Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2016 / Paper 1 / 14 / a / iv / Solution

The inverse fails to exist exactly when 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.