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Let $v$ and $w$ be non-zero vectors in $\mathbb R^n$, $n\ge3$ such that $w$ is not a scalar multiple of $v$. Prove that there exist a linear transformation $T:\mathbb R^n \to \mathbb R^n$ such that $T^3=T$, $T(v)=w$ and $T$ has at least $3$ distinct eigen values.

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No need to shout. –  lhf Feb 5 '13 at 10:50
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Welcome to SE. I edited your post to use latex codes for the maths symbols. Feel free to look again at your question (i.e., edit it) to learn the latex codes. If this is homework please tag is as such. If you want to get good answers you should include relevant background for the question as well as what you tried so far and where you got stuck. –  Ittay Weiss Feb 5 '13 at 10:53

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Hints: The constraints $T^3=T$ and $Tv=w$ suggest that $Tw=v$ would be a natural choice for $Tw$ (note: this is only a choice or an educated guess, but not a necessity; in fact, in the sequel, we will see that there are other legitimate choices of $Tw$ as well). What are the two eigenvalues of $T$ when it is restricted to the subspace $V=\operatorname{span}\{v,w\}$? Now, note that if $T^3=T$ and $\lambda$ is an eigenvalue of $T$, we must have $\lambda^3=\lambda$. (Why?) So, having found the two eigenvalues of $T|_V$ (the restriction of $T$ to $V$), if $T$ has at least three distinct eigenvalues, what is the only possibility for the third distinct eigenvalue? How to extend $T|_V$ to a linear transformation $T$ defined on the whole $\mathbb{R}^n$?

I hope you can find the most natural choice of $T$ from the above hints. Since I don't want to do the homework for you, I should stop giving more details here. However, to enhance your understanding on the problem, you may study the following alternative solution, which I believe is not what the question setter expected. Let $u_1=v$ and $u_2=w$. Extend $\{u_1,u_2\}$ to a basis $\{u_1,\ldots,u_n\}$ of $\mathbb{R}^n$. Define \begin{align*} Tu_1 &= Tu_2 = u_2,\\ Tu_3 &= 2u_1+u_2-u_3,\\ Tu_4 &= \ldots = Tu_n = 0. \end{align*} Then this $T$ will satisfy the problem's requirements. (Why?) Note that we define $Tv=Tw=w$ here. So, this $T$ is not the aforementioned "natural" choice of $T$.

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Thank u Sir.... it was of great help. –  user61062 Feb 5 '13 at 19:14

Hint: $\{v_1,\dots,v_n\}$ be a basis for $V$ and $\{w_1,\dots,w_n\}$ be a basis for $W$ then there exist a unique Linear Transformation $T:V\rightarrow W$ such that $T(v_i)=w_i$

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please provide me with some more hints –  user61062 Feb 5 '13 at 12:30

Hint: $T^3=T$ means that $T^2=1$, so $T(v)=w\Rightarrow TT(v)=1(v)=T(w)\Rightarrow T(w)=v$ Can you think of a transformation of space that does that?: If you have two independent vectors, it swaps them, and if you apply it again, it must swap them back to ther beginning position?

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please provide me with some more hints –  user61062 Feb 5 '13 at 12:29
    
@user61062 mirrors –  MyUserIsThis Feb 5 '13 at 12:30

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