A real invertible matrix with no real logarithm

Direct calculation using [F3] gives $\det A=1$, so $A\in {\mathrm{GL}}_2^{+}(\mathbb{R})$.

By [F1], the positive-determinant open subgroup is a Lie group; it is path connected. Indeed, for any $B=(b_1{b}_2)$ in it, put $u=b_1\mathrm{/}\parallel b_1\parallel$, let $v$ be the positive quarter-turn of $u$, and set $Q=(uv)\in \mathrm{SO}(2)$. Then $Q^{T}B=R=\left(\begin{array}{cc}r& s\\ 0& t\end{array}\right)$ with $r=\parallel b_1\parallel >0$ and $t=\det (B)\mathrm{/}r>0$. The path $R_{\lambda }=\left(\begin{array}{cc}(1-\lambda )r+\lambda & (1-\lambda )s\\ 0& (1-\lambda )t+\lambda \end{array}\right)$ joins $R$ to $I$ through positive-determinant matrices, while writing the fixed $Q$ as a rotation through some angle $\theta$ gives the path of rotations from $Q$ to $I$. Concatenating $B=QR$ first to $Q$ and then to $I$ proves path connectedness.

Assume for contradiction that a real matrix $X$ satisfies $e^X=A$. The defining power series commutes with $X$, so $XA=AX$. Since $A$ has the two distinct eigenspaces $\mathbb{R}{e}_1$ and $\mathbb{R}{e}_2$, commutation makes each of them $X$-invariant. Hence $X{e}_1=x{e}_1$ for some real $x$, and the power series gives $e^X{e}_1=e^x{e}_1$ with $e^x>0$, whereas $A{e}_1=-2e_1$. This is impossible.

Thus $A$ has no real matrix logarithm. By [F2], it is not in the image of the Lie exponential of the connected group established in step 1.2, disproving surjectivity. The witness is nonsingular and two-dimensional; no claim is made in dimensions zero or one. There is no boundary, metric, interval endpoint, or iff issue. The logarithm obstruction and path construction are choice-free; ${\mathrm{A}\mathrm{C}}_{\omega }$ is present only because the current Lie-exponential interface [F2] carries it.