Is matrix exponential injective and surjective?

Question

Exponential of real numbers exp: $\mathbb{R}\rightarrow\mathbb{R}$ is injective. Does the same hold for exponential of real matrices exp: $\mathbb{R}^{2\times2}\rightarrow\mathbb{R}^{2\times2}$? What about surjectivity, can every regular real matrix be written as $X=$ exp$A$?

Answer 1

It is neither injective nor surjective. Recall that $e^{2\pi i}=e^0=1$. We can replicate this example by letting $$ J:=\begin{bmatrix} 0 & -1 \\ 1 &0 \end{bmatrix},$$ which satisfies $J^2=-I$, so $$ \exp(2\pi J)=I.$$ Since $I=\exp(0)$, the exponential is not injective.

The exponential is also not surjective as a map of the space of all matrices in itself, because the equation $\exp(X)=0$ has no solution. Even if one restricts the range to nonzero matrices, the exponential is still not surjective; one way to see it is to note that $$\tag{1} \det \exp A= e^{\operatorname*{trace}(A)}\ne 0, $$ so if $Y$ is a matrix with $\det Y=0$, then the equation $\exp(X)=Y$ has no solution. (See the Wikipedia page on the matrix logarithm for a precise description of the range of the exponential function).

Remark. Here's a proof of (1). If $A$ is diagonalizable, which means that $$ A=U\begin{bmatrix} \lambda_1 & 0 & \ldots \\ 0& \lambda_2 & \ldots \\ \ldots & \ldots &\ldots\end{bmatrix}U^{-1}, $$ for some invertible matrix $U$, then (1) is immediate, because $$ \exp(A)=U \begin{bmatrix} e^{\lambda_1} & 0 & \ldots \\ 0& e^{\lambda_2} & \ldots \\ \ldots & \ldots &\ldots\end{bmatrix}U^{-1}, $$ so $\det \exp (A) = \prod e^{\lambda_j}=e^{\sum \lambda_j}$. The general case follows from this computation by continuity; indeed, the set of diagonalizable matrices is dense in the space of all matrices, both in the real and in the complex case.