User:Egm6321.f12.team4.mishra/Homework5 R5.1 & R5.2

Homework 5 Problem R*5.1 – Proof of Exponentiation of a matrix
sec20-2b

Statement
Show that $$ exp [A^T] = exp[A]^T $$

Solution
We begin by assuming $$ A^T $$ as a 2x2 matrix ,


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$$ A^T= \begin{bmatrix} 0 & -t\\ t & 0 \end{bmatrix} $$     (5.1.1)
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Since,
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$$ exp A^T = \sum_{k=0}^{\infty} \frac{(A^T)^k}{k!} $$     (5.1.2)
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$$ A^T*A^T= \begin{bmatrix} 0 & -t\\ t & 0 \end{bmatrix}*\begin{bmatrix} 0 & -t\\ t & 0 \end{bmatrix} = \begin{bmatrix} -t^2 & 0\\ 0 & -t^2 \end{bmatrix} $$     (5.1.3)
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$$ A^T*A^T*A^T= \begin{bmatrix} -t^2 & 0\\ 0 & -t^2 \end{bmatrix}*\begin{bmatrix} 0 & -t\\ t & 0 \end{bmatrix}= \begin{bmatrix} 0 & t^3\\ -t^3 & 0 \end{bmatrix} $$     (5.1.4)
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Similarly,


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$$ (A^T)^k= \begin{bmatrix} -t^k & 0\\ 0 & -t^k \end{bmatrix} $$ if k is even (5.1.5)
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$$ =\begin{bmatrix} 0 & t^k\\ -t^k & 0 \end{bmatrix} $$ if k is odd (5.1.6)
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Putting the above in Eq. (5.1.2),
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$$ \begin{bmatrix} 1-\frac{t^2}{2!}+\frac{t^4}{4!}-\frac{t^6}{6!}+.. & -\frac{t^1}{1!}+\frac{t^3}{3!}-\frac{t^5}{5!}+..\\ +\frac{t^1}{1!}-\frac{t^3}{3!}+\frac{t^5}{5!}+.. & 1-\frac{t^2}{2!}+\frac{t^4}{4!}-\frac{t^6}{6!}+.. \end{bmatrix} $$ (5.1.7)
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Since,
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$$ sin t = t-\frac{t^3}{3!} + \frac{t^5}{5!} + \frac{t^7}{7!} + ... $$ (5.1.8)
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$$ cos t = 1 - \frac{t^2}{2!} + \frac{t^4}{4!} + \frac{t^6}{6!} + ... $$    (5.1.9)
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Eq. (5.1.7) can be rewritten in terms of (5.1.7) and (5.1.8) as ,


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$$ exp[A^T]=\begin{bmatrix} cos t & -sin t\\ sin t & cos t \end{bmatrix} $$ (5.1.8)
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Taking transpose of $$ A^T $$


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$$ A= \begin{bmatrix} 0 & t\\ -t & 0 \end{bmatrix} $$     (5.1.9)
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Since,
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$$ exp A = \sum_{k=0}^{\infty} \frac{(A)^k}{k!} $$     (5.1.10)
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$$ A*A= \begin{bmatrix} 0 & t\\ -t & 0 \end{bmatrix}*\begin{bmatrix} 0 & t\\ -t & 0 \end{bmatrix} = \begin{bmatrix} -t^2 & 0\\ 0 & -t^2 \end{bmatrix} $$     (5.1.11)
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$$ A^T*A^T*A^T= \begin{bmatrix} -t^2 & 0\\ 0 & -t^2 \end{bmatrix}*\begin{bmatrix} 0 & t\\ -t & 0 \end{bmatrix}= \begin{bmatrix} 0 & -t^3\\ t^3 & 0 \end{bmatrix} $$     (5.1.12)
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Similarly,


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$$ (A^T)^k= \begin{bmatrix} -t^k & 0\\ 0 & -t^k \end{bmatrix} $$ if k is even (5.1.13)
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$$ =\begin{bmatrix} 0 & -t^k\\ t^k & 0 \end{bmatrix} $$ if k is odd (5.1.14)
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Putting the above in Eq. (5.1.2),
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$$ \begin{bmatrix} 1-\frac{t^2}{2!}+\frac{t^4}{4!}-\frac{t^6}{6!}+.. & \frac{t^1}{1!}-\frac{t^3}{3!}+\frac{t^5}{5!}+..\\ -\frac{t^1}{1!}+\frac{t^3}{3!}-\frac{t^5}{5!}+.. & 1-\frac{t^2}{2!}+\frac{t^4}{4!}-\frac{t^6}{6!}+.. \end{bmatrix} $$ (5.1.15) which can be rewritten as ,
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$$ exp[A]=\begin{bmatrix} cos t & sin t\\ -sin t & cos t \end{bmatrix} $$ (5.1.16)
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From Eq. (5.1.8) & (5.1.16), it is obvious that


