# Prove that there exists a linear transformation $T: R^n \rightarrow R^n$ such that $T^3 = T$, and T has at least three distinct eigenvalues.

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?

• 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. – 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

$Tv_{1}=v_{2}$

$Tv_{2}=v_{1}$

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

Then $T^{3}=T$.

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

$T(v_{1}+v_{2})=(v_{1}+v_{2})$

$T(v_{2}-v_{1})=-(v_{2}-v_{1})$

• 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. – Parveen Chhikara Jun 22 '15 at 13:52
• 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 – Parveen Chhikara Jun 22 '15 at 13:59
• If we transform some $v_i$ for $i > 3$ to zero, and some to their multiples. Then our work is DONE. – Parveen Chhikara Jun 22 '15 at 14:03
• @ParveenChhikara: I already defined $Tv_{i}=0$ if $i>3$. – Matematleta Jun 22 '15 at 14:16
• @ParveenChhikara: there are no extra eigenvalues because the vectors corresponding to the $\lambda \neq 0$, span Im$T$ – 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 $\lambda$?

• that $1/\lambda$ is also an eigenvalue. But the exercise was to produce examples. – Matematleta Jun 22 '15 at 15:28
• @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. – John Gowers Jun 22 '15 at 15:34