Science:Math Exam Resources/Courses/MATH307/December 2005/Question 04 (a)

MATH307 December 2005

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Question 04 (a)
Show that $\displaystyle e^{A+B}\neq e^{A}e^{B}$ where $\displaystyle A={\begin{bmatrix}0&0\\1&0\end{bmatrix}}$ and $\displaystyle B={\begin{bmatrix}0&1\\0&0\end{bmatrix}}$. Hint: $\displaystyle {\begin{bmatrix}0&1\\1&0\end{bmatrix}}={\frac {1}{2}}{\begin{bmatrix}1&1\\1&-1\end{bmatrix}}{\begin{bmatrix}1&0\\0&-1\end{bmatrix}}{\begin{bmatrix}1&1\\1&-1\end{bmatrix}}$

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Let $A={\begin{bmatrix}0&0\\1&0\end{bmatrix}},\quad B={\begin{bmatrix}0&1\\0&0\end{bmatrix}}$ Remember Taylor series: $e^{A}=\sum _{k=0}^{\infty }{\frac {A^{k}}{k!}}=A^{0}+A+{\frac {1}{2}}A^{2}...=I+A={\begin{bmatrix}1&0\\0&1\end{bmatrix}}+{\begin{bmatrix}0&0\\1&0\end{bmatrix}}={\begin{bmatrix}1&0\\1&1\end{bmatrix}}$ Note that $\displaystyle A^{2}=0,A^{3}=0,A^{4}=0$ etc. A similar computation gives us that $e^{B}={\begin{bmatrix}1&1\\0&1\end{bmatrix}}$ Hence $M=e^{A}e^{B}={\begin{bmatrix}1&1\\1&2\end{bmatrix}}$ Next, let $C=A+B={\begin{bmatrix}0&1\\1&0\end{bmatrix}}$ We now use the hint. Let $T={\begin{bmatrix}1&1\\1&-1\end{bmatrix}}$ and $P={\begin{bmatrix}1&0\\0&-1\end{bmatrix}}$ then as $T^{-1}={\frac {1}{2}}{\begin{bmatrix}1&1\\1&-1\end{bmatrix}}$ we have via the hint, $\displaystyle C=A+B=TPT^{-1}$ By the rules of matrix exponentiation, we have $\displaystyle e^{A+B}=e^{C}=e^{TPT^{-1}}=Te^{P}T^{-1}$ As $\displaystyle P^{2}=I$, we have $\displaystyle {\begin{aligned}e^{P}&=P^{0}+P^{1}+1/2P^{2}+1/6P^{3}+.....\\&=I+P+(1/2)I+(1/6)P+(1/24)I+...\\&={\begin{bmatrix}1+1+1/2+1/6+...&0\\0&d\end{bmatrix}}\\&={\begin{bmatrix}e&0\\0&d\end{bmatrix}}\end{aligned}}$ where e = e1 is the Euler number, and d is some other finite value. Thus, $\displaystyle e^{A+B}=e^{C}=e^{TPT^{-1}}=Te^{P}T^{-1}={\frac {1}{2}}{\begin{bmatrix}1&1\\1&-1\end{bmatrix}}{\begin{bmatrix}e&0\\0&d\end{bmatrix}}{\begin{bmatrix}1&1\\1&-1\end{bmatrix}}={\frac {1}{2}}{\begin{bmatrix}e+d&e-d\\e-d&e+d\end{bmatrix}}$ Comparing with M = eAeB we would need that ${\begin{aligned}1&={\frac {e+d}{2}}\\1&={\frac {e-d}{2}}\\1&={\frac {e-d}{2}}\\-1&={\frac {e+d}{2}}\end{aligned}}$ which is impossible. Hence, $\displaystyle e^{A}e^{B}\not =e^{A+B}$.

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