Saved Bookmarks
| 1. |
Let θ be a real number. Then check whether the matrices [ cos θ − sin θ sin θ cos θ ] and [e iθ −0 0 e −iθ ] are similar over the field of complex numbers |
|
Answer» Let A = \(\begin{bmatrix} cos \theta & -sin \theta \\[0.3em] sin \theta& \cos \theta\\[0.3em] \end{bmatrix}\) B = \(\begin{bmatrix} e^{i\theta} &0 \\[0.3em] 0 & e^{-i \theta} \end{bmatrix}\) Let eigen values of matrix A is \(\lambda\)1 & \(\lambda\)2. And also let eigen values of matrix B is \(\mu\)1 & \(\mu\)2. \(\because\) Trace of any matrix is equal to sum of diagonal elements. \(\therefore\) Trace of matrix A is \(\lambda\)1 + \(\lambda\)2 = cos\(\theta\) + cos\(\theta\) = 2 cos\(\theta\) (\(\because\) Trace of matrix = sum of eigen values of that matrix) And trace of matrix B is \(\mu\)1 + \(\mu\)2 = e\(i \theta\) + e\(-i \theta\) \(\mu\)1 + \(\mu\)2 = ( cos\(\theta\) + i sin\(\theta\)) + (cos\(\theta\) - i sin\(\theta\)) = 2cos\(\theta\) Thus, \(\lambda\)1 + \(\lambda\)2 = \(\mu\)1 + \(\mu\)2 .........(1) Now |A| = \(\begin{bmatrix} cos \theta & -sin \theta \\[0.3em] sin \theta& \cos \theta\\[0.3em] \end{bmatrix}\) = cos2\(\theta\) + sin2\(\theta\) = 1. Thus \(\lambda\)1 \(\lambda\)2 = 1 (\(\because \)|A| = Product of eigen values of matrix A) And |B| = \(\begin{bmatrix} e^{i\theta} &0 \\[0.3em] 0 & e^{-i \theta} \end{bmatrix}\) = e\(i \theta\) . e\(-i \theta\) - 0 = \(e^{i\theta - i \theta}\) = e0 = 1. Thus, \(\mu\)1 \(\mu\)2 = 1 Then, \(\lambda\)1 \(\lambda\)2 = \(\mu\)1 \(\mu\)2. Now, ( \(\lambda\)1 - \(\lambda\)2)2 = ( \(\lambda\)1 + \(\lambda\)2)2 + 4 \(\lambda\)1\(\lambda\)2 = \(\big(\)\(\mu\)1 + \(\mu\)2\(\big)\)2 + 4 \(\mu\)1\(\mu\)2 = \(\big(\)\(\mu\)1 - \(\mu\)2\(\big)\)2 \(\Rightarrow\) \(\lambda\)1 - \(\lambda\)2 = \(\mu\)1 - \(\mu\)2 ..........(2) or \(\lambda\)1 - \(\lambda\)2 = -( \(\mu\)1 - \(\mu\)2) = \(\mu\)2 - \(\mu\)1 ......... (3) Case - I By adding equation (1) and (2), we get. 2\(\lambda\)1 = 2 \(\mu\)1 \(\Rightarrow\) \(\lambda\)1 = \(\mu\)1 Then form equation (1) we get \(\lambda\)2 = \(\mu\)2. Case II By adding equations (1) and (3), we get 2 \(\lambda\)1 = 2\(\mu\)2 \(\Rightarrow\) \(\lambda\)1 = \(\mu\)2. Then from equation (1), we get \(\lambda\)2 = \(\mu\)1 Hence, eigen values of both matrices A and B are same. Therefore, both matrices A and B are similar matrices Hence, both given matrices are similar over the field of complex numbers for any real value of \(\theta\). |
|