Solution
= Solution
The inverse fails to exist exactly when \b[at least one <eigenvalue> is zero]. In that case a nonzero <eigenvector> lies in the <kernel of a linear map>, so $A$ is not one-to-one. Equivalently, its <determinant> $\prod_n\lambda_n$ is zero. Conversely, if every <eigenvalue> is nonzero, the preceding construction gives an inverse.