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Let $v_{1}$ and $v_{2}$ be non-zero in $R^n$, $n\ge3$ such that $v_{2}$ is not a scalar multiple of $v_1$. Prove that there exists a linear transformation $T:R^n\to R^n$ such that $T^3=T$, $Tv_1=v_2$, and $T$ has atleast three distinct eigenvalues.

Before creating such a transformation $T$, if $\lambda$ is an eigen value of $T$ then it must satisfy $\lambda^3=\lambda$ giving $\lambda=0,1,-1$. Whatever the transformation is it must have its eigen values as $0,-1,1$.

Let $\{v_1,v_2...v_n\}$be the basis for $R^n$. Since a transformation will be completely defined by the images of basis elements, we start defining the transformation keeping this in mind.

So Starting the process of creation , by the given condition

$Tv_1=v_2$. Let's define $Tv_2=v_1$ so that $T^3=T$ is satisfied. Then $Tv_3=0$ so that $\lambda=0$ is satisfied. Also define $Tv_4=-v_4$. We fix the rest of the vectors i.r $Tv_k=v_k$ for $k=5,6..n$.

Will this be good?

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$Tv_4=-v_4$ is unnecessary, and also doesn't work for $n=3$. Note that $T(v_1+v_2)=v_1+v_2$ and $T(v_1-v_2)=-(v_1-v_2)$ –  Callus Feb 6 '14 at 3:02

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