Conceptual question about equivalence of eigenvectors Suppose for a matrix the eigenvalue is 1 and the eigenvector is (2,-3). Then does that mean (-4,6) and (4,-6) are equivalent eigenvectors as the ratios are the same? 
 A: If by two eigenvectors being "equivalent" you mean that they have the same eigenvalue, then yes: given an eigenvalue $\lambda$, the set of all eigenvectors with eigenvalue $\lambda$ is a vector space. In particular, means that it is closed under scalar multiplication: if $v$ has eigenvalue $\lambda$ then so does $cv$ for each scalar $c$. So if $(2,-3)$ is an eigenvector with eigenvalue $\lambda$ then so are $(-4,6) = 2(2,-3)$ and $(4,-6) = (-2)(2,-3)$.
A: Let T be a transformation, and let $\lambda$ be an eigenvalue with eigenvector $v$, ie. $T(v)=\lambda v$. Then if $c$ is any scalar, $cv$ is also an eigenvector with eigenvalue $\lambda$, since $T(cv)=cT(v)=c\lambda v=\lambda(cv)$
A: Let $\vec{x} = [2, -3]^T$ be an eigenvector of the matrix $A$ with eigenvalue $\lambda$. Then
$A \vec{x} = \lambda \vec{x}$.
The vector you are suggesting to look at is $\vec{y} = [4, -6]^T = 2 \vec{x}$. So note that
$A \vec{y} = 2A \vec{x} = 2 \lambda \vec{x} = \lambda \vec{y}$,
so $\vec{y}$ is also an eigenvector corresponding to the same eigenvalue.
In general, any constant multiple would be, and also if $\vec{x},\vec{y}$ are eigenvectors corresponding to the same eigenvalue, so would be $a \vec{x} + b\vec{y}$ for any $a,b \in \mathbb{R}$.
