Matrix exponential commutes with conjugation

Statement

Suppose $K$ is a field and $A,B$ are $n\times n$ matrices over $K$ such that the matrix exponential of $A$, denoted $\exp(A)$, exists. Then, the matrix exponential of $BAB^{-1}$ also exists, and:

$\!\exp(BAB^{-1})=B\exp(A)B^{-1}$

Related facts

• Exponential map commutes with adjoint action is a version for Lie groups and Lie algebras.
• Exponential of adjoint map corresponds to conjugation by group element for Lazard Lie ring is the version for the Lazard correspondence.