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Question:

let matrix $$X=\begin{bmatrix} a&b\\ c&d \end{bmatrix} ,e^{X}=\begin{bmatrix} -1&2\\ 0&-1 \end{bmatrix}$$ and such $a+d=0$,

Find the matrix $X$ my idea $$e^{Tr{(X)}}=det{(e^{X})}$$ so $$1=1$$ is true

I think we can find the matrix function: $e^X$,but I have find this is ver ugly,so I think this problem have other methods

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Firstly, since the eigenvalues of $e^X$ are the exponentials of the eigenvalues of $X$, $X$ can only have eigenvalues of the form $i(\pi+2n\pi)$. Since $e^X$ is not diagonalizable, neither is $X$, so $X$ only has one eigenvalue, say $i\pi$. So we can try $X=\begin{bmatrix} i\pi&z\\ 0&i\pi\end{bmatrix}$. Then, $e^X=e^{i\pi I}e^{\begin{bmatrix} 0&z\\0&0\end{bmatrix}}=-I\left(I+\begin{bmatrix} 0&z\\ 0&0\end{bmatrix}\right)=\begin{bmatrix} -1&-z\\ 0&-1\end{bmatrix}$. So, we can let $z=-2$, and then we see that $\exp\left({\begin{bmatrix} i\pi&-2\\ 0&i\pi\end{bmatrix}}\right)=\begin{bmatrix} -1&2\\ 0&-1\end{bmatrix}$.

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    $\begingroup$ I like your approach a lot. It's definitely cleaner than mine. I'd made the diagonalizability connection but didn't piece it together the way you did. Very nice! $\endgroup$ – Cameron Williams Oct 1 '14 at 3:46
  • $\begingroup$ Thanks. I'm not sure if the OP wanted a real $X$, but I suspect there is no real matrix logarithm of the given matrix. $\endgroup$ – Nishant Oct 1 '14 at 13:22
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Here's a sketch. Since $d = -a$, you can write $X = \left(\begin{array}{rr} a & b \\ c & -a\end{array}\right)$, which gives

$$X^2 = \left(\begin{array}{cc} a^2 + bc & 0 \\ 0 & a^2 + bc\end{array}\right) = (a^2+bc)I.$$

This allows you to write a general expression for $X^n$: $X^{2n} = (a^2+bc)^n I$ and likewise $X^{2n+1} = (a^2+bc)^nX$.

With these two in hand, we can evaluate the exponential:

$$\exp X = \sum_{n=0}^{\infty} \frac{1}{n!} X^n = \sum_{n=0}^{\infty} \frac{1}{(2n)!} X^{2n} + \sum_{n=0}^{\infty} \frac{1}{(2n+1)!}X^{2n+1}.$$

Making use of our above observation:

$$\exp X = \sum_{n=0}^{\infty} \frac{(a^2+bc)^n}{(2n)!}\, I+ \sum_{n=0}^{\infty} \frac{(a^2+bc)^n}{(2n+1)!}\,X.$$

Since $\exp X = \left(\begin{array}{rr} -1 & 2 \\ 0 & -1\end{array}\right)$, we see that $c= 0$, so that

$$\exp X = \sum_{n=0}^{\infty} \frac{a^{2n}}{(2n)!}\, I+\sum_{n=0}^{\infty} \frac{a^{2n}}{(2n+1)!}\,X.$$

From here, if you equate terms, you will find that $a = i\pi+2k\pi$ for some $k\in\Bbb Z$ and $b = -2$ as Nishant derived.

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