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As we know, the equation $$e^x=-1,\quad x\in\mathbb{C}$$ has no real solution (in fact $x=i\pi+2ki\pi$, $k\in\mathbb{Z}$). I am now considering the generalization of this question to $2\times 2$ matrices:

Question: Is there a real matrix $X\in M_2(\mathbb{R})$ such that $$\exp(X)=-I,$$ where $\exp$ is the matrix exponential?

I found that the (unreal) matrix $$X=\begin{pmatrix}i\pi & 0 \\ 0 & i\pi\end{pmatrix}$$ satisfy the equation. But I have no idea on how to show whether there are other real solutions.

Scott
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3 Answers3

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There is an embedding $\mathbb C \to M_2 (\mathbb R)$ given by $$a + bi \mapsto \left( \begin{matrix} a & b \\ -b & a \end{matrix} \right),$$ which you can check preserves multiplication and addition. Thus it also preserves the exponential; so one solution is given by the matrix $\left(\begin{matrix} 0 & \pi \\ -\pi & 0\end{matrix}\right)$ corresponding to $i \pi$.

Another way to understand this if you're familiar with Lie groups is that $-I$ is a rotation by $\pi$, so it can be achieved by exponentiating the infinitesimal rotation $\left(\begin{matrix} 0 & \pi \\ -\pi & 0\end{matrix}\right)$.

Anthony Carapetis
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Try with $$\left(\begin{matrix} 0 & \pi \\ -\pi & 0 \end{matrix}\right)$$

Crostul
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More generally, consider the equation $e^X=A$ where $A\in GL_n(\mathbb{C})$ is a diagonalizable matrix.

Necessarily $X$ is a diagonalizable matrix (why ?) and $XA=AX$. Then $A,X$ are simultaneously diagonalizable over $\mathbb{C}$ ; we may assume that $A=diag(a_i),X=diag(x_i)$ where $a_i\not=0$ and $e^{x_i}=a_i$. In this way, we obtain all the complex solutions of our equation.

The real case ($A,X\in M_n(\mathbb{R}))$ is more complicated. There is a real solution in $X$ iff $A$ is the square of an invertible real matrix. For example, the equation in $M_3(\mathbb{R})$: $e^X=-I_3$ has no real solutions.