Solution
= Solution
For a square <matrix> $M$, an <eigenvalue>–<eigenvector> pair consists of a <scalar> $\lambda$ and a \b[nonzero] <vector> $v$ satisfying
$$
\boxed{Mv=\lambda v,\qquad v\ne0.}
$$
The <linear map> represented by $M$ therefore scales $v$ by $\lambda$. The exclusion of the zero <vector> matters: otherwise every <scalar> would satisfy the equation trivially.