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[hide] Question 10
Let AH = -A be a skew-Hermitian complex matrix. Show that the matrix eAt is unitary. Note: The superscript H denotes the Hermitian conjugate $(\mathbf {A} ^{H})_{ij}={\overline {\mathbf {A} _{ji}}}$, which is often also called adjoint matrix or conjugate transpose and written as A* = AH.

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 hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

[show] Hint 1
In order for eAt to be unitary, we need $\left(e^{At}\right)^{H}e^{At}=I.$

[show] Hint 2
Use the Maclaurin Expansion $e^{x}=\sum _{k=0}^{\infty }{\frac {x^{k}}{k!}}$

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[show] Solution
Calculating the Maclaurin expansion we find that $\left(e^{At}\right)^{H}=\left(\sum _{k=0}^{\infty }{\frac {(At)^{k}}{k!}}\right)^{H}=\sum _{k=0}^{\infty }{\frac {(A^{H}t)^{k}}{k!}}=e^{\left(A^{H}\right)t}$ Thus, $\left(e^{At}\right)^{H}e^{At}=\left(e^{A^{H}t}\right)e^{At}=\left(e^{-At}\right)e^{At}=e^{0}=I.$

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