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$$ \begin{bmatrix} cos t & -sin t\\ sin t & cos t \end{bmatrix} = \begin{bmatrix} cos t & sin t\\ -sin t & cos t \end{bmatrix}^T $$ (5.1.17)
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Hence,
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$$ exp [A^T] = exp[A]^T $$ (5.1.18)
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Author and References

 * Solved and Typed by -- Pushkar Mishra

Homework 5 Problem R*5.2 – Proof of Exponentiation of a diagonal matrix
sec20-2b

Statement
Consider a diagonal matrix $$ [D]=\begin{bmatrix} d_1 & 0 & 0 & .. &0 \\ 0& d_2 &  &  & \\ .& & d_3 &  & \\ .& &  & . & \\  0&  &  &  &d_n \end{bmatrix} $$ where $$ d_1, d_2 , ... , d_n $$ are complex numbers.

We have to show that $$ exp([D])=\begin{bmatrix} e^d_1 & 0 & 0 & .. &0 \\ 0& e^d_2 &  &  & \\ .& & e^d_3 &  & \\ .& &  & . & \\  0&  &  &  &e^d_n \end{bmatrix} $$

Solution
We start by considering


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$$ exp D = \sum_{k=0}^{\infty} \frac{(D)^k}{k!} $$     (5.2.1)
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Evaluating individual terms ,


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$$ D*D = \begin{bmatrix} d_1 & 0 & 0 & .. &0 \\ 0& d_2 &  &  & \\ .& & d_3 &  & \\ .& &  & . & \\  0&  &  &  &d_n \end{bmatrix} * \begin{bmatrix} d_1 & 0 & 0 & .. &0 \\ 0& d_2 &  &  & \\ .& & d_3 &  & \\ .& &  & . & \\  0&  &  &  &d_n \end{bmatrix} = \begin{bmatrix} d_1^2 & 0 & 0 & .. &0 \\ 0& d_2^2 &  &  & \\ .& & d_3^2 &  & \\ .& &  & . & \\  0&  &  &  &d_n^2 \end{bmatrix} $$     (5.2.2)
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$$ D*D*D = \begin{bmatrix} d_1^2 & 0 & 0 & .. &0 \\ 0& d_2^2 &  &  & \\ .& & d_3^2 &  & \\ .& &  & . & \\  0&  &  &  &d_n^2 \end{bmatrix}* \begin{bmatrix} d_1 & 0 & 0 & .. &0 \\ 0& d_2 &  &  & \\ .& & d_3 &  & \\ .& &  & . & \\  0&  &  &  &d_n \end{bmatrix} = \begin{bmatrix} d_1^3 & 0 & 0 & .. &0 \\ 0& d_2^3 &  &  & \\ .& & d_3^3 &  & \\ .& &  & . & \\  0&  &  &  &d_n^3 \end{bmatrix} $$     (5.2.3)
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Similarly ,


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$$ D^k = \begin{bmatrix} d_1^k & 0 & 0 & .. &0 \\ 0& d_2^k &  &  & \\ .& & d_3^k &  & \\ .& &  & . & \\  0&  &  &  &d_n^k \end{bmatrix} $$    (5.2.4)
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Putting Eq (5.2.2), Eq (5.2.3), Eq (5.2.4) in Eq. (5.2.1) ,


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$$ exp (D) = \begin{bmatrix} 1+d_1+\frac{d_1^2}{2!}+\frac{d_1^3}{3!}+...+\frac{d_1^k}{k!} & 0 & 0 & .. &0 \\ 0& 1+d_2+\frac{d_2^2}{2!}+\frac{d_2^3}{3!}+...+\frac{d_2^k}{k!} &  &  & \\ .& & 1+d_3+\frac{d_3^2}{2!}+\frac{d_3^3}{3!}+...+\frac{d_3^k}{k!} &  & \\ .& &  & . & \\  0&  &  &  &1+d_n+\frac{d_n^2}{2!} + \frac{d_n^3}{3!}+...+\frac{d_n^k}{k!} \end{bmatrix} $$    (5.2.5)
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Since we know,


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$$ exp (x) = 1+ x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + .. + \frac{x^n }{n!} $$    (5.2.6)
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Using the above in Eq. (5.2.5)


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$$ exp (D) = \begin{bmatrix} e^{d_1} & 0 & 0 & .. &0 \\ 0& e^{d_2} &  &  & \\ .& & e^{d_3} &  & \\ .& &  & . & \\  0&  &  &  &e^{d_n} \end{bmatrix} $$    (5.2.7)
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Hence proved.

Author and References

 * Solved and Typed by -- Pushkar Mishra