# Linear transformation and eigenvalues

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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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