##### A check on our work

When finding eigenvalues and their associated eigenvectors in this way, we first find eigenvalues \(\lambda\) by solving the characteristic equation. If \(\lambda\) is a solution to the characteristic equation, then \(A-\lambda I\) is not invertible and, consequently, the reduced row echelon form of \(A-\lambda I\) must contain a row without a pivot.

This serves as a check on our work. If we row reduce \(A-\lambda I\) and find the identity matrix, then we have made an error either in solving the characteristic equation or in finding \(\nul(A-\lambda I)\text{.}\)