When and how can we solve the matrix equation \(\exp(A) = B\)? Do it over the complex numbers and over the real numbers. Give a counterexample with real entries.
Solution
Over \(\mathbb C\), the equation \(e^A=B\) has a solution exactly when \(B\) is invertible. Necessity follows from \[ \det(e^A)=e^{\operatorname{tr}A}\neq0. \] For sufficiency, put \(B\) in Jordan form. For a block \[ J=\lambda(I+N),\qquad \lambda\neq0,\quad N^m=0, \] choose any complex logarithm \(\ell\) of \(\lambda\) and set \[ \log J=\ell I+\sum_{r=1}^{m-1}\frac{(-1)^{r+1}}rN^r. \] The sum is finite, and exponentiating gives \(J\). Conjugating the block logarithms back yields a logarithm of \(B\).
Over \(\mathbb R\), invertibility and positive determinant are necessary but not sufficient. Culver’s criterion says that a real invertible matrix has a real logarithm iff, for every negative real eigenvalue, the Jordan blocks of each size occur with even multiplicity. Positive real Jordan blocks use the same finite nilpotent logarithm as above; nonreal conjugate blocks are paired, and equal negative blocks are paired to obtain real logarithm blocks.
For example, \[ B=\begin{pmatrix}-1&0\\0&-2\end{pmatrix} \] has positive determinant but no real logarithm: each negative eigenvalue occurs in a single \(1\times1\) Jordan block, violating Culver’s criterion.