I wonder how the following identity makes sense?
$\exp(x^\top A x) = \exp(A x x^\top)$
My approach:
$\exp(x^\top A x) = \exp(\operatorname{tr}(x^\top A x)) = \exp(\operatorname{tr}(Axx^\top)) = \det(\exp(Axx^\top))$
But what should it formally mean to drop the determinant? This is used here to identify an exponential form; but there is no determinant..
Note that $$ \det\exp(Axx^T) $$ is the product of the eigenvalues of this matrix (assuming complex $A$). However, the rank of matrix $Axx^T$ is one (and zero if $x=0$). So the determinant is just $e^\lambda$ with $\lambda$ being the only nonzero eigenvalue of $Axx^T$. Its eigenvector is $Ax$ (assuming $Ax\ne0$): $$ (Axx^T)Ax= (x^TAx)Ax, $$ hence $\lambda = x^TAx$, and $$ \det\exp(Axx^T) = \exp(x^TAx). $$