An EIGENVECTOR $v$ of $A$ satisfies
A$Av = 0$ for some non-zero $v$
B$Av = \lambda v$, $v \neq 0$
C$v$ is some column of matrix $A$
D$\det(v) = 0$ for vector $v$
Answer & Solution
Correct answer: B. $Av = \lambda v$, $v \neq 0$
1. An EIGENVECTOR of $A$ is a NON-ZERO vector $v$ such that $Av$ is a scalar multiple of $v$.
2. Formally: $Av = \lambda v$ for some scalar $\lambda$ (the corresponding EIGENVALUE).
3. The EXCLUSION $v \neq 0$ is essential — otherwise the trivial vector would satisfy the equation for any $\lambda$.
4. Option A would mean $v$ is in the kernel (eigenvalue 0 IS allowed, but the equation is the special case $\lambda=0$).
5. Options C, D are unrelated.
_Source: Sergei Treil, "Linear Algebra Done Wrong", §4.5 (Eigenvalues and eigenvectors)._
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