Solution
= Solution
A <scalar> $\lambda$ is an <eigenvalue> of an $n\times n$ <matrix> $P$ if there is a nonzero <vector> $\mathbf v$ such that
$$
\boxed{P\mathbf v=\lambda\mathbf v,\qquad\mathbf v\ne\mathbf0.}
$$
Any such <vector> is an <eigenvector> corresponding to $\lambda$. The requirement that an <eigenvector> be nonzero is essential: the zero <vector> would satisfy the equation for every <scalar> and would convey no information.