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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?

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Yes. The eigenvalue doesn't even need to be $1$. –  Gregor Bruns Dec 3 '12 at 17:55

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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 $(4,-6) = (-2)(2,-3)$.

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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 $T(cv)=cT(v)=c\lambda v=\lambda(cv)$

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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 $a,b \in \mathbb{R}$.

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@BrettFrankel thank you, corrected. –  gt6989b Dec 4 '12 at 15:13

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