##### 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, $$A-\lambda I$$ must contain a row without a pivot position.

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 $$\nul(A-\lambda I)\text{.}$$

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