User:Egm6321.f12.team6.bang/report4/problem 5.4

=Report 5=

Problem Statement

 * Show
 * a) $$ \displaystyle \mathbf B t = \begin{bmatrix} 1 & 1 \\ -i & i \end{bmatrix} \begin{bmatrix} {it} & 0 \\ 0 & {-it} \end{bmatrix} \begin{bmatrix} i & -1 \\ i & 1 \end{bmatrix} \frac{1}{2i}$$,
 * and
 * b) $$ \displaystyle \exp [\mathbf B t] = \begin{bmatrix} \cos t & -\sin t \\ \sin t & \cos t \end{bmatrix}$$

Nomenclature

 * $$ \displaystyle \exp [\mathbf B] = e^{\mathbf B} $$
 * $$ \displaystyle \text{Diag} [\lambda_1,\dots,\lambda_n] = \begin{bmatrix} \lambda_1 & \ldots & 0 \\ \vdots & \ddots & \vdots \\ 0 & \ldots & \lambda_n \end{bmatrix} $$
 * $$ \displaystyle \sqrt {-1} = i $$

Part a): Decomposition of Bt

 * First of all, in order to find matrix of $$ \displaystyle \mathbf \Phi $$ and $$ \displaystyle \mathbf \Lambda $$, we have to calculate for eigenvalues and eigenvectors. After finding eigenvalues and eigenvectors, we can decompose matrix B using diagonalization. We know that $$ \displaystyle \mathbf B = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} $$.


 * Therefore eigenvalues are $$ \displaystyle i$$, and $$ \displaystyle -i$$.
 * When $$ \displaystyle \lambda = i$$, then an eigenvector is $$ \displaystyle \boldsymbol \phi_1 = \begin{bmatrix} 1 \\ -i \end{bmatrix} $$.
 * Also for $$ \displaystyle \lambda = -i$$, an eigenvector is $$ \displaystyle \boldsymbol \phi_2 = \begin{bmatrix} 1 \\ i \end{bmatrix} $$.
 * Since there are 2 linearly independent eigenvectors, we know that $$ \displaystyle \mathbf \Phi = [\boldsymbol \phi_1, \boldsymbol \phi_2] $$ from Equation 4.1. Therefore, $$ \displaystyle \mathbf \Phi $$ expression can be derived as following.


 * Lastly, from the given we know that $$ \displaystyle \mathbf \Lambda $$ is the diagonal matrix of eigenvalues. This means that we can express $$ \displaystyle \mathbf \Lambda $$ in the following.


 * Using Equation 4.3, combining terms of Equation 4.8, Equation 4.9, and Equation 4.10 give decomposed expression of matrix B.


 * Lastly, adding t variable, we have the same expression.


 * {| style="width:100%" border="0"|-

$$ \displaystyle \mathbf \mathbf B t = \begin{bmatrix} 1 & 1 \\ -i & i \end{bmatrix} \begin{bmatrix} it & 0 \\ 0 & -it \end{bmatrix} \begin{bmatrix} i & -1 \\ i & 1 \end{bmatrix}\frac{1}{2i} $$
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * (4.12)
 * }

Part b): Proof of exp[Bt]

 * Using Equation 4.4, and Equation 4.12, we can transform Bt into exp[Bt]. Furthermore, please refer to Problem 5.2 for detailed derivation.


 * Using Euler's Equation $$ e^{it} = \cos t + i\sin t $$, we can express Equation 4.16 into sine and cosine terms.


 * Therefore, we can concluded that
 * {| style="width:100%" border="0"|-

$$ \displaystyle \exp [\mathbf B t] = \begin{bmatrix} \cos t & -\sin t \\ \sin t & \cos t \end{bmatrix} $$
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * (4.20)
 * }