\alpha satisfying the condition Aα+AαT=IA_\alpha + A_\alpha^T = I.
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Part (i) is a direct matrix multiplication followed by the cosine/sine addition formulas; part (ii) uses Aα+AαT=2cos⁡α IA_\alpha+A_\alpha^T=2\cos\alpha\, I and solves cos⁡α=12\cos\alpha=\tfrac12.

Step 1. (i) Multiply AαAβA_\alpha A_\beta.

AαAβ=(cos⁡α−sin⁡αsin⁡αcos⁡α)(cos⁡β−sin⁡βsin⁡βcos⁡β)=(cos⁡αcos⁡β−sin⁡αsin⁡β−cos⁡αsin⁡β−sin⁡αcos⁡βsin⁡αcos⁡β+cos⁡αsin⁡β−sin⁡αsin⁡β+cos⁡αcos⁡β)A_\alpha A_\beta=\begin{pmatrix}\cos\alpha&-\sin\alpha\\ \sin\alpha&\cos\alpha\end{pmatrix}\begin{pmatrix}\cos\beta&-\sin\beta\\ \sin\beta&\cos\beta\end{pmatrix}=\begin{pmatrix}\cos\alpha\cos\beta-\sin\alpha\sin\beta & -\cos\alpha\sin\beta-\sin\alpha\cos\beta\\ \sin\alpha\cos\beta+\cos\alpha\sin\beta & -\sin\alpha\sin\beta+\cos\alpha\cos\beta\end{pmatrix}

Step 2. (i) Apply the angle-addition identities. Using cos⁡αcos⁡β−sin⁡αsin⁡β=cos⁡(α+β)\cos\alpha\cos\beta-\sin\alpha\sin\beta=\cos(\alpha+\beta) and sin⁡αcos⁡β+cos⁡αsin⁡β=sin⁡(α+β)\sin\alpha\cos\beta+\cos\alpha\sin\beta=\sin(\alpha+\beta),

AαAβ=(cos⁡(α+β)−sin⁡(α+β)sin⁡(α+β)cos⁡(α+β))=Aα+β.A_\alpha A_\beta=\begin{pmatrix}\cos(\alpha+\beta) & -\sin(\alpha+\beta)\\ \sin(\alpha+\beta) & \cos(\alpha+\beta)\end{pmatrix}=A_{\alpha+\beta}.

This proves part (i).

Step 3. (ii) Compute AαTA_\alpha^T and add.

AαT=(cos⁡αsin⁡α−sin⁡αcos⁡α)A_\alpha^T=\begin{pmatrix}\cos\alpha & \sin\alpha\\ -\sin\alpha & \cos\alpha\end{pmatrix}, so …

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