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I am solving a problem: $$A^3=\alpha^2A\implies \exp(A)=E+\frac{\mathrm{sinh}\alpha}{\alpha}A+\frac{\mathrm{cosh}\alpha-1}{\alpha^2}A^2;\, \alpha\in\mathbb{C},\,A\in\mathbb{M}_{n\times n}(\mathbb{C})$$ and I am stuck. I did prove the statement, but the proof turned out to be faulty and now I have to begin from a scratch(well, not from a scratch, the eigenvalues of the matrix are rather obvious) and I'd need a little guidance.

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  • $\begingroup$ What's the matrix E? $\endgroup$ – JPi Apr 14 '14 at 10:55
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Hint. In this case I wouldn't use the "smart" method of calculating the exponential, but go back to the definition via Taylor series. If $A^3=\alpha^2A$ then $A^4=\alpha^2A^2$ and $A^5=\alpha^2A^3=\alpha^4A$ and so on; therefore $$\eqalign{\exp(A)&=I+A+\frac{1}{2!}A^2+\frac{1}{3!}A^3+\frac{1}{4!}A^4+\cdots\cr &=I+A+\frac{1}{2!}A^2+\frac{1}{3!}\alpha^2A+\frac{1}{4!}\alpha^2A^2+\cdots\cr &=I+(\cdots)A+(\cdots)A^2\ .\cr}$$ I'm sure you can now fill in the gaps. Good luck!

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  • $\begingroup$ Thank you! I did it two ways - first was the "smart" method that led me nowhere, then I thought of Taylor polynomial, but came nowhere, because I did not separate A and A^2. Sometimes I have the solution right before my eyes, but my teachers tought me to always try to do the problem the hard/complicated way, so I am totaly blind to easy and simple solutions ^^ Thank you again. $\endgroup$ – user74200 Apr 14 '14 at 12:04

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