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So obviously one solution to $\exp(A)=I$ is $A=0$, however is it the only solution? And also If $\exp(A)$ is diagonalizable does this mean $A$ is diagonalizable?

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    $\begingroup$ Well, I expect that the equation in within square matrices. Is it $\mathbb{R}^{n\times n}$ or $\mathbb{C}^{n\times n}$ ? $\endgroup$ Dec 5, 2016 at 17:45
  • $\begingroup$ @Ilcolosso please consider my comment under Emilio's answer. $\endgroup$
    – Squirtle
    Dec 5, 2016 at 17:56

2 Answers 2

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No, they are infinitely many solutions of $\exp(A)=I$. For all integer $k$ consider the following matrix: $$A_k:=\begin{pmatrix}0&-2k\pi\\2k\pi&0\end{pmatrix}.$$ In fact, $\exp(A)=I$ if and only if $A$ is diagonalizable over $\mathbb{C}$ and $\textrm{Sp}(A)\subseteq 2i\pi\mathbb{Z}$. The converse is easy to prove and the direct implication follows from Jordan-Chevalley decomposition.

Regarding your other question, the answer is yes. This also follows from Jordan-Chevalley decomposition.

For both of your questions, the key observation is that if $A=D+N$ is the Jordan-Chevalley decomposition of $A$, then $\exp(A)=\exp(D)+\exp(D)(\exp(N)-I)$ is the Jordan-Chevalley decomposition of $\exp(A)$.

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    $\begingroup$ I really like this answer! That last observation is very neat and has a couple of nice little exercises like showing that if $N$ is nilpotent then $\exp(N)$ is unipotent, and also the checking that $\exp(D)(\exp(N)-I)$ is nilpotent and commutes and so on. The emphasis in the wikipedia reference on the more abstract operator formalism rather than matrix operators may be off putting for the OP? Anyway great answer. $\endgroup$
    – Nadiels
    Dec 5, 2016 at 18:37
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For the first question: not true in general. As a counter example consider the matrix with complex entries: $$ A=2k\pi i\begin{bmatrix} m&0\\0&n \end{bmatrix} $$ with $k,m,n$ integers.

or the matrix with real entries : $$ A=2k\pi \begin{bmatrix} 0&-1\\1&0 \end{bmatrix} $$

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  • $\begingroup$ I thought it was assumed that we were working only in the real case, since this is obviously not even true in if we restrict ourselves to one by one matrices. Since if we take $\mathbb{C}$ to be our field for such matrices then $e^{2\pi i}=e^0=1$ but this is obvious. If the field is $\mathbb{R}$, however, then $e^x=1$ if and only if $x=0$. I thought the question was generalizing from there... maybe not. $\endgroup$
    – Squirtle
    Dec 5, 2016 at 17:55

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