Let $v_1$ and $v_2$ be nonzero vectors in $R^n, n>2,$ such that $v_2$ is not a scalar multiple of $v_1$. Prove that there exists a linear transformation $T: R^n \rightarrow R^n$ such that $T^3 = T, Tv_1 = v_2,$ and $T$ has at least three distinct eigenvalues.

I thought it like: Clearly $v_1$ and $v_2$ are linearly independent. So by extension $B = {v_1, v_2, . . . , v_n}$ is an ordered basis for $R^n.$ I defined $T: R^n \rightarrow R^n$ such that $Tv = [Tv ]_B.$ Then $A^3 = A, Av_1 = v_2$, $$A = [Tv]_B.$$

If we define $A$ in a convenient way as $A = \left( \begin{array}{cc} B & 0 & \\ 0 & C \end{array} \right) $, where B is chosen as $3 - by - 3 $ matrix such that $B^3 = B,$ then would it work?

  • $\begingroup$ Your instincts are right. We might as well take $C$ to be the identity matrix and let $B$ be $3\times3$ satisfying the conditions in the question. This means we can reduce immediately to the $3\times3$ case. $\endgroup$ – John Gowers Jun 22 '15 at 15:42

Let $\left \{ v_{1},\cdots , v_{n}\right \}$ be an ordered basis for $R^{n}$

Define $T:R^{n}\rightarrow R^{n}$ by



$Tv_{i}=0$ if $i\geq 3$

Then $T^{3}=T$.

Now, clearly $0$ is an eigenvalue. But so are $1$ and $-1$ for



  • $\begingroup$ thanx @Chilango. I think it is not a desired linear transformation. Because there may exist some eigenvalues other than 0, 1 and -1. What about them? The transformation, you consider, has only 0, 1 and -1 as its eigenvalues, I think. $\endgroup$ – Parveen Chhikara Jun 22 '15 at 13:52
  • $\begingroup$ I think. We should include some more "generality"? I mean to say that we have to take a transformation, that may have some "extra" eigenvalues. @Chilango $\endgroup$ – Parveen Chhikara Jun 22 '15 at 13:59
  • $\begingroup$ If we transform some $v_i$ for $i > 3$ to zero, and some to their multiples. Then our work is DONE. $\endgroup$ – Parveen Chhikara Jun 22 '15 at 14:03
  • $\begingroup$ @ParveenChhikara: I already defined $Tv_{i}=0$ if $i>3$. $\endgroup$ – Matematleta Jun 22 '15 at 14:16
  • $\begingroup$ @ParveenChhikara: there are no extra eigenvalues because the vectors corresponding to the $\lambda \neq 0$, span Im$T$ $\endgroup$ – Matematleta Jun 22 '15 at 15:27

Hint: Let $\lambda$ be an eigenvalue for $T$. What does the condition $T^3=T$ tell us about $\lambda$?

  • $\begingroup$ that $1/\lambda $ is also an eigenvalue. But the exercise was to produce examples. $\endgroup$ – Matematleta Jun 22 '15 at 15:28
  • $\begingroup$ @Chilango That's not what I had in mind. I'm aware that the exercise is to produce examples; that's why this is labelled as a hint, rather than a complete answer. $\endgroup$ – John Gowers Jun 22 '15 at 15:34

